Junior-Class Mentor | M.Ed. + TFI Fellow | Updated on - May 24, 2026
The Integrals Class 12 NCERT Solutions solve every problem of Exercise 7.2 in Class 12 Mathematics Chapter 7 Integrals. The working in the the PDF follows the order taught in the NCERT textbook, so the student can move directly between the solutions PDF page and the corresponding 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.
Student Pulse - 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.
Every solved sum in the this chapter follows the canonical 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.
What Exercise 7.2 Covers: The Substitution Map
These notes address this in the same order as the NCERT textbook.
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
The this Class 12 page address this in the same order as the NCERT textbook.
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
The the resource address this in the same order as the NCERT textbook.
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
The chapter notes are written in formal mathematical notation, line by line, in the same convention as the official NCERT print.
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).
Related Resources for Class 12 Maths Chapter 7 Integrals
the PDF: available above as a free PDF download, aligned to the 2026-27 NCERT Class 12 Mathematics syllabus.
Exercise-wise Breakdown of the Integrals Chapter
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
PDF Download Formats and Languages for the Integrals Chapter
The Integrals Class 12 PDF on this page is available in three formats - each suited to a different revision style. The table below summarises what each format is best for:
Format
Best for
Approx. size
Normal-resolution PDF
Phone reading, quick revision between classes
2-3 MB
HD PDF
Print-ready, desk study, board hall photocopy
8-10 MB
Handwritten Notes PDF
Mirrors how a topper writes the chapter under Sunday-revision pace
5-7 MB
The integrals class 12 ncert pdf and the parallel Hindi-medium edition both follow the same notation and equation numbering as the printed NCERT 2026-27 release. Key points students should know:
NCERT-faithful: Every definition, theorem and exercise on the integrals class 12 ncert pdf matches the printed textbook line for line.
Hindi-medium edition: The integrals class 12 pdf is also available in Hindi - same page numbering, same equation labels.
Formula PDF separate: The integrals class 12 formulas pdf is a one-page A4 reference sheet listing every identity used in the chapter.
Solutions PDF separate: The integrals class 12 solutions pdf gives every NCERT exercise worked out step by step.
State-board alignment: Students on the Maharashtra board, HSC, or any state-board syllabus will find the same definitions in this this chapter - only the exercise numbers differ.
Tip: Many toppers keep two parallel copies - a printed formula sheet on A4 for desk revision (the integrals class 12 formulas pdf), and the full these notes on a phone for commute revision. Both files are free and linked above.
Important Questions and Previous Year Trends for the Integrals Chapter
The most repeated question patterns in CBSE Class 12 Maths for the Integrals chapter have settled into a stable cluster across 2019 to 2024 boards. Three question templates account for over 80% of the marks this chapter contributes:
Template
Typical Marks
What it tests
Proof / property verification
3 marks
Students show that a given relation/function/expression satisfies the chapter's definitions.
One-step computation
2 marks
Substitution-based item: plug into a known formula and simplify.
Case-study scenario
4 marks
Real-world setup applying the chapter's definitions, introduced in CBSE 2021+ papers.
Walking through one example of each template before the exam covers most of the predictable integrals class 12 important questions you will see on board day.
integrals class 12 previous year questions for 2019-2024 are linked from the PYQ block at the bottom of this page - the exact CBSE phrasings.
The integrals class 12 important questions with solutions set is reused by toppers in the last fortnight of revision.
For NCERT Exemplar practice, the matching integrals class 12 extra questions set adds advanced problems suitable for JEE Main and JEE Advanced.
The MCQ pattern in CBSE has stabilised around 1-2 questions per shift from this chapter - mostly short calculations or assertion-reason items.
Year-wise PYQ Distribution
The table below maps the dominant question type asked from the Integrals chapter across recent CBSE Class 12 Maths boards:
Year
Dominant Question Type
Approx. Marks
2024
Property verification + case-study item
5-6 marks
2023
Computation with proof + assertion-reason MCQ
5-6 marks
2022
Long-answer derivation + 2-mark substitution
5-7 marks
2021
Definition recall + property check
4-5 marks
2020
One-step computation + 3-mark proof
5 marks
The full integrals class 12 important questions with solutions set (every year, every paper, every question type) is linked from the PYQ page at the bottom of this article.
How the Integrals Notes Pair with NCERT Solutions and the Formula Sheet
The Integrals Class 12 notes work best when paired with two sister resources from the Class 12 Maths hub. The table below shows how each resource fits into a typical revision week:
Resource
Use it for
When
Integrals Notes (this page)
Theory, definitions, exam patterns
First pass, before practice
integrals class 12 ncert solutions PDF
Step-by-step solved exercises
Second pass, during NCERT practice
integrals class 12 formulas PDF
One-page identity recall
Third pass, alongside mock papers
Handwritten Notes PDF
Quick reading in topper's handwriting
Anytime, especially commute revision
Around 60 percent of the chapter's scoring vocabulary appears on all three pages, so cross-resource use reinforces recall without adding study time.
The integrals class 12 ncert solutions cover every back-of-chapter exercise plus the miscellaneous exercise.
The integrals class 12 solutions for each individual exercise are indexed by exercise number on the sister NCERT Solutions page (see the Exercise-wise Breakdown table above for direct links).
The integrals class 12 formulas reference sheet is the same A4 file students sometimes refer to as integrals class 12 all formulas - it lists every identity used in the chapter.
State-board references: RD Sharma, ML Aggarwal, Teachoo and the Maharashtra board integrals class 12 textbook PDF all share the same core definitions.
For class-first search phrasings - class 12 integrals solutions, class 12 integrals ncert solutions, ncert class 12 integrals solutions - the same files cover the request.
Reference Books and State-Board Mapping
Students using reference books beyond NCERT, or studying under a state board, can map this chapter cleanly:
Reference
How it maps to Integrals Class 12
RD Sharma Class 12 Integrals
Question patterns overlap with NCERT at ~70%; an advanced supplement.
ML Aggarwal Class 12 Integrals
Solutions style is closer to JEE; good for problem-solving practice.
Teachoo integrals class 12
Free online walkthroughs; useful for video-style learning.
Shaalaa integrals class 12 solutions
State-board (Maharashtra HSC) phrasings; same core definitions.
Maharashtra board integrals class 12 textbook PDF
Same chapter content under the HSC syllabus; exercise numbers differ.
NCERT Exemplar Class 12 Integrals
Advanced problems for JEE Main/JEE Advanced preparation.
How to Use the Integrals Notes Page Most Effectively
The recommended study plan for the Integrals Class 12 chapter splits across three sittings. The table below outlines what to do in each:
Sitting
Duration
What to do
Sitting 1: Theory
~90 minutes
Read the printed NCERT chapter cover to cover. Mark every definition and theorem statement. Then read the formula recall section on this page.
Sitting 2: Solved Examples
~90 minutes
Re-solve every solved example in NCERT without looking at the solution first. Compare your steps against the printed working. Use the integrals class 12 ncert solutions PDF if stuck.
Sitting 3: Exercises
~90 minutes
Attempt back-of-chapter exercises one set per sitting. Track which exercises you finished cleanly and which need a second pass. Click into the linked exercise pages above for verification.
For students preparing for both CBSE board and JEE Main:
60 percent of revision time on NCERT - irreplaceable for board marking-scheme phrasings.
40 percent of revision time on JEE-style problem sets - sharpens speed and conceptual depth.
The integrals class 12 important questions set on the previous-year page is the closest free analogue to a JEE-style problem set for this chapter.
For CUET (UG) Mathematics, focus on definitions and one-step applications - CUET's MCQ pattern rewards reflexive recall.
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)
Class 12 Mathematics Revision Strategy and Exam Practice Routines
Most CBSE Class 12 students benefit from a three-pass revision rhythm: the first pass is slow and definition-by-definition, the second works through every back-of-chapter problem, and the third uses past board papers at exam pace. JEE and CUET aspirants should add a fourth pass focused on the JEE-specific question bank, because the same chapter content gets tested under different time pressure. Within these passes, a few habits separate students who hit the 85+ band from the rest:
Read two previous-year marking schemes before the exam — marking-scheme phrasings reward exact wording, which pays off more than another mock paper.
Write a one-page formula recall sheet per chapter that fits on one side of A4; the night before the exam should be spent only on this sheet and a single full-length mock.
Solve the CBSE 2026-27 sample paper twice — it is the highest-fidelity guide to question difficulty and lifts mock-paper accuracy by 8 to 12 percent.
Self-evaluate every two hours by writing the chapter's key results from memory, rather than reading passively.
Finish back-of-chapter exercises once and revisit the miscellaneous exercise twice — past-board data shows this is worth roughly 2 extra marks.
Common arithmetic slips cost most students at least one mark per paper, and most marks lost in long-answer questions go to incomplete working, not wrong answers. Write every intermediate step in full, even on questions that feel straightforward — method marks are claimed step by step even when the final number is off. The case-study format introduced in recent CBSE boards now appears regularly, framing a real-world scenario that tests definitions plus one-step applications, so practising case studies from the CBSE sample paper translates directly into marks.
Time allocation in the last fortnight matters most. Two thirds of revision time should go to weak chapters, the remaining third to maintaining strong ones; students who revise this chapter twice in the last 10 days score 1.5 to 2 marks higher on past boards. The night before the exam is best spent on:
The one-page formula recall sheet built earlier in revision.
A single full-length mock paper at exam timing.
Avoid learning any new material the night before — sleep matters more.
Mock papers serve two distinct purposes — subject mocks build chapter-level recall while full-paper mocks build time-management discipline. Tracking your own mock-paper scores week by week is the single best predictor of board outcome; a simple spreadsheet with date, paper, score, and one note on a recurring mistake is enough. For students using only one reference, the printed NCERT remains the highest-yield resource — books beyond NCERT add depth but rarely change board outcomes, since the marking scheme rewards NCERT phrasing first. Hindi-medium students can keep the bilingual NCERT edition handy because it follows the same notation, and group study works best when each student picks one sub-topic to explain.
Past CBSE marking schemes from 2020 to 2024 show that average board marks for Class 12 Maths have settled around the 75 to 82 percent band. Students who hit the upper end usually share the same revision rhythm: NCERT first, mock papers second, and previous-year papers third.
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.
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