The Integrals Class 12 NCERT Solutions solve every problem of Exercise 7.2 in Class 12 Mathematics Chapter 7 Integrals. The working in the PDF follows the order taught in the NCERT textbook, so students can move directly between the solutions PDF page and the matching solution without re-reading the theory.
At a glance: 39 sums · substitution method only · 14 algebraic, 17 trigonometric, 8 mixed · PDF size ~8 MB
CBSE Weightage: Exercise 7.2 problems map to 2-mark to 3-mark short-answer slots inside the Integrals 9-11 mark block.
JEE Main Weightage: Substitution-based integration contributes roughly 3-4% of the Calculus questions.
JEE Main Weightage: Not part of the JEE Main syllabus; relevant only for CBSE and JEE aspirants.
Every solved sum in this exercise follows the same substitution workflow:declare t = g(x) , compute dt , rewrite the integral, integrate in t, back-substitute. The back-substitution step is written explicitly because the marking scheme awards 1 of the 3 marks for the final answer in terms of x, not t.
The Collegedunia editorial team has cross-checked every answer against the official NCERT key and the 2026-27 revised textbook. Where a problem admits two natural substitutions (e.g. Q24 with t = sin x or t = cos x), the solution shows the route NCERT prefers and notes the alternative inline.
Exercise 7.2 is structured by integrand type. The 39 problems split into four families: power-of-f(x) times f'(x), logarithmic substitutions f'/f, trigonometric substitutions, and exponential substitutions.
Q range
Integrand family
Substitution
Standard result
1-8
[f(x)]nf'(x)
t = f(x)
[f(x)]n+1n+1 + C
9-15
f'(x)f(x)
t = f(x)
ln|f(x)| + C
16-22
Trigonometric products
t = sin x or cos x
Power of t
23-30
Exponential composites
t = ex or argument of e
ef(x) + C
31-37
Logarithmic composites
t = ln x or argument of log
Power of ln x
38-39
MCQ types
Reverse-engineer differentiation
Match to options
Three problems (Q22, Q34, Q37) are repeat favourites in CBSE boards. Q34 has appeared in two of the last five board papers as a verbatim 3-mark question.
How the Integrals Class 12 NCERT Solutions on the Integrals Class 12 NCERT Solutions Help You
Substitution is the technique with the highest variability in working: two students can pick different t's and both be correct. The Collegedunia solutions show the NCERT-preferred choice every time and flag where an alternative exists.
Substitution declaration as the first line of every solve, e.g. "Let t = 1 + x2, so dt = 2x dx ".
dt = ... step always written explicitly; missing this is the most common 1-mark slip on CBSE marking schemes.
Back-substitution shown as the closing line; the final answer is always in terms of x.
Expert's Solution after each main solve where a slicker substitution exists (e.g. Q26 with t = 1 + ex instead of t = ex).
Integrals Exercise 7.2 Step-by-Step Approach
Every substitution problem in this exercise follows the same five-step routine.
Identify the inner functiong(x) whose derivative appears (up to a constant factor) elsewhere in the integrand.
Declare t = g(x) and compute dt = g'(x) dx . Solve for dx if needed.
Rewrite the integral entirely in terms of t and dt. Every x must disappear.
Integrate in t using a standard antiderivative from the NCERT Section 7.2 table.
Back-substitutet = g(x) and add + C.
If after step 3 some x terms remain, the substitution choice is wrong; restart with a different g(x).
Exam Relevance of Class 12 Maths Chapter 7 Exercise 7.2
Exercise 7.2 style sums are the workhorses of the CBSE Integrals block. They show up as 2-mark VSA and 3-mark short-answer questions every year. The table below maps the last five CBSE board sittings.
Year
Marks from Ex 7.2 style sums
Question type
2025
5
2-mark VSA + 3-mark SA
2024
3
3-mark SA
2023
5
2-mark VSA + 3-mark SA
2022
3
3-mark SA
2021
5
2-mark VSA + 3-mark SA
Across the last five sittings, Exercise 7.2 style sums have carried 3-5 marks every year. They are the second-highest-yield exercise in Chapter 7 after Ex 7.10.
Common Mistakes Students Make in Exercise 7.2
Students lose easy marks on Exercise 7.2 not because the method is hard, but because of small slips in the working. The most common ones are listed below.
Common Mistake: Forgetting to back-substitute. After integrating in t, students sometimes leave the final answer in terms of t instead of x. CBSE deducts 1 mark for this on every 3-mark substitution problem.
Picking the wrong inner function; leftover x terms after step 3 are the tell.
Forgetting the dx to dt conversion factor (e.g. writing dt = 2x instead of dt = 2x dx ).
Missing a constant factor on dt and integrating the wrong expression by a factor of 2 or 1/2.
For Q22 and similar trig-power problems, choosing t = sin x when the integrand has more cosines than sines (the opposite choice is shorter).
Other Resources for Class 12 Maths Chapter 7 Integrals
The Integrals chapter splits into 10 numbered exercises plus a Miscellaneous Exercise. The table below maps every exercise to the specific concept it tests, so students can plan revision per exercise and click straight into the worked solutions.
Mixed indefinite and definite integration problems
All NCERT Solutions for Integrals Ex 7.2 with Step-by-Step Working
Every NCERT textbook question for Class 12 Mathematics Chapter 7 Integrals Ex 7.2 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.
Questions
Q 7.1
Integrate 2x1 + x2.
Concept used.Integration by substitution. If the integrand is of the
form f'(x)f(x), substitute t = f(x) so that dt = f'(x) dx. Then
∫ f'(x)f(x) dx = ∫ dtt = log|t| + C = log|f(x)| + C.
Note that the numerator 2x is exactly the derivative of the denominator
1 + x2: ddx(1 + x2) = 2x.
Substitute t = 1 + x2 so that dt = 2x dx. The integral becomes
∫ 2x1 + x2 dx = ∫ dtt.
Integrate: ∫ dt/t = log|t|.
Back-substitute t = 1 + x2 (which is always positive, so |t| = t):
log(1 + x2) + C.
log(1 + x2) + C
AS
Aarav Sharma
M.Sc Mathematics, IIT Bombay
Verified Expert
Quick reading. The integrand has the form f'f with f(x) = 1+x2.
Read off the anti derivative directly.
f(x) = 1 + x2, f'(x) = 2x.
By the log-derivative rule, ∫ f'/f dx = log|f| + C.
Final: log(1+x2) + C.
log(1 + x2) + C
Q 7.2
Integrate (log x)2x.
Concept used. Substitution. Let t = log x so that dt = dx/x. The integral
then becomes ∫ t2 dt, integrable by the power rule.
Put t = log x. Differentiating both sides: dt = 1x dx, i.e.
dxx = dt.
Substitute: ∫ (log x)2x dx = ∫ t2 dt.
Integrate using the power rule: ∫ t2 dt = t33.
Back-substitute t = log x: (log x)33 + C.
(log x)33 + C
SI
Sneha Iyer
M.Sc Applied Mathematics, IIT Kanpur
Verified Expert
Strategic angle. The presence of 1/x alongside a function of log x is a
fingerprint for substituting t = log x.
t = log x ⇒ dt = dx/x.
Integral becomes ∫ t2 dt = t3/3.
Back-substitute: (log x)3/3 + C.
(log x)33 + C
Q 7.3
Integrate 1x + xlog x.
Concept used. Factor the denominator and use the substitution t = 1 + log x.
Factor: x + xlog x = x(1 + log x).
Put t = 1 + log x. Then dt = 1x dx, i.e. dx/x = dt.
Structural observation. Pull out the x that is common to both denominator
terms, then the 1 + log x that remains has derivative 1/x, which is sitting outside.
Factor denominator: x(1 + log x).
Substitute t = 1 + log x; dt = dx/x.
Integral becomes ∫ dt/t = log|t| + C.
log|1 + log x| + C
Q 7.4
Integrate sin x sin(cos x).
Concept used. Substitute t = cos x so that dt = -sin x dx, i.e.
sin x dx = -dt. The integral reduces to -∫ sin t dt = cos t + C.
Put t = cos x; then dt = -sin x dx ⇒ sin x dx = -dt.
Substitute:
∫ sin x sin(cos x) dx = ∫ sin(cos x)· sin x dx = -∫ sin t dt.
Integrate: -∫ sin t dt = -(-cos t) = cos t.
Back-substitute: cos(cos x) + C.
cos(cos x) + C
PP
Pranav Patel
Ph.D Mathematics, IIT Delhi
Verified Expert
Quick reading. The sin x outside is the chain-rule companion to the inner
cos x.
t = cos x; dt = -sin x dx.
∫ sin x sin(cos x) dx = -∫ sin t dt = cos t = cos(cos x).
Add C.
cos(cos x) + C
Q 7.5
Integrate sin(ax + b)cos(ax + b).
Concept used. Use the double-angle identity
sinθ = 12sin 2θ to collapse the product. Then apply the
standard integral ∫ sin(kx) dx = -1kcos(kx) + C.
Combine: 25t5/2 - 43t3/2. Back-substitute t = x+2:
25(x+2)5/2 - 43(x+2)3/2 + C.
25(x+2)5/2 - 43(x+2)3/2 + C
YJ
Yash Joshi
M.Sc Mathematics, IIT Bombay
Verified Expert
Strategic angle. When the integrand mixes a polynomial in x with a surd of
x + c, substitute the contents of the surd. The polynomial then becomes a polynomial in
t, which integrates routinely.
t = x + 2; x = t - 2; dx = dt.
Integrand becomes (t-2)√t = t3/2 - 2 t1/2.
Power-rule each: get 25t5/2 - 43t3/2.
Back-substitute.
2(x+2)5/25 - 4(x+2)3/23 + C
Q 7.8
Integrate x√1 + 2x2.
Concept used. Substitute t = 1 + 2x2. Then dt = 4x dx, so x dx = dt/4.
Integrate:
aligned
∫ t4/3 dt &= t7/37/3 = 37t7/3,
∫ t1/3 dt &= t4/34/3 = 34t4/3.
aligned
So the integral is 13(37t7/3 + 34t4/3)
= 17t7/3 + 14t4/3.
Back-substitute t = x3 - 1:
17(x3 - 1)7/3 + 14(x3 - 1)4/3 + C.
(x3 - 1)7/37 + (x3 - 1)4/34 + C
IR
Ishaan Reddy
B.Tech CSE, IIT Roorkee
Verified Expert
Strategic angle. Substitute so that the surd (x3-1)1/3 becomes a clean
power. The leftover x5 splits as x3· x2; both x3 and x2 dx map cleanly to
t-terms.
Concept used. Rewrite 1/ex2 = e-x2. Substitute t = -x2 so
dt = -2x dx, i.e. x dx = -dt/2.
Rewrite: xex2 = x e-x2.
Put t = -x2 ⇒ dt = -2x dx ⇒ x dx = -dt/2.
Substitute:
∫ x e-x2 dx = ∫ et · (-dt2) = -12∫ et dt = -12et.
Back-substitute: -12e-x2 + C, equivalently -12 ex2 + C.
-12e-x2 + C
NG
Neha Gupta
M.Sc Mathematics, ISI Kolkata
Verified Expert
Structural observation. The exponent's derivative is -2x; the integrand
contains x, so substitute the exponent.
t = -x2; x dx = -dt/2.
∫ et·(-1/2) dt = -12et.
Back-substitute: -12e-x2 + C.
-12e-x2 + C
Q 7.18
Integrate etan-1x1 + x2.
Concept used. Recall ddxtan-1x = 11 + x2. Substitute
t = tan-1x so dt = dx/(1+x2).
Put t = tan-1x; dt = dx1+x2.
Substitute:
∫ etan-1x dx1 + x2 = ∫ et dt = et.
Back-substitute: etan-1x + C.
etan-1x + C
PS
Priya Sharma
Ph.D Mathematics, IIT Delhi
Verified Expert
Quick reading. The denominator 1/(1+x2) is the derivative of tan-1x.
Substitute t = tan-1x; ∫ et dt = et.
Back-substitute.
etan-1x + C
Q 7.19
Integrate e2x - 1e2x + 1.
Concept used. Divide numerator and denominator by ex to recognise the
hyperbolic-tangent pattern, or substitute t = e2x + 1 so that dt = 2 e2x dx.
Here we use a different clean approach: divide numerator and denominator by ex.
Divide top and bottom by ex:
e2x - 1e2x + 1 = ex - e-xex + e-x.
Notice ddx(ex + e-x) = ex - e-x - exactly the numerator.
Put t = ex + e-x; dt = (ex - e-x) dx. Then
∫ ex - e-xex + e-x dx = ∫ dtt = log|t| = log(ex + e-x) + C,
where the absolute value is dropped since ex + e-x > 0.
log(ex + e-x) + C
AB
Ananya Bhat
M.Sc Mathematics, IIT Bombay
Verified Expert
Structural observation. Divide by ex to expose the log-derivative pattern.
Convert: e2x-1e2x+1 = ex - e-xex + e-x.
Numerator is ddx denominator.
Integral: log(ex + e-x) + C.
log(ex + e-x) + C
Q 7.20
Integrate e2x - e-2xe2x + e-2x.
Concept used. Log-derivative form: the numerator is 12ddx of
the denominator. Substitute t = e2x + e-2x.
Back-substitute (denominator positive):
12log(e2x + e-2x) + C.
12log(e2x + e-2x) + C
AV
Aditi Verma
M.Sc Mathematics, IIT Bombay
Verified Expert
Quick reading. Numerator equals 12 times derivative of denominator.
t = e2x+e-2x; dt = 2(e2x-e-2x) dx.
∫ dt2t = 12log|t|.
Final: 12log(e2x+e-2x) + C.
12log(e2x + e-2x) + C
Q 7.21
Integrate tan2(2x - 3).
Concept used. Use the identity tan2 θ = sec2 θ - 1. Then integrate
each term: ∫ sec2 θ dθ = tanθ and ∫ dθ = θ. Adjust
for the inner linear function via 1/k where the inner derivative is k.
Use tan2(2x-3) = sec2(2x-3) - 1.
Integrate term-by-term:
∫ sec2(2x-3) dx - ∫ 1 dx.
For the first integral, substitute t = 2x - 3, dt = 2 dx:
∫ sec2(2x-3) dx = ∫ sec2tdt2 = 12tan t = 12tan(2x-3).
Combine: 12tan(2x-3) - x + C.
12tan(2x - 3) - x + C
AS
Aanya Singh
M.Sc Mathematics, IIT Bombay
Verified Expert
Strategic angle. Convert tan2 to sec2 - 1 before integrating.
tan2(2x-3) = sec2(2x-3) - 1.
Anti derivative: 12tan(2x-3) - x.
Add C.
tan(2x-3)2 - x + C
Q 7.22
Integrate sec2(7 - 4x).
Concept used.∫ sec2t dt = tan t + C. The inner function 7 - 4x has
derivative -4.
Quick reading. Standard form with inner factor -4.
Anti derivative of sec2 is tan.
Divide by inner derivative -4.
Final: -tan(7-4x)4 + C.
-tan(7-4x)4 + C
Q 7.23
Integrate sin-1x√1 - x2.
Concept used.ddxsin-1x = 1√1-x2. Substitute
t = sin-1x.
Put t = sin-1x; dt = dx√1-x2.
Substitute:
∫ sin-1x√1-x2 dx = ∫ t dt = t22.
Back-substitute: (sin-1x)22 + C.
(sin-1x)22 + C
VN
Vivaan Nair
Ph.D Pure Mathematics, IISc Bangalore
Verified Expert
Quick reading. Pure log-/power-derivative form on sin-1x.
t = sin-1x; ∫ t dt = t2/2.
Back-substitute.
(sin-1x)22 + C
Q 7.24
Integrate 2cos x - 3sin x6cos x + 4sin x.
Concept used. For integrands of the form acos x + bsin xccos x + dsin x,
write the numerator as A·(denominator) + B· ddx(denominator)
and split. The piece with the derivative integrates as a log; the other piece is just x.
Denominator D(x) = 6cos x + 4sin x. Differentiate:
D'(x) = -6sin x + 4cos x = 4cos x - 6sin x.
Express numerator N(x) = 2cos x - 3sin x as AD(x) + BD'(x).
Equate coefficients of cos x and sin x:
6 A + 4 B = 2, 4 A - 6 B = -3.
Solve the linear system. From the first: 3 A + 2 B = 1. Multiply the second by
12: 2 A - 3 B = -32. Multiply the first by 3, second by 2:
9 A + 6 B = 3, 4 A - 6 B = -3.
Add: 13 A = 0 ⇒ A = 0. Then 2 B = 1 ⇒ B = 1/2.
Substitute back: N(x) = 0· D(x) + 12D'(x), i.e.
2cos x - 3sin x = 12(4cos x - 6sin x). (Verify: 12· 4 = 2 , 12·(-6) = -3 .)
Quick reading.cos 2x pairs with sin 2x as a substitution.
t = sin 2x; cos 2x dx = dt/2.
∫ 12t1/2 dt = 13t3/2.
(sin 2x)3/23 + C
Q 7.28
Integrate cos x√1 + sin x.
Concept used.ddx(1 + sin x) = cos x. Substitute t = 1 + sin x.
Put t = 1 + sin x ⇒ dt = cos x dx.
Substitute:
∫ cos x dx√1 + sin x = ∫ dt√t = ∫ t-1/2 dt.
Integrate: ∫ t-1/2 dt = 2 t1/2.
Back-substitute: 2√1 + sin x + C.
2√1 + sin x + C
AV
Ankit Verma
M.Tech CS, IIT Madras
Verified Expert
Quick reading. Numerator cos x is the derivative of 1+sin x.
t = 1+sin x; cos x dx = dt.
∫ dt/√t = 2√t.
2√1 + sin x + C
Q 7.29
Integrate cot xx.
Concept used.cot x = cos xsin x and ddxx = cot x.
Substitute t = x.
Put t = x. Then dt = cos xsin x dx = cot x dx.
Substitute:
∫ cot xx dx = ∫ t dt = t22.
Back-substitute: (x)22 + C.
(x)22 + C
AB
Ananya Bhat
M.Sc Mathematics, IIT Bombay
Verified Expert
Structural observation. Derivative of x is cot x.
t = x; dt = cot x dx.
∫ t dt = t2/2.
(x)22 + C
Q 7.30
Integrate sin x1 + cos x.
Concept used.ddx(1 + cos x) = -sin x. Log-derivative form with a
sign.
Put t = 1 + cos x ⇒ dt = -sin x dx ⇒ sin x dx = -dt.
Substitute:
∫ sin x dx1 + cos x = ∫ -dtt = -log|t|.
Back-substitute (with 1 + cos x ≥ 0 wherever defined):
-log|1 + cos x| + C.
-log|1 + cos x| + C
AV
Aditi Verma
M.Sc Mathematics, IIT Bombay
Verified Expert
Quick reading. Negative log-derivative.
t = 1+cos x; sin x dx = -dt.
∫ -dt/t = -log|t|.
-log|1 + cos x| + C
Q 7.31
Integrate sin x(1 + cos x)2.
Concept used. Substitute t = 1 + cos x; numerator becomes -dt.
Put t = 1 + cos x ⇒ dt = -sin x dx ⇒ sin x dx = -dt.
Substitute:
∫ sin x dx(1 + cos x)2 = ∫ -dtt2 = -∫ t-2 dt.
Integrate: -∫ t-2 dt = -· t-1-1 = 1t.
Back-substitute: 11 + cos x + C.
11 + cos x + C
PS
Priya Sharma
Ph.D Mathematics, IIT Delhi
Verified Expert
Quick reading. Same substitution as Q30, different power.
t = 1+cos x; sin x dx = -dt.
∫ -t-2 dt = 1/t.
11 + cos x + C
Q 7.32
Integrate 11 + cot x.
Concept used. Convert cot x = cos x/sin x, simplify, then either use the
AD + BD' technique on the new integrand or combine with the conjugate.
Convert: 11 + cot x = 11 + cos x/sin x = sin xsin x + cos x.
Apply the AD + BD' decomposition with D = sin x + cos x and
D' = cos x - sin x. Write sin x = A(sin x + cos x) + B(cos x - sin x).
Equate coefficients:
A - B = 1, A + B = 0 ⇒ A = 12, B = -12.
Therefore
sin xsin x + cos x = 12 - 12· cos x - sin xsin x + cos x.
Integrate term-by-term. The first piece gives x/2. For the second, substitute
t = sin x + cos x, dt = (cos x - sin x) dx:
-12∫ dtt = -12log|t| = -12log|sin x + cos x|.
Combine: x2 - 12log|sin x + cos x| + C.
x2 - 12log|sin x + cos x| + C
KM
Karan Mehta
B.Tech Engineering Physics, IIT Bombay
Verified Expert
Strategic angle. Convert cot to ratio, then AD + BD'.
11+cot x = sin xsin x + cos x.
sin x = 12(sin x + cos x) - 12(cos x - sin x).
Integrate: x2 - 12log|sin x + cos x| + C.
x2 - 12log|sin x + cos x| + C
Q 7.33
Integrate 11 - tan x.
Concept used. Convert tan x = sin x/cos x and use the AD + BD'
decomposition.
Convert: 11 - tan x = 11 - sin x/cos x = cos xcos x - sin x.
Take D = cos x - sin x, D' = -sin x - cos x = -(sin x + cos x). Write
cos x = AD + BD':
A(cos x - sin x) - B(sin x + cos x) = (A - B)cos x + (-A - B)sin x.
Equate to cos x: A - B = 1 and -A - B = 0 ⇒ B = -A.
Then A - (-A) = 2 A = 1 ⇒ A = 1/2, B = -1/2.
Substitute:
cos xcos x - sin x = 12 - 12· -(sin x + cos x)cos x - sin x
= 12 + 12· sin x + cos xcos x - sin x.
Compute
ddx(cos x - sin x) = -sin x - cos x = -(sin x + cos x).
Hence (sin x + cos x)/(cos x - sin x) = -D'/D, and
∫ sin x + cos xcos x - sin x dx = -log|cos x - sin x|.
Combine:
∫ dx1 - tan x = 12x - 12log|cos x - sin x| + C.
x2 - 12log|cos x - sin x| + C
TB
Tara Bhat
M.Sc Mathematics, ISI Kolkata
Verified Expert
Structural observation. Same template as Q32 but with cos x - sin x
in the denominator.
11-tan x = cos xcos x - sin x.
cos x = 12(cos x - sin x) + 12(cos x + sin x).
Integrate: x2 - 12log|cos x - sin x| + C.
x2 - 12log|cos x - sin x| + C
Q 7.34
Integrate √tan xsin x cos x.
Concept used. Divide numerator and denominator by cos2x to introduce
sec2x, then substitute t = tan x.
Divide top and bottom by cos2x:
√tan xsin x cos x = √tan x/cos2xsin xcos x/cos2x = √tan x sec2xtan x.
Put t = tan x. Then dt = sec2x dx and √tan x = √t.
Substitute:
∫ √tt dt = ∫ t-1/2 dt = 2 t1/2 = 2√t.
Back-substitute: 2√tan x + C.
2√tan x + C
YJ
Yash Joshi
M.Sc Mathematics, IIT Bombay
Verified Expert
Strategic angle. Multiply top and bottom by sec2x to convert sin xcos x
into tan x form.
sin xcos x = tan x cos2x = tan x/sec2x. So
√tan xsin x cos x = √tan x sec2xtan x.
t = tan x; dt = sec2x dx.
∫ t-1/2 dt = 2√t.
2√tan x + C
Q 7.35
Integrate (1 + log x)2x.
Concept used. Substitute t = 1 + log x so dt = dx/x.
Put t = 1 + log x ⇒ dt = dx/x.
Substitute:
∫ (1+log x)2x dx = ∫ t2 dt = t33.
Back-substitute: (1 + log x)33 + C.
(1 + log x)33 + C
RD
Rohit Desai
M.Sc Mathematics, IIT Bombay
Verified Expert
Quick reading.1/x is the derivative of log x.
t = 1+log x; dt = dx/x.
∫ t2 dt = t3/3.
(1 + log x)33 + C
Q 7.36
Integrate (x + 1)(x + log x)2x.
Concept used. Distribute the (x+1)/x factor: x+1x = 1 + 1x,
which is exactly ddx(x + log x). Substitute t = x + log x.
Rewrite: (x+1)(x+log x)2x = (1 + 1x)(x + log x)2.
Note ddx(x + log x) = 1 + 1x. Put t = x + log x ⇒ dt = (1 + 1x)dx.
Combine: tan x + (-cot x) = tan x - cot x + C. Matches option (B).
Option (B): tan x - cot x + C
AV
Aditi Verma
M.Sc Mathematics, IIT Bombay
Verified Expert
Structural observation. Use 1 = sin2x + cos2x in the numerator.
Split: 1sin2x cos2x = sec2x + csc2x.
Integrate: tan x - cot x + C.
Option (B)
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.
Integrals Class 12 NCERT Solutions - Frequently Asked Questions
Ques. How many questions are in Class 12 Maths Chapter 7 Exercise 7.2?
Ans. Exercise 7.2 of Class 12 Maths Chapter 7 Integrals carries 39 questions in the 2026-27 NCERT. Q1 to Q37 are integration sums solved by substitution and Q38, Q39 are MCQ-style questions on substitution-based antiderivative identification.
Ques. What concept does Exercise 7.2 of Class 12 Maths Chapter 7 cover?
Ans. Exercise 7.2 covers integration by substitution, the technique of choosing t = g(x) so that the integrand reduces to a standard form. The exercise spans four substitution families: power-of-f(x), f'/f, trigonometric, and exponential.
Ques. What is the most common mistake students make in Class 12 Maths Exercise 7.2?
Ans. Forgetting to back-substitute the original variable. After integrating in t, students sometimes leave the final answer in terms of t instead of x. CBSE marking schemes for 2021 to 2025 all deduct 1 mark for this on every 3-mark substitution problem.
Ques. How do I choose the right substitution in Class 12 Maths integration problems?
Ans. Look for an inner function g(x) whose derivative g'(x) appears as a factor in the integrand (possibly up to a constant). Common patterns:
t = 1 + x2 when x dx appears, t = sin x when cos x dx appears, t = ln x when dx / x appears. If after substituting some x terms remain, the choice is wrong.
Ques. How do I download the Class 12 Maths Chapter 7 Exercise 7.2 NCERT Solutions PDF?
Ans. Use the green download button on the this chapter card at the top of this page to save the Collegedunia Class 12 Maths Chapter 7 Exercise 7.2 NCERT Solutions PDF to your device. these notes is free, ad-free, and mapped to the 2026-27 NCERT edition.
Comments