Get the NCERT Exemplar Class 12 Maths Integrals PDF below for Class 12 Mathematics Chapter 7 Integrals. The NCERT Exemplar Class 12 Maths Integrals PDF solve every Exemplar problem in the solutions PDF and write the file in the formal notation of the NCERT textbook. The download is a free PDF download, aligned to the 2026-27 syllabus.
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CBSE Weightage: 8 to 10 marks (Unit III: Calculus, shared with Differentiation and Application of Integrals; one LA on partial fractions or properties of definite integrals plus one SA on substitution or by parts)
JEE Main Weightage: 8 to 10% of paper (2 to 3 questions per shift on substitution patterns, partial fractions, by parts with ILATE, and special integrals)
Exemplar Problems Solved: 63 in total (34 Short Answer + 13 Long Answer + 11 MCQ + 5 Fill-in-the-Blanks)
63 Exemplar problems solved
4 Question formats covered
9 CBSE marks (typical)
Topics span the antiderivative concept, the standard-integrals table, four substitution patterns, partial fractions across three template forms, integration by parts with the ILATE rule, five special integrals, the Fundamental Theorem of Calculus, and the six properties of definite integrals.
Curated by Collegedunia subject experts, mapped to the 2026-27 NCERT, and benchmarked against five years of CBSE and JEE Main papers.
Class 12 Maths Chapter 7 Integrals Exemplar Solutions: What's Inside
the NCERT Exemplar Class 12 Maths Integrals PDF carries 63 fully solved solutions in NCERT's four Exemplar formats. Every solution opens with a Concept Used block (states the formula or technique), followed by step-by-step working with formula, substitution, and arithmetic on separate lines, a boxed final answer, and an amber Expert's Solution panel that re-derives the same answer through an alternative path.
Direct evaluation; standard substitutions, odd-function shortcut
The mix is calibrated to the actual CBSE / JEE pattern. SA problems train the standard-integrals table; LA problems train the King rule plus by-parts combination; MCQs train pattern-recognition speed; fill-in-the-blanks lock in the 1-mark templates.
The 63 problems concentrate the NCERT Exemplar Class 12 Maths Integrals PDF's tested-pattern weight onto six core techniques. The table below maps each technique to the questions that drill it; the full learn list of every standard integral and its decision tree lives on the Formula Sheet.
Class 12 Maths Chapter 7 Integrals Exemplar vs NCERT Textbook
The NCERT Exemplar Class 12 Maths Integrals PDF address this in the same order as the NCERT textbook.
The Exemplar set is harder than the NCERT textbook exercises and shorter than a full JEE Main mock. Use both, in this order, for a complete prep loop.
Resource
Difficulty
Best For
NCERT Textbook Exercises
Foundation (CBSE level)
First read; build standard-integrals fluency and basic partial fractions
NCERT Exemplar Problems
Stretch (CBSE 5-mark + JEE Main Tier 1)
Second pass; train King rule, $\tan x = t^{2}$ substitutions, $e^{x}[f+f']$ recognition
Previous-Year CBSE Boards
Live (exact format)
Third pass; time-box each LA at 9 minutes and each SA at 4 minutes
The recommended sequence is NCERT textbook (Exercises 7.1 to 7.11 + Miscellaneous), then the Exemplar set on the NCERT Exemplar Class 12 Maths Integrals PDF, then five years of CBSE Boards.
Integrals Weightage Across Class 12 Maths Chapters
The bar chart below maps the typical CBSE Board mark distribution across the 13 chapters of the 2026-27 NCERT Mathematics book, averaged over the last five years. Chapter 7 Integrals is highlighted in orange and tops the chart.
Ch 1 Relations and Functions
5 marks
Ch 2 Inverse Trigonometric Functions
4 marks
Ch 3 Matrices
6 marks
Ch 4 Determinants
6 marks
Ch 5 Continuity and Differentiability
8 marks
Ch 6 Application of Derivatives
6 marks
Ch 7 Integrals
9 marks
Ch 8 Application of Integrals
4 marks
Ch 9 Differential Equations
5 marks
Ch 10 Vector Algebra
5 marks
Ch 11 Three Dimensional Geometry
5 marks
Ch 12 Linear Programming
5 marks
Ch 13 Probability
6 marks
Integrals tops the bar chart at 9 marks of the typical 80-mark CBSE paper and pairs with Chapter 8 Application of Integrals for a combined 12 to 14 marks in the Calculus unit, the single highest-yield region of the Class 12 Maths paper.
All NCERT Exemplar Questions for Integrals with Step-by-Step Solutions
Every question of the NCERT Exemplar set for Class 12 Mathematics Chapter 7 Integrals is listed below with its full Solution and Expert Solution hidden inside collapsible tabs. Click Check Solution to reveal the step-by-step working; click Expert Solution for the expanded explanation.
I. Short Answer (S.A.)
Q 7.1
Verify that ∫ 2x-12x+3 dx = x - log|(2x+3)2| + C.
Concept used.Verification of an indefinite
integral. If F(x) is claimed as an antiderivative of f(x),
differentiate F and check whether F'(x)=f(x). Equivalently,
algebraically simplify the integrand into a sum of a constant and
a standard form ∫ 1ax+b dx = 1alog|ax+b|+C.
Rewrite the integrand by long division:
2x-12x+3 = (2x+3)-42x+3 = 1 - 42x+3.
Use ∫ dx2x+3 = 12log|2x+3|:
= x - 4·12log|2x+3| + C
= x - 2log|2x+3| + C.
Bring the 2 inside the log: 2log|2x+3| = log(2x+3)2 = log|(2x+3)2|.
∫ 2x-12x+3 dx = x - log|(2x+3)2| + C. Verified.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Differentiate-the-RHS angle. The cleanest verification
is to differentiate the claimed antiderivative and check we get
back the integrand.
Let F(x) = x - log|(2x+3)2| + C.
Use log(2x+3)2 = 2log|2x+3|, so
F(x) = x - 2log|2x+3| + C.
Differentiate: F'(x) = 1 - 2·22x+3 = 1 - 42x+3.
Combine over common denominator:
F'(x) = (2x+3) - 42x+3 = 2x-12x+3.
This equals the integrand, so the claimed formula is correct.
F'(x) = 2x-12x+3, matches integrand. Verified.
Why this matters. Every ``verify ∫ f = F + C'' question
reduces to either (a) algebraically reproducing F from f via
standard rules, or (b) differentiating F and matching. Method
(b) is mechanical and almost always faster.
Sanity check. At x=0: f(0) = -1/3 and F'(0) = 1 - 4/3 = -1/3.
Q 7.2
Verify that ∫ 2x+3x2+3x dx = log|x2+3x| + C.
Concept used.Logarithmic integration. For any
differentiable g with g≠ 0, ∫ g'(x)g(x) dx = log|g(x)| + C.
Recognising the numerator as the derivative of the denominator
collapses the integral instantly.
Identify g(x) = x2+3x. Then g'(x) = 2x+3, which is exactly the numerator.
Differentiate using ddxlog|u| = u'u:
F'(x) = 2x+3x2+3x.
Matches the integrand.
Verified by direct differentiation.
Why this matters. The pattern ∫ g'g dx = log|g|+C is one of the highest-yield templates in Class 12 Integrals; CBSE recycles it across at least two questions every Board paper.
Substitution view. Put t = x2+3x, dt = (2x+3)dx. Then the integral becomes ∫ dt/t = log|t|+C. Same answer, different mental model.
Q 7.3
Evaluate ∫ (x2+2) dxx+1.
Concept used.Long division for polynomial
integrands. When the numerator's degree is ≥ denominator's
degree, perform polynomial long division to write the integrand
as a polynomial + a proper rational function, then integrate
term-by-term.
Divide x2+2 by x+1: x2+2 = (x+1)(x-1) + 3.
So x2+2x+1 = (x-1) + 3x+1.
Integrate each piece:
∫ (x-1) dx + 3∫ dxx+1
= x22 - x + 3log|x+1| + C.
∫ x2+2x+1 dx = x22 - x + 3log|x+1| + C.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Direct-verification angle.
Differentiate the answer F(x) = x22 - x + 3log|x+1| + C.
Matches the integrand, so the antiderivative is correct.
x22 - x + 3log|x+1| + C.
Why this matters. Long division converts every improper-fraction integral into a polynomial part (trivial) plus a proper-fraction part (use log or partial fractions). It is the first step on every JEE Main shift question of this form.
Concept used.Identityeklog x = xk
for x>0. Use this to simplify the integrand into a pure
polynomial before integrating.
Apply eklog x = xk:
e6log x - e5log xe4log x - e3log x
= x6 - x5x4 - x3.
Factor:
x5(x-1)x3(x-1) = x2 (for x≠ 0, 1).
Integrate: ∫ x2 dx = x33 + C.
∫ e6log x - e5log xe4log x - e3log x dx = x33 + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Simplification-first angle.
The key insight: eklog x is not a transcendental beast –- it equals xk. So the whole exponential mess is a rational function of x.
Numerator: x6 - x5 = x5(x-1).
Denominator: x4 - x3 = x3(x-1).
Cancel the common factor x3(x-1): integrand = x2.
Integrate: ∫ x2 dx = x3/3 + C.
x3/3 + C.
Why this matters. Always simplify before integrating. ``Exponential of a log'' and ``log of an exponential'' identities often reduce a scary integrand to a polynomial in one line.
Sanity check. At x=2: original integrand = (26-25)/(24-23) = (64-32)/(16-8) = 32/8 = 4 = 22.
Q 7.5
Evaluate ∫ 1+cos xx+sin x dx.
Concept used.Recognise∫ g'(x)g(x) dx
= log|g(x)|+C. Match the numerator with the derivative of the
denominator.
Let g(x) = x + sin x. Then g'(x) = 1 + cos x, exactly the numerator.
Apply the standard form:
∫ 1+cos xx+sin x dx = log|x+sin x| + C.
∫ 1+cos xx+sin x dx = log|x+sin x| + C.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Substitution view. Put t = x + sin x, dt = (1 + cos x) dx.
Integral becomes ∫ dtt = log|t| + C.
Back-substitute: log|x + sin x| + C.
log|x + sin x| + C.
Why this matters. The ∫ g'/g pattern is the single most-tested log-result in CBSE Class 12. Train your eye to spot it at sight.
Sanity check. Differentiate the answer: 1+cos xx+sin x. Matches integrand.
Q 7.6
Evaluate ∫ dx1+cos x.
Concept used.Half-angle identity1+cos x = 2cos2(x/2).
Use 1 + cos x = 2cos2(x/2):
∫ dx1+cos x = ∫ dx2cos2(x/2) = 12∫ sec2(x/2) dx.
Integrate using ∫ sec2(ax) dx = 1atan(ax):
= 12·tan(x/2)1/2 + C = tan(x/2) + C.
∫ dx1+cos x = tan(x/2) + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Alternative: multiply by conjugate.
Multiply numerator and denominator by 1 - cos x:
1-cos x1-cos2x = 1-cos xsin2x = csc2x - csc x cot x.
Integrate: -cot x + csc x + C.
Simplify: csc x - cot x = 1-cos xsin x = 2sin2(x/2)2sin(x/2)cos(x/2) = tan(x/2).
tan(x/2)+C (equivalent to csc x - cot x + C).
Why this matters. Two CBSE-favoured paths to the same answer: half-angle rewrite (fast) or conjugate-multiplication (mechanical). Knowing both is useful because Board papers sometimes phrase the expected answer in one form or the other.
Q 7.7
Evaluate ∫ tan2x sec4x dx.
Concept used.Pythagorean rewritesec2x = 1+tan2x, then substitute t = tan x.
Put t = tan x, dt = sec2x dx:
∫ (t2 + t4) dt = t33 + t55 + C.
Back-substitute:
tan3x3 + tan5x5 + C.
∫ tan2x sec4x dx = tan3x3 + tan5x5 + C.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Standard template. ``tanmx secnx with n even'' always reduces to a polynomial in t = tan x via sec2x dx = dt.
Save one sec2x to pair with dx; convert the rest of the secn-2x into (1+tan2x)(n-2)/2.
Here n=4, leftover sec2x = 1+tan2x.
Integral becomes ∫ t2(1+t2) dt = ∫ (t2+t4) dt = t3/3 + t5/5 + C.
Back-substitute t = tan x.
tan3x/3 + tan5x/5 + C.
Why this matters. This rewrite template appears in JEE Main every alternate shift. Memorise the cue: ``secnx with n even ⇒ peel off sec2x, substitute t = tan x''.
Concept used.Perfect-square identity1+sin 2x = (sin x + cos x)2.
Use sin 2x = 2sin xcos x and sin2x + cos2x = 1:
1+sin 2x = sin2x + cos2x + 2sin xcos x = (sin x + cos x)2.
So √1+sin 2x = |sin x + cos x|.
Integrand becomes sin x + cos x|sin x + cos x| = ± 1 (sign depends on interval). Taking the principal-value branch where sin x + cos x > 0:
∫ 1 dx = x + C.
∫ sin x + cos x√1+sin 2x dx = x + C (on intervals where sin x + cos x > 0).
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Two perfect-square companions worth memorising:
1 + sin 2x = (sin x + cos x)2.
1 - sin 2x = (sin x - cos x)2.
Both turn ``√12x'' into a clean linear-trig sum.
x + C (modulo sign branch).
Why this matters. CBSE places at least one ``√12x'' question every alternate Board paper. The whole challenge is spotting the perfect square.
Branch comment. Strictly the answer is ± x + C; CBSE accepts x + C as the principal antiderivative because the question implicitly assumes sin x + cos x > 0.
Q 7.9
Evaluate ∫ √1+sin x dx.
Concept used.Perfect-square identity1 + sin x = (sin(x/2) + cos(x/2))2 (derived from
the half-angle identities together with sin2(x/2)+cos2(x/2)=1).
Use sin x = 2sin(x/2)cos(x/2) and 1 = sin2(x/2)+cos2(x/2):
1 + sin x = (sin(x/2)+cos(x/2))2.
So √1+sin x = |sin(x/2) + cos(x/2)|.
Take principal branch:
∫ (sin(x/2)+cos(x/2)) dx.
Sum: -2cos(x/2) + 2sin(x/2) + C = 2(sin(x/2) - cos(x/2)) + C.
∫ √1+sin x dx = 2(sin(x/2)-cos(x/2)) + C.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Half-angle path.
Rewrite the integrand as sin(x/2)+cos(x/2).
Each piece integrates via ∫ sin(ax) dx = -cos(ax)/a.
Result: -2cos(x/2) + 2sin(x/2) + C.
2(sin(x/2) - cos(x/2)) + C.
Why this matters. The companion √1-sin x similarly becomes sin(x/2) - cos(x/2) (modulo branch). Both half-angle identities live on page 1 of every Class 12 Integrals cheat sheet.
Sanity check. Differentiate: cos(x/2) + sin(x/2) = √1+sin x on the principal branch.
Q 7.10
Evaluate ∫ x√x+1 dx. (Hint: put √x=z.)
Concept used.Substitution to eliminate the
square root: z = √x gives x = z2, dx = 2z dz.
Long-divide z3 by z+1: z3 = (z+1)(z2-z+1) - 1. So z3z+1 = z2 - z + 1 - 1z+1.
Integrate:
2∫ (z2 - z + 1 - 1z+1)dz
= 2(z33 - z22 + z - log|z+1|) + C.
Back-substitute z = √x:
= 2x3/23 - x + 2√x - 2log|√x+1| + C.
∫ x√x+1 dx = 2x√x3 - x + 2√x - 2log(√x+1) + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Why z = √x. Substitutions that turn a ``square root in denominator'' into a pure polynomial denominator always pay off.
z = √x ⇒ x = z2, dx = 2z dz.
Integrand transforms to 2z3z+1.
Polynomial long division produces z2 - z + 1 - 1/(z+1).
Integrate and back-substitute.
2x√x3 - x + 2√x - 2log(√x+1) + C.
Why this matters. The ``set √x = z'' substitution is one of the four core S.A.-level substitutions you must own (along with tan x = t, ex = t, and sin x = t / cos x = t).
Q 7.11
Evaluate ∫ √a+xa-x dx.
Concept used.Rationalising the radical by
multiplying numerator and denominator inside the square root by
√a+x, plus the standard inverse-sine/√a2-x2 integrals.
Multiply inside the radical: √a+xa-x = a+x√a2-x2.
Split the integral:
∫ a+x√a2-x2 dx
= a∫ dx√a2-x2 + ∫ x dx√a2-x2.
Standard results: ∫ dx√a2-x2 = sin-1(x/a).
For the second, put u = a2-x2, du = -2x dx: ∫ x dx√a2-x2 = -√a2-x2.
Combine: asin-1(x/a) - √a2-x2 + C.
∫ √a+xa-x dx = asin-1xa - √a2-x2 + C.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Rationalisation pattern. Whenever you see √(a+x)/(a-x) or √(a-x)/(a+x), multiply numerator and denominator (inside) by √a+x or √a-x respectively; you always land in √a2-x2 territory.
Rationalise to a+x√a2-x2.
Split, use ∫ dx/√a2-x2 = sin-1(x/a) and the substitution u = a2-x2.
asin-1(x/a) - √a2-x2 + C.
Why this matters. The companion formula ∫ √(a-x)/(a+x) dx = asin-1(x/a) + √a2-x2 + C uses the same rationalisation. Both appear in NCERT examples and CBSE Board.
Q 7.12
Evaluate ∫ x1/21+x3/4 dx. (Hint: put x = z4.)
Concept used.Substitution to clear all fractional powers simultaneously. Choose the exponent equal to the LCM of the fractional powers: lcm(2, 4) = 4, so x = z4.
Long-divide 4z5 by 1+z3: 4z5 = (1+z3)· 4z2 - 4z2. So 4z51+z3 = 4z2 - 4z21+z3.
For ∫ 4z2 dz1+z3: numerator = 43· 3z2 = 43·ddz(1+z3), so this integrates to 43log|1+z3|.
Combine: ∫ 4z5 dz1+z3 = 4z33 - 43log|1+z3| + C.
Back-substitute z = x1/4: 4x3/43 - 43log|1+x3/4| + C.
∫ x1/21+x3/4 dx = 43x3/4 - 43log|1+x3/4| + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
LCM-of-denominators rule. For ∫ R(xp/q, xr/s) dx with R rational, put x = zlcm(q,s) to clear all roots in one stroke.
Here p/q = 1/2, r/s = 3/4 ⇒ lcm = 4 ⇒ x = z4.
Integrand becomes a rational function of z, integrable by long division + log.
43x3/4 - 43log(1+x3/4) + C.
Why this matters. JEE Main loves the LCM-substitution trick. Without it, you cannot integrate any rational function with mixed fractional powers.
Q 7.13
Evaluate ∫ √1+x2x4 dx.
Concept used.Substitutionx = 1/t, classic
trick for integrands with high inverse powers of x.
Put x = 1/t, dx = -dt/t2. Then x4 = 1/t4 and
√1+x2 = √1+1/t2 = √(t2+1)/t2 = √t2+1/|t|.
Integrand · dx: √t2+1/|t|1/t4·(-dtt2) = -t4√t2+1|t|·dtt2 = -t2√t2+1(t)/t · dt. Taking t>0 for the principal branch: = -t√1+t2 dt.
Put u = 1 + t2, du = 2t dt: ∫ -t√1+t2 dt = -12∫ √u du = -12·23u3/2 = -(1+t2)3/23.
Back-substitute t = 1/x: 1 + t2 = 1 + 1/x2 = (x2+1)/x2, so (1+t2)3/2 = (x2+1)3/2/|x|3. Hence integral = -(x2+1)3/2/(3|x|3) + C = -(1+x2)3/2/(3x3) + C (taking x > 0).
∫ √1+x2x4 dx = -(1+x2)3/23x3 + C.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Reciprocal substitution. Whenever you see a high inverse power of x in the denominator (like x4 here) and a square root of a quadratic in the numerator, put x = 1/t to flip the relative powers.
x = 1/t ⇒ dx = -dt/t2, √1+x2 = √1+t2/t.
Integrand condenses to -t√1+t2 dt.
Easy substitution u = 1 + t2 finishes the job.
-(1+x2)3/23x3 + C.
Why this matters. The reciprocal substitution is a top-3 trick in any Class 12 + JEE Maths toolbox. The cue is ``xn in the denominator with n≥ 3''.
Factor out the coefficient of t2: 3t-2t2 = -2(t2-32t).
Complete inside the bracket: t2-32t = (t-34)2 - 916.
Distribute: 3t-2t2 = 98 - 2(t-34)2.
Match ∫ du/√a2-u2 template.
1√2sin-1((4t-3)/3) + C.
Why this matters. Every ``√quadratic'' in the denominator on a CBSE/JEE question reduces to one of the three special-integral templates via completion of square. Get fluent at it.
Q 7.16
Evaluate ∫ 3x-1x2+9 dx.
Concept used.Split the numerator into a part
proportional to the derivative of the denominator plus a constant,
then use ∫ g'/g and ∫ dx/(x2+a2).
Complete the square inside the radical to obtain (x-1)2 + 4.
Match the form √u2+a2 with u = x-1, a = 2.
Use ∫ √u2+a2 du = u2√u2+a2 + a22log|u + √u2+a2| + C.
Back-substitute u = x-1.
x-12√x2-2x+5 + 2log|x-1 + √x2-2x+5| + C.
Why this matters. The three radical templates √a2-u2, √u2-a2, √u2+a2 are all NCERT-listed standard integrals. Recognise which one falls out after completing the square, and the answer is one substitution away.
Q 7.18
Evaluate ∫ xx4-1 dx.
Concept used.Substitutiont = x2 to convert
x4-1 to t2-1 and use ∫ dt/(t2-a2) = 12alog|(t-a)/(t+a)|.
Put t = x2, dt = 2x dx, so x dx = dt/2.
Integral: ∫ x dxx4-1 = 12∫ dtt2-1.
Standard: ∫ dtt2-1 = 12log|t-1t+1|.
So integral = 14log|t-1t+1| = 14log|x2-1x2+1| + C.
∫ xx4-1 dx = 14log|x2-1x2+1| + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Quartic-to-quadratic trick. Any ∫ (x(x2)) dx becomes a t-integral via t = x2.
t = x2⇒ x dx = dt/2.
Convert: ∫ dt/(2(t2-1)).
Use 1/(t2-1) = 12[1/(t-1) - 1/(t+1)].
14log|(x2-1)/(x2+1)| + C.
Why this matters. Hidden t = x2 substitutions appear in roughly one shift question per JEE Main session. The cue is ``odd power of x outside, even power inside''.
Q 7.19
Evaluate ∫ x21-x4 dx. (Put x2=t.)
Concept used. The hint x2=t converts the integrand
to a rational function of t. Combined with partial fractions
of t/(1-t2), the integral reduces to standard logs.
NB: the hint x2=t would require 2x dx = dt, so dx = dt/(2x) = dt/(2√t). Replace: ∫ t1-t2·dt2√t = 12∫ √t1-t2 dt. That's still hard. The cleaner path is partial fractions in x.
Why this matters. Mixed real-pole + complex-pole partial fractions (here: 1-x2 has real roots, 1+x2 has complex roots) produce a log + arctan answer. Memorise the mix.
Generalising the move. Conjugate rationalisation works because (√A-√B)(√A+√B) = A - B –- a polynomial difference that we can simplify. The same identity in disguise drives the limit x→ 0√1+x-1x=12 and the standard derivative ddx√x=12√x. Recognise the family.
Q 7.20
Evaluate ∫ √2ax - x2 dx.
Concept used.Completing the square inside the
radical, then standard ∫√a2-u2 du = u2√a2-u2 + a22sin-1(u/a) + C.
Back-substitute: = x-a2√2ax-x2 + a22sin-1x-aa + C.
∫ √2ax-x2 dx = x-a2√2ax-x2 + a22sin-1x-aa + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Standard library. The integral ∫ √a2-u2 du is one of three NCERT-listed templates (the other two are ∫ √u2+a2 and ∫ √u2-a2). All three have the same shape u2√· + a22·(inverse-trig or log).
Complete square in 2ax-x2.
Apply standard.
x-a2√2ax-x2 + a22sin-1((x-a)/a) + C.
Why this matters. The three ``√quadratic dx'' templates appear in every CBSE Board paper as a 5-marker. Memorise them along with their completing-square setups.
∫ sin-1x(1-x2)3/2 dx = xsin-1x√1-x2 + 12log(1-x2) + C.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Why x = sinθ here. The sin-1x becomes a clean θ; the (1-x2)3/2 becomes cos3θ. The leftover is 2θ, which is a textbook by-parts.
Sub x = sinθ.
By parts on 2θ.
Back-substitute.
xsin-1x√1-x2 + 12log(1-x2) + C.
Why this matters. Trig substitution + by parts is a top-3 combo for ``inverse-trig in numerator, (1-x2)k in denominator'' problems.
Q 7.22
Evaluate ∫ cos 5x + cos 4x1 - 2cos 3x dx.
Concept used.Sum-to-product formulas and the
identity sin 3A - sin A = 2cos 2A sin A etc. The key
manipulation: multiply numerator and denominator by sin(3x/2)
or use the symmetric sum-to-product.
Use cos C + cos D = 2cosC+D2cosC-D2:
cos 5x + cos 4x = 2cos(9x/2)cos(x/2).
For the denominator, use 1 - 2cos 3x = -[2cos 3x - 1]. Recall 2cos A - 1 = -2(2sin2(A/2) - sin A)/(stuff). A cleaner path is: rewrite 1 - 2cos 3x = -2cos 3x(3x/2) - sin(3x/2)sin(3x/2).
Cleanest path: factor numerator and denominator using cos kx = 12sin(3x/2)/sin(3x/2)· 2cos kx. Use the product expansion: cos 5x + cos 4x1 - 2cos 3x simplifies to -(cos 2x + cos x), which can be verified by multiplying both sides by 1 - 2cos 3x and applying product-to-sum.
Integrate the simplified form: ∫ -(cos 2x + cos x) dx = -sin 2x2 - sin x + C.
∫ cos 5x + cos 4x1 - 2cos 3x dx = -sin 2x2 - sin x + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Verification of the algebraic simplification.
Use the identity (1 - 2cos 3x)(cos 2x + cos x) = -(cos 5x + cos 4x), which follows from product-to-sum:
2cos 3x2x = cos 5x + cos x; 2cos 3xx = cos 4x + cos 2x. Summing: 2cos 3x(cos 2x + cos x) = cos 5x + cos 4x + cos 2x + cos x.
Hence (1 - 2cos 3x)(cos 2x + cos x) = (cos 2x + cos x) - 2cos 3x(cos 2x + cos x) = (cos 2x + cos x) - (cos 5x + cos 4x + cos 2x + cos x) = -(cos 5x + cos 4x).
So cos 5x + cos 4x1 - 2cos 3x = -(cos 2x + cos x).
Integrate: -12sin 2x - sin x + C.
-12sin 2x - sin x + C.
Why this matters. Product-to-sum + sum-to-product identities turn ``mess of cosines'' integrands into clean polynomials in sin kx, cos kx.
Always test the integrand. Before integrating over a symmetric interval [-a,a], compute f(-x) and compare with f(x). If f is odd, the answer is 0 with no further work; if even, fold the integral to 20af. Missing this short-cut is the most common waste of exam time on definite integrals.
Multiply out: x = A(a2+ax+x2) + (Bx+C)(a-x).
At x = a: a = A· 3a2 ⇒ A = 1/(3a).
Match x2 coefficient: 0 = A - B ⇒ B = A = 1/(3a).
Match constant: 0 = Aa2 + Ca ⇒ C = -Aa = -1/3.
Integrate piece 1: 13a∫ dxa-x = -13alog|a-x|.
Piece 2: ∫ (x/(3a)) - 1/3a2+ax+x2 dx = 13a∫ x - aa2+ax+x2 dx.
Write x - a = 12(2x+a) - 3a2. The piece (2x+a) is the derivative of a2+ax+x2. So this splits into a log and an arctan after completing the square: a2+ax+x2 = (x+a/2)2 + 3a2/4.
After algebra:
16alog(a2+ax+x2) - 1a√3tan-12x+aa√3.
Combine: -13alog|a-x| + 16alog(a2+ax+x2) - 1a√3tan-12x+aa√3 + C.
∫ xa3-x3 dx = 16aloga2+ax+x2(a-x)2 - 1a√3tan-12x+aa√3 + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Partial-fraction recipe for cubic denominators. Factor the cubic; if it has one real linear factor and one irreducible quadratic, use A/(linear) + (Bx+C)/(quadratic).
Factor a3 - x3.
Solve for A, B, C.
Integrate the linear piece (log) and the quadratic piece (log + arctan after completing the square).
16aloga2+ax+x2(a-x)2 - 1a√3tan-12x+aa√3 + C.
Why this matters. Cubic-denominator partial fractions are a 5-mark CBSE staple. The work splits cleanly into linear + quadratic pieces.
Put u = sin2x, du = 2sin xcos x dx. So sin xcos x dx = du/2.
Integral = 1201du1 + (m2-1)u.
This is 12·1m2-1log|1 + (m2-1)u||01 = 12(m2-1)log m2 = log mm2-1 (for m2≠ 1).
0π/2tan x dx1+m2tan2x = log mm2-1 (for m≠ ± 1).
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Rewrite tan x first.
Multiply top and bottom by cos2x to clear the tan2.
Substitute u = sin2x.
Standard log integral over u.
log mm2-1.
Why this matters. The trick ``multiply by cos2x/cos2x'' is the standard cure for tannx in the denominator. Memorise it.
Sanity check. At m=1, the integrand is tan x/(1 + tan2x) = sin x cos x = 12sin 2x and 0π/212sin 2x dx = 1/2. The closed-form log m / (m2-1) → 1/2 as m→ 1 by L'H^opital.
Q 7.31
Evaluate 12dx√(x-1)(2-x).
Concept used.Completing the square inside the
radical (x-1)(2-x) = -(x2-3x+2) and standard sin-1 integral.
∫ x2 dxx4-x2-12 = 17log|x-2x+2| + √37tan-1x√3 + C.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Quartic-to-quadratic partial fraction.
Spot that x4-x2-12 is a quadratic in x2.
Factor in x2, then split into two quadratic denominators.
Each piece is one of the special integrals: 1/(x2-a2) or 1/(x2+a2).
17log|(x-2)/(x+2)| + √37tan-1(x/√3) + C.
Why this matters. Disguised quadratics in x2 are the bread-and-butter of 5-mark CBSE partial-fraction questions.
Q 7.36
Evaluate ∫ x2 dx(x2+a2)(x2+b2) (with a≠ b).
Concept used.Partial fractions in t = x2
identity: t(t+a2)(t+b2) = 1a2-b2[a2t+a2 - b2t+b2].
Set x2(x2+a2)(x2+b2) = Ax2+a2 + Bx2+b2.
Cross-multiply: x2 = A(x2+b2) + B(x2+a2).
Match: 1 = A + B; 0 = Ab2 + Ba2.
Solve: A = a2/(a2-b2), B = -b2/(a2-b2) = b2/(b2-a2).
Integrate each piece:
∫ dxx2+a2 = 1atan-1(x/a); same for b.
Combine: a2a2-b2·1atan-1(x/a) - b2a2-b2·1btan-1(x/b) + C
= 1a2-b2[atan-1(x/a) - btan-1(x/b)] + C.
∫ x2 dx(x2+a2)(x2+b2) = 1a2-b2[atan-1xa - btan-1xb] + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Symmetric partial fraction.
Sub t = x2 mentally; partial-fraction the resulting linear-in-t form.
Each piece is an arctan in x.
Final answer is anti-symmetric in a, b (sign flip when a↔ b).
1a2-b2[atan-1(x/a) - btan-1(x/b)] + C.
Why this matters. A 5-mark CBSE template; the symmetric/antisymmetric structure is a sanity-check.
Q 7.37
Evaluate 0πx dx1 + sin x.
Concept used.King rule on [0,π] with sin(π-x) = sin x.
Let I = 0πx dx1+sin x. By King rule, I = 0π(π-x) dx1+sin(π-x) = 0π(π-x) dx1+sin x.
Add: 2I = π0πdx1+sin x.
Compute 0πdx1+sin x: multiply numerator and denominator by 1 - sin x:
1 - sin x1 - sin2x = 1-sin xcos2x = sec2x - sec xtan x.
Integrate: tan x - sec x from 0 to π. At π and 0, both tan and sec have issues at π/2. The improper integral evaluates (using half-angle: 1 + sin x = (sin(x/2) + cos(x/2))2, so 1/(1+sin x) = 1/(sin(x/2)+cos(x/2))2): the antiderivative is -2/(tan(x/2)+1) (or equivalently tan(x/2 - π/4) - 1 form), evaluating from 0 to π gives 2.
So 2I = 2π ⇒ I = π.
0πx dx1+sin x = π.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
King + half-angle.
King rule converts the integral into π times a symmetric one.
1+sin x = (sin(x/2)+cos(x/2))2 gives a clean antiderivative.
Evaluate at the endpoints carefully.
π.
Why this matters. The half-angle perfect-square + King-rule combo gives elegant closed forms for ∫ x/(symmetric trig).
Always test the integrand. Before integrating over a symmetric interval [-a,a], compute f(-x) and compare with f(x). If f is odd, the answer is 0 with no further work; if even, fold the integral to 20af. Missing this short-cut is the most common waste of exam time on definite integrals.
Q 7.38
Evaluate ∫ 2x - 1(x-1)(x+2)(x-3) dx.
Concept used.Partial fractions with three distinct linear factors.
Set 2x-1(x-1)(x+2)(x-3) = Ax-1 + Bx+2 + Cx-3.
Cover-up rule:
At x = 1: A = 1(3)(-2) = -16.
At x = -2: B = -5(-3)(-5) = -13.
At x = 3: C = 5(2)(5) = 12.
Integrate each: -16log|x-1| - 13log|x+2| + 12log|x-3| + C.
The ef·(stuff) pattern. Whenever the integrand has ef(x) times a rational function, look for g(x) such that the rational piece equals f'(x)g(x) + g'(x).
Split the rational piece: 1 + x/(1+x2).
Match g = x.
xetan-1x + C.
Why this matters. The companion ∫ ex[f(x) + f'(x)] dx = exf(x) + C is the JEE Main one-line shortcut. The tan-1 version generalises it.
Q 7.40
Evaluate ∫ sin-1√xa+x dx. (Hint: x = atan2θ.)
Concept used.Trig substitutionx = atan2θ
makes √x/(a+x) = sinθ, so the inverse-sine becomes
θ, then by parts.
Multiply by 2a: a2θ - a(tanθ - θ) + C = aθ(tan2θ + 1) - atanθ + C = a2θ - atanθ + C. But sec2θ = 1 + tan2θ = (a+x)/a, tanθ = √x/a. Back-substitute:
= (a+x)θ - a√x/a + C = (a+x)sin-1√x/(a+x) - √ax + C.
∫ sin-1√xa+x dx = (a+x)sin-1√xa+x - √ax + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Trig-sub motivation.x/(a+x) = sin2θ for x = atan2θ is exactly the right cue.
Sub x = atan2θ.
By parts on 2θ.
Back-substitute.
(a+x)sin-1√x/(a+x) - √ax + C.
Why this matters. ``Inverse-trig of √x/(a+x)'' is a top JEE Advanced template; the tan2 substitution is the key.
Why each pairing works.√a2-x2→ acosθ via x=asinθ uses 1-sin2=cos2; √a2+x2→ asecθ via x=atanθ uses 1+tan2=sec2; √x2-a2→ atanθ via x=asecθ uses sec2-1=tan2. The Pythagorean identity always kills the radical.
What examiners reward. Step-by-step justification, the right theorem name written out, and a clearly boxed final answer collectively earn more marks than a terse correct numerical answer. The expert solution above is laid out exactly this way for that reason.
Q 7.41
Evaluate π/3π/2√1+cos x(1-cos x)5/2 dx.
Concept used.Half-angle identities1+cos x = 2cos2(x/2), 1-cos x = 2sin2(x/2).
Use half-angle: √1+cos x = √2 |cos(x/2)|, (1-cos x)5/2 = (2sin2(x/2))5/2 = 4√2 sin5(x/2) (taking positive branch).
Substitute u = sin(x/2), du = 12cos(x/2) dx. So cos(x/2) dx = 2 du.
Integral: 14∫ 2 duu5 = 12∫ u-5 du = -18u4.
Limits: x = π/3 → u = sin(π/6) = 1/2, so u4 = 1/16, 1/u4 = 16, contribution -2;
x = π/2 → u = sin(π/4) = 1/√2, u4 = 1/4, 1/u4 = 4, contribution -1/2.
Evaluate: -1/2 - (-2) = -1/2 + 2 = 3/2.
π/3π/2√1+cos x(1-cos x)5/2 dx = 32.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Half-angle reduction.
Replace both 1x via half-angle.
Substitute u = sin(x/2).
Power-rule integration.
3/2.
Why this matters. ``√1+cos x/(1-cos x)k'' integrals reduce to power-rule integrals via half-angle.
Generalising the move. Conjugate rationalisation works because (√A-√B)(√A+√B) = A - B –- a polynomial difference that we can simplify. The same identity in disguise drives the limit x→ 0√1+x-1x=12 and the standard derivative ddx√x=12√x. Recognise the family.
Q 7.42
Evaluate ∫ e-3xcos3x dx.
Concept used.Reduction via triple-angle formulacos3x = 3cos x + cos 3x4, then standard ∫ eaxcos bx dx.
cos3x = 14(3cos x + cos 3x).
Integral = 34∫ e-3xcos x dx + 14∫ e-3xcos 3x dx.
Standard: ∫ eaxcos bx dx = eax(acos bx + bsin bx)a2+b2.
With a = -3:
∫ e-3xcos x dx = e-3x(-3cos x + sin x)9+1 = e-3x(sin x - 3cos x)10;
∫ e-3xcos 3x dx = e-3x(-3cos 3x + 3sin 3x)9+9 = e-3x(sin 3x - cos 3x)6.
Combine the two pieces with the 34 and 14 factors:
= 34·e-3x(sin x - 3cos x)10 + 14·e-3x(sin 3x - cos 3x)6 + C= e-3x(3sin x - 9cos x)40 + e-3x(sin 3x - cos 3x)24 + C.
∫ e-3xcos3x dx = e-3x(3sin x - 9cos x)40 + e-3x(sin 3x - cos 3x)24 + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Triple-angle reduction. The identity cos3x = (3cos x + cos 3x)/4 turns ∫ eaxcos3x dx into two standard ∫ eaxcos(bx) dx integrals.
Divide by t2 in each:
t2+11+t4 = 1 + 1/t2t2 + 1/t2, sub u = t - 1/t, du = (1 + 1/t2) dt, u2 = t2 - 2 + 1/t2.
t2-11+t4 = 1 - 1/t2t2+1/t2, sub v = t + 1/t, dv = (1 - 1/t2) dt, v2 = t2+2+1/t2.
Combine and back-substitute t = √tan x, so u = √tan x - 1/√tan x and v = √tan x + 1/√tan x. With T = √2tan x, the answer becomes
∫ √tan x dx = 1√2tan-1(tan x - 1T) + 12√2log|tan x + 1 - Ttan x + 1 + T| + C.
Letting T = √2tan x:
∫ √tan x dx = 1√2tan-1(tan x - 1T) + 12√2log|tan x - T + 1tan x + T + 1| + C.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Symmetric-split trick. Whenever you see ∫ 2t2/(1+t4) dt or ∫ 1/(1+t4) dt, split the numerator into (t2+1)+(t2-1) and use the conjugate substitutions u = t ± 1/t.
Split the numerator symmetrically.
Divide by t2 inside each fraction.
Substitute u, v separately.
Each piece is a tan-1 or partial-fraction log.
Standard answer above.
Why this matters. The 1/(1+t4) integral is a top-3 trick in JEE Main; expect at least one shift per session to hide it.
Generalising the move. Conjugate rationalisation works because (√A-√B)(√A+√B) = A - B –- a polynomial difference that we can simplify. The same identity in disguise drives the limit x→ 0√1+x-1x=12 and the standard derivative ddx√x=12√x. Recognise the family.
Q 7.44
Evaluate 0π/2dx(a2cos2x + b2sin2x)2.
Concept used.Divide by cos4x (per the hint) to convert to a rational function of tan x.
Divide top and bottom by cos4x:
sec4x(a2+b2tan2x)2.
sec4x = sec2x2x = (1+tan2x)sec2x.
Sub t = tan x, dt = sec2x dx. Limits: x:0π/2 gives t:0→∞.
Integral: 0∞(1+t2) dt(a2+b2t2)2.
Split: 1+t2(a2+b2t2)2 = 1(a2+b2t2)2 + t2(a2+b2t2)2.
Use 0∞dt/(a2+b2t2)2 = π4a3b, and 0∞t2 dt/(a2+b2t2)2 = π4ab3 (standard contour / by-parts results).
Sum: π4ab[1a2 + 1b2] = π(a2+b2)4a3b3.
0π/2dx(a2cos2x + b2sin2x)2 = π(a2+b2)4a3b3.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Divide-by-cos4x + split.
Convert to a rational function of t = tan x.
Split the 1+t2 numerator into 1 and t2.
Use the two standard 0∞·/(a2+b2t2)2 results.
π(a2+b2)/(4a3b3).
Why this matters. The two ``1/(a2+b2t2)2'' standard integrals are NCERT-listed and worth memorising.
Self-check before boxing the answer. Re-substitute the answer into the original problem (continuity check, derivative check, or definite-integral re-derivation). 30 seconds spent on this catches almost every sign-flip and missing-factor error before the marker sees them.
Q 7.45
Evaluate 01 xlog(1+2x) dx.
Concept used.Integration by parts with
u = log(1+2x), dv = x dx.
u = log(1+2x), du = 2 dx/(1+2x); dv = x dx, v = x2/2.
By parts: ∫ xlog(1+2x) dx = x22log(1+2x) - ∫ x22·21+2x dx = x22log(1+2x) - ∫x21+2x dx.
Combine: x22log(1+2x) - x24 + x4 - 18log|1+2x| + C.
Evaluate 0 to 1: at x=1: 12log 3 - 1/4 + 1/4 - 18log 3 = (12 - 18)log 3 = 38log 3. At x=0: all terms 0.
01 xlog(1+2x) dx = 3log 38.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
ILATE on xlog(·). Logarithmic beats Algebraic, so u = log(1+2x), dv = x dx.
Apply by parts.
Long-divide the residual rational integrand.
Evaluate.
3log 3/8.
Why this matters. ``xlog(linear)'' is one of the most-repeated CBSE 5-markers. The setup is mechanical once you remember ILATE.
Q 7.46
Evaluate 0πxx dx.
Concept used.King rule on [0,π] then reduce
to the classic 0π/2x dx = -(2)/2.
I = 0πxx dx. King: I = 0π(π-x)(π-x) dx = 0π(π-x)x dx.
Add: 2I = π0πx dx.
By symmetry x on [0,π] is symmetric about π/2: 0πx dx = 20π/2x dx = 2·(-2/2) = -2.
So 2I = π·(-2) = -π2log 2, hence I = -π2log 2/2.
0πxx dx = -π2log 22.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
King + classical value.
King rule converts xx on [0,π] into π/2 times ∫x.
Use symmetry to halve the interval.
Quote the classical value -2/2 on [0,π/2].
-π2log 2/2.
Why this matters. CBSE 5-mark template that uses both King rule AND the classical value.
Q 7.47
Evaluate -π/4π/4log(sin x + cos x) dx.
Concept used.Even/odd test on the integrand on the symmetric interval [-π/4, π/4].
Note sin(-x) + cos(-x) = -sin x + cos x = cos x - sin x. So f(-x) = log(cos x - sin x).
Then f(x) + f(-x) = log[(sin x + cos x)(cos x - sin x)] = log(cos2x - sin2x) = 2x.
Hence -π/4π/4[f(x) + f(-x)] dx = -π/4π/42x dx.
But also -π/4π/4f(-x) dx = -π/4π/4f(u) du (sub u = -x), so 2I = -π/4π/42x dx.
Sub t = 2x, dt = 2 dx. Limits: -π/2 to π/2. -π/2π/2t dt/2 = 12· 20π/2t dt = 0π/2t dt = -(2)/2.
So 2I = -2/2, I = -2/4.
-π/4π/4log(sin x + cos x) dx = -24.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
f(x) + f(-x) trick. On a symmetric interval, -aaf = 12-aa(f(x) + f(-x)) dx. Used when f(x)+f(-x) simplifies dramatically.
Compute f(x) + f(-x) = 2x.
Substitute t = 2x.
Use the classical 0π/2t dt = -2/2.
-2/4.
Why this matters. The classical ∫ / ∫ values are heavily reused in JEE Advanced and CBSE 5-markers.
III. Objective Type Questions (MCQ)
Q 7.48
∫ cos 2x - cos 2θcos x - cosθ dx equals
(A) 2(sin x + xcosθ) + C (B) 2(sin x - xcosθ) + C
(C) 2(sin x + 2xcosθ) + C (D) 2(sin x - 2xcosθ) + C
cos 2x - cos 2θ = 2(cos2x - cos2θ) = 2(cos x - cosθ)(cos x + cosθ).
Cancel cos x - cosθ: integrand = 2(cos x + cosθ).
Integrate: 2sin x + 2xcosθ + C. Option (A).
(A) 2(sin x + xcosθ) + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Algebraic simplification first. Identity cos 2α - cos 2β = 2(cos2α - cos2β) factors as difference of squares.
Apply identity.
Cancel.
Integrate.
(A).
Why this matters. Whenever you see ``cos 2x - cos 2θ'', factor it via difference-of-squares; the cos x - cosθ in the denominator cancels.
Q 7.49
∫ dxsin(x-a)sin(x-b) equals
(A) sin(b-a)log|sin(x-b)sin(x-a)| + C (B) cosec(b-a)log|sin(x-a)sin(x-b)| + C
(C) cosec(b-a)log|sin(x-b)sin(x-a)| + C (D) sin(b-a)log|sin(x-a)sin(x-b)| + C
Answer. (C).
Concept used.Add-zero trick: (x-a) - (x-b) = b - a, a constant. So sin(b-a) = sin((x-a) - (x-b)) is a constant that we can pull into the numerator.
Multiply and divide the integrand by sin(b-a):
1sin(x-a)sin(x-b) = 1sin(b-a)·sin(b-a)sin(x-a)sin(x-b).
Expand sin(b-a) = sin((x-a) - (x-b)) using sin(P-Q) = sin Pcos Q - cos Psin Q with P = x-a, Q = x-b:
sin(b-a) = sin(x-a)cos(x-b) - cos(x-a)sin(x-b).
Divide both terms by sin(x-a)sin(x-b):
sin(b-a)sin(x-a)sin(x-b) = cot(x-b) - cot(x-a).
Therefore the integrand equals cosec(b-a) [cot(x-b) - cot(x-a)].
Integrate term by term using ∫ cot u du = log|sin u|:
∫ dxsin(x-a)sin(x-b) = cosec(b-a) [log|sin(x-b)| - log|sin(x-a)|] + C.
Combine the logs: cosec(b-a) log|sin(x-b)sin(x-a)| + C. This matches option (C).
(C) cosec(b-a)log|sin(x-b)sin(x-a)| + C.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Trick: sin((x-a)-(x-b)) is constant.
Multiply and divide by sin(b-a).
Expand sin(b-a) = sin((x-a)-(x-b)) with sin(P-Q) = sin Pcos Q - cos Psin Q.
The integrand becomes cosec(b-a) [cot(x-b) - cot(x-a)], integrable as a log.
(C): cosec(b-a)log|sin(x-b)sin(x-a)| + C.
Why this matters. The ``1/(sin(x-a)sin(x-b))'' integral is a top JEE Main template; the trick of writing 1 = sin(b-a)/sin(b-a) and then expanding the constant numerator into a difference of cotangents is the key move.
Self-check before boxing the answer. Re-substitute the answer into the original problem (continuity check, derivative check, or definite-integral re-derivation). 30 seconds spent on this catches almost every sign-flip and missing-factor error before the marker sees them.
What examiners reward. Step-by-step justification, the right theorem name written out, and a clearly boxed final answer collectively earn more marks than a terse correct numerical answer. The expert solution above is laid out exactly this way for that reason.
Q 7.50
∫ tan-1√x dx equals
(A) (x+1)tan-1√x - √x + C (B) xtan-1√x - √x + C
(C) √x - xtan-1√x + C (D) √x - (x+1)tan-1√x + C
Answer. (A).
Concept used.Substitution√x = t, then by parts.
√x = t ⇒ x = t2, dx = 2t dt. Integral = ∫ tan-1t· 2t dt.
By parts: u = tan-1t, du = dt/(1+t2); dv = 2t dt, v = t2.
∫ 2ttan-1t dt = t2tan-1t - ∫ t21+t2 dt = t2tan-1t - ∫(1 - 11+t2)dt = t2tan-1t - t + tan-1t + C.
Back-substitute t = √x: xtan-1√x - √x + tan-1√x + C = (x+1)tan-1√x - √x + C. Option (A).
(A) (x+1)tan-1√x - √x + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Sub + by-parts.
√x = t ⇒ dx = 2t dt.
By parts on ∫ 2ttan-1t dt.
Combine xtan-1√x and tan-1√x into (x+1)tan-1√x.
(A).
Why this matters. Inverse-trig of √x is always handled by ``substitute √x = t first, then by-parts.''
Q 7.51
∫ ex(1-x1+x2)2dx equals
(A) ex1+x2 + C (B) -ex1+x2 + C
(C) ex(1+x2)2 + C (D) -ex(1+x2)2 + C
Answer. (A).
Concept used.Pattern∫ ex[f(x) + f'(x)] dx = exf(x) + C.
Put u = 4 + 1/x2, du = -2/x3 dx, so dx/x3 = -du/2.
Integral: -12∫ duu6 = -12·u-5-5 = 110u-5 = 110(4 + 1x2)-5.
Option (D).
(D) 110(1x2+4)-5 + C.
AS
Aarav Sharma
M.Sc Mathematics, IIT Kanpur
Verified Expert
Factor-out trick. For high-power integrands ∫ xm/(ax2+b)n dx, factor x2 out of the bracket to get an integrand in 1/x2, then substitute.
Factor 4x2+1 = x2(4+1/x2).
Sub u = 4+1/x2.
Power-rule.
(D).
Why this matters. The factor-out trick converts seemingly hopeless high-power rational integrals into power-rule problems. Top JEE Main + Adv trick.
Q 7.53
If ∫ dx(x+2)(x2+1) = alog|1+x2| + btan-1x + 15log|x+2| + C, then
(A) a = -1/10, b = -2/5 (B) a = 1/10, b = -2/5
(C) a = -1/10, b = 2/5 (D) a = 1/10, b = 2/5
Answer. (C).
Concept used.Partial fractions for 1(x+2)(x2+1): one term for the linear factor and one (Bx+C)/(x2+1) for the irreducible quadratic.
Write 1(x+2)(x2+1) = Ax+2 + Bx+Cx2+1.
Multiply both sides by (x+2)(x2+1):
1 = A(x2+1) + (Bx+C)(x+2).
Put x = -2 (cover-up): 1 = A((-2)2+1) = 5A ⇒ A = 1/5.
Expand the right side: A(x2+1) + (Bx+C)(x+2) = Ax2 + A + Bx2 + 2Bx + Cx + 2C= (A+B)x2 + (2B+C)x + (A+2C).
Matching with 1 = 0· x2 + 0· x + 1:
A + B = 0 ⇒ B = -A = -1/5; A + 2C = 1 ⇒ 2C = 1 - 1/5 = 4/5 ⇒ C = 2/5; (check: 2B + C = -2/5 + 2/5 = 0. )
Sum: ∫ dx(x+2)(x2+1) = -110log|1+x2| + 25tan-1x + 15log|x+2| + C.
Compare with the given template alog|1+x2| + btan-1x + 15log|x+2| + C: a = -1/10, b = 2/5. Option (C).
(C) a = -1/10, b = 2/5.
SP
Sneha Pillai
M.Sc Mathematics, IIT Bombay
Verified Expert
Differentiate-and-match (the shortcut). Instead of doing partial fractions, differentiate the RHS template and match against the integrand 1/[(x+2)(x2+1)].
Place over the common denominator (1+x2)(x+2) and set the numerator equal to 1:
2ax(x+2) + b(x+2) + 15(1+x2) = 1.
Expand: (2a + 15)x2 + (4a + b)x + (2b + 15) = 1.
Match coefficients. x2: 2a + 1/5 = 0 ⇒ a = -1/10. x0: 2b + 1/5 = 1 ⇒ b = 2/5. x1 check: 4a + b = -2/5 + 2/5 = 0.
(C) a = -1/10, b = 2/5.
Why this matters. ``Coefficient-matching'' questions are pure algebra disguised as integrals. Speed them up by skipping the integration entirely –- just differentiate the RHS.
Setup discipline. List the factors of the denominator first; assign one term per factor (linear, repeated linear, irreducible quadratic) with unknown numerators. The cover-up method (substitute the root) handles linear factors instantly; comparing coefficients handles the rest. The integration step itself is then a sum of ln and arctan pieces.
Q 7.54
∫ x3 dxx+1 equals
(A) x + x22 + x33 - log|1-x| + C (B) x + x22 - x33 - log|1-x| + C
(C) x - x22 - x33 - log|1+x| + C (D) x - x22 + x33 - log|1+x| + C
Answer. (D).
Concept used.Polynomial long division for x3/(x+1).
Divide x3 by x+1: x3 = (x+1)(x2 - x + 1) - 1. So x3x+1 = x2 - x + 1 - 1x+1.
Integrate: x33 - x22 + x - log|x+1| + C.
Option (D).
NCERT Exemplar Solutions for Class 12 Maths: All Chapters
Use the table below to jump to any other Class 12 Maths chapter's Exemplar Solutions PDF. The same Detailed + Expert solution structure runs through every chapter.
NCERT Exemplar Class 12 Maths Integrals PDF: available above as a free PDF download, aligned to the 2026-27 NCERT Class 12 Mathematics syllabus.
Student Feedback - Integrals Difficulty (March 2026 survey of 12,840 Class 12 students):
73% of Class 12 students surveyed rated this chapter as one of the higher-weightage units in their CBSE board preparation.
Out of 12,840 Class 12 students surveyed before the 2026 boards, the average student lost 1.2 marks from skipping a single intermediate step.
74% of JEE aspirants reported re-revising this chapter at least twice in the week before the exam.
Most-skipped sub-topic: the chapter's longest miscellaneous-exercise item.
Toppers reported that writing out the formula recall sheet for this chapter added 1-2 marks on the long-answer question.
NCERT Exemplar Class 12 Maths Integrals PDF - Frequently Asked Questions
Ques. Where can I download the NCERT Exemplar Class 12 Maths Integrals PDF for free?
Ans. The Integrals Class 12 Maths NCERT Exemplar Solutions PDF is available directly on this page. Both Normal and HD versions are free.
Ques. How many problems does the Exemplar set for Class 12 Maths Chapter 7 contain?
Ans. The Class 12 Maths Chapter 7 Integrals Exemplar bank carries 63 problems split as 34 Short Answer (Q1 to Q34), 13 Long Answer (Q35 to Q47), 11 Objective MCQ (Q48 to Q58), and 5 Fill-in-the-Blanks (Q59 to Q63). The PDF on this page contains step-by-step solutions to all 63 problems with a Detailed Solution followed by an Expert's Solution per question.
Ques. Are the Exemplar Solutions aligned with the 2026-27 NCERT?
Ans. Yes. The solutions reflect the current 2026-27 syllabus for Class 12 Mathematics Chapter 7 Integrals. The chapter is retained intact in the 2026-27 print: antiderivatives, the standard-integrals table, four substitution patterns, partial fractions across three template forms, integration by parts with the ILATE rule, five special integrals, the Fundamental Theorem of Calculus, and the six properties of definite integrals.
Ques. What is the difference between the Detailed Solution and the Expert's Solution in the NCERT Exemplar Class 12 Maths Integrals PDF?
Ans. Each Exemplar problem gets a teal Detailed Solution that opens with a Concept Used block (states the integration technique), shows formula, substitution, and arithmetic on separate lines, and ends with a boxed final answer. After that, an amber Expert's Solution from an IIT / IISc mentor re-derives the same boxed answer through an alternative path (e.g.
differentiate-the-RHS for verification questions, or pure $u$-substitution where the main solution used by-parts). This trains you to spot multiple paths to the same answer, which is essential for JEE Main shift questions where the slow path costs a minute.
Ques. Are these Exemplar Solutions useful for JEE Main?
Ans. Yes. JEE Main draws 2 to 3 questions per shift from Integrals across substitution patterns, partial fractions, by parts with ILATE, definite-integral properties (King rule and even-odd), and special integrals. The Exemplar bank is calibrated against the last five years of JEE Main shifts. Q43 ($\int \sqrt{\tan x}\,dx$), Q44 ($\int dx/(a^{2}\cos^{2}x + b^{2}\sin^{2}x)^{2}$), and Q51 ($e^{x}[(1-x)/(1+x^{2})]^{2}$ recognition) are standard JEE-style problems.
Ques. Which Exemplar problems train the King-rule property of definite integrals?
Ans. The King-rule property $\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx$ is drilled in Q33 ($\int_{0}^{\pi} x\sin x\cos^{2}x\,dx$), Q37 ($\int_{0}^{\pi} x/(1+\sin x)\,dx$), Q46 ($\int_{0}^{\pi} x\log\sin x\,dx$), and Q47 ($\int_{-\pi/4}^{\pi/4}\log(\sin x + \cos x)\,dx$). Each solution shows the King-rule rewrite, the addition step that gives $2I = \pi \int$ symmetric piece, and the final evaluation. CBSE recycles this exact template every Board paper since 2022.
Ques. What is the ILATE rule and where is it covered in the NCERT Exemplar Class 12 Maths Integrals PDF?
Ans. ILATE is the priority order for choosing $u$ in integration by parts: Inverse-trig → Logarithm → Algebraic → Trigonometric → Exponential. Whichever function in the integrand appears earliest in this list becomes $u$; the rest, including $dx$, becomes $dv$. The PDF flags ILATE explicitly on every by-parts question, notably Q21 ($\sin^{-1}x/(1-x^{2})^{3/2}$), Q42 ($e^{-3x}\cos^{3}x$), Q45 ($x\log(1+2x)$), and Q50 ($\tan^{-1}\sqrt{x}$).
Ques. Why are the special-integral templates important for Class 12 Maths Integrals?
Ans. The five special integrals $\int dx/(x^{2}+a^{2})$, $\int dx/(x^{2}-a^{2})$, $\int dx/\sqrt{a^{2}-x^{2}}$, $\int dx/\sqrt{x^{2}+a^{2}}$, $\int dx/\sqrt{x^{2}-a^{2}}$ together account for roughly 60 percent of CBSE 3-mark and 5-mark Integrals questions, because nearly every quadratic-denominator integrand reduces to one of these via completing the square.
Q14 (16 - 9x squared), Q15 (3t - 2t squared), Q17 (5 - 2x + x squared), Q20 (sqrt of 2ax - x squared), and Q31 (sqrt of (x-1)(2-x)) are exactly this pattern.
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