Class 12 Maths NCERT Solutions

~28 min read

Integrals Class 12 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 7, Integrals — 123 questions from Ex 7.1 to Ex 7.11, each worked through step by step in the CBSE marking pattern. Indefinite integrals by substitution, partial fractions and trigonometric methods, definite integrals, and the fundamental theorem of calculus.

Class:12Subject:MathsChapter:7

4 Key Formulas

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 12 Maths Chapter 7?

Chapter 7 carries 11 exercise questions, numbered Ex 7.1 to Ex 7.11. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents12 sections

Revision & Practice

Chapter Overview

Integration is one of the two fundamental operations of calculus, the other being differentiation. Since differentiation and integration are inverse processes, every differentiation formula gives a corresponding integration formula. This chapter begins with antiderivatives, progresses through techniques — substitution, trigonometric identities, partial fractions, and integration by parts — and culminates in definite integrals, the Fundamental Theorem of Calculus, and properties that make evaluation tractable. The exercises below carry every question of the NCERT textbook with short, exam-pattern working.

Board pattern

Integration questions award marks for the technique chosen, intermediate steps, and the final answer with an arbitrary constant C. Always write the substitution explicitly (e.g. Let u = x^2), show the transformed integral, integrate, and back-substitute. For definite integrals, show the antiderivative evaluated at both limits.

$Ex 7.1 — Integration as inverse of differentiation: basic antiderivatives (22 Q)$
$Ex 7.2 — Integration by substitution (22 Q)$
$Ex 7.3 — Integration using trigonometric identities (14 Q)$
$Ex 7.4 — Integration of particular functions: partial fractions I (11 Q)$
$Ex 7.5 — Integration by partial fractions (continued) (15 Q)$
$Ex 7.6 — Integration by parts (15 Q)$
$Ex 7.7 — Definite integrals: fundamental theorem (8 Q)$
$Ex 7.8 — Evaluation of definite integrals by substitution (8 Q)$
$Ex 7.9 — Some properties of definite integrals (9 Q)$
$Ex 7.10 — More definite integral problems (8 Q)$
$Ex 7.11 — Properties and evaluation (10 Q)$
$Misc — Mixed problems on integration (20 Q)$

Exercise 7.1 — Integration as Inverse of Differentiation

18Exercise questions

Final answer

(i) - \frac{1}{2} cos 2 x + C; (ii) \frac{1}{3} sin 3 x + C; (iii) \frac{1}{2} e^{2 x} + C; (i v) - \frac{1}{π} cos π x + C; (v) \frac{1}{π} sin π x + C

Final answer

(i) \frac{1}{4} sin 4 x + C; (ii) ln ∣ cos x ∣ + C

Final answer

(i) \frac{4}{3} e^{3 x} + x + C; (ii) \frac{x ^{3}}{3} - x + C; (iii) \frac{a}{3} x^{3} + \frac{b}{2} x^{2} + c x + C

Step-by-step solution

1(i) Use $sin^{2} θ = \frac{1 - c o s 2 θ}{2}$ :
2Let $θ = 2 x + 5$ : $\int \frac{1 - c o s ( 4 x + 10 )}{2} d x = \frac{x}{2} - \frac{s i n ( 4 x + 10 )}{8} + C$
3(ii) Use $2 sin A cos B = sin (A + B) + sin (A - B)$

Final answer

(i) \frac{x}{2} - \frac{s i n ( 4 x + 10 )}{8} + C; (ii) - \frac{1}{14} cos 7 x + \frac{1}{2} cos x + C

Step-by-step solution

1Multiply numerator and denominator by conjugate, then separate and integrate.

Final answer

(i) x - tan x + sec x + C; (ii) tan x - x + tan \frac{x}{2} + C

Step-by-step solution

1Perform polynomial long division, then integrate term by term.

Final answer

(i) \frac{x ^{3}}{3} + C; (ii) \frac{x ^{2}}{2} + 3 x - 2 ln ∣ x + 2∣ + C; (iii) \frac{x ^{2}}{2} - 2 x + 3 ln ∣ x + 1∣ + C

Final answer

\frac{x ^{2}}{2} + 3 sin x + 2 cos x - 2 ln ∣ x ∣ + C

Final answer

\frac{x ^{3}}{3} - x + ln ∣ x ∣ + C

Step-by-step solution

1Note $sin^{2} x + cos^{2} x = 1$ and $2 sin x cos x = sin 2 x$
2So $1 - sin 2 x = (sin x - cos x)^{2}$ , hence
3 $\frac{s i n x + c o s x}{∣ s i n x - c o s x ∣}$
4For $sin x > cos x$ : $\frac{s i n x + c o s x}{s i n x - c o s x}$ ; integrate by letting $u = sin x - cos x$

Final answer

- ln ∣ sin x - cos x ∣ + C

Step-by-step solution

1Let $u = 1 + sin x$ , $d u = cos x d x$

Final answer

2 1 + sin x + C

Final answer

sec x + tan x + C

Step-by-step solution

1Split: $\frac{1}{c o s ^{2} x} - \frac{1}{s i n ^{2} x} = sec^{2} x - csc^{2} x$

Final answer

tan x + cot x + C

Step-by-step solution

1Rewrite as $1 - (csc^{2} x - 1) = 2 - csc^{2} x$

Final answer

2 x + cot x + C

Step-by-step solution

1Use $tan^{2} x = sec^{2} x - 1$

Final answer

tan x - x + C

Step-by-step solution

1This is $\frac{1/ c o s ^{2} x}{1/ s i n ^{2} x} = tan^{2} x$

Final answer

tan x - x + C

Step-by-step solution

1Rewrite as $\frac{c o s x}{s i n x}$

Final answer

ln ∣ sin x ∣ + C

Final answer

ln ∣ csc x - cot x ∣ + C

Step-by-step solution

1Standard form: multiply by $\frac{s e c x + t a n x}{s e c x + t a n x}$

Final answer

ln ∣ sec x + tan x ∣ + C

Exercise 7.2 — Integration by Substitution

20Exercise questions

Step-by-step solution

1Let $u = x + 1$ , so $x = u - 1$ , $d x = d u$
2 $\int \frac{u - 1}{u} d u = u - ln ∣ u ∣ + C = x + 1 - ln ∣ x + 1∣ + C$

Final answer

x - ln ∣ x + 1∣ + C

Step-by-step solution

1Multiply by $\frac{x + x - 1}{x + x - 1}$ :
2 $\int (x + x - 1) d x = \frac{2}{3} x^{3/2} + \frac{2}{3} (x - 1)^{3/2} + C$

Final answer

\frac{2}{3} (x^{3/2} + (x - 1)^{3/2}) + C

Step-by-step solution

1Partial fractions: $\frac{1}{x ( x ^{2} + 1 )} = \frac{1}{x} - \frac{x}{x ^{2} + 1}$

Final answer

ln ∣ x ∣ - \frac{1}{2} ln (x^{2} + 1) + C

Step-by-step solution

1Write $sin^{3} x = sin x (1 - cos^{2} x)$ and let $u = cos x$

Final answer

- cos x + \frac{cos ^{3} x}{3} + C

Step-by-step solution

1Let $u = cos x$ , $d u = - sin x d x$

Final answer

- cos (cos x) + C

Step-by-step solution

1Let $u = x - 1$ , so $x = u^{2} + 1$ , $d x = 2 u d u$
2 $\int \frac{( u ^{2} + 1 ) ^{2}}{u} \cdot 2 u d u = 2 \int (u^{2} + 1)^{2} d u$

Final answer

\frac{2}{5} (x - 1)^{5/2} + \frac{4}{3} (x - 1)^{3/2} + 2 x - 1 + C

Step-by-step solution

1Let $u = sin x$ , $d u = cos x d x$
2Partial fractions: $\frac{1}{( 1 + u ) ( 2 + u )} = \frac{1}{1 + u} - \frac{1}{2 + u}$

Final answer

ln \frac{1 + sin x}{2 + sin x} + C

Step-by-step solution

1Let $u = x^{2} + 1$ , $d u = 2 x d x$

Final answer

x^{2} + 1 + C

Step-by-step solution

1Let $u = x^{2} + 1$ , $x^{2} = u - 1$ , $x d x = \frac{d u}{2}$

Final answer

\frac{1}{3} (x^{2} + 1)^{3/2} - (x^{2} + 1)^{1/2} + C

Step-by-step solution

1Use $cos 2 x = 2 cos^{2} x - 1$

Final answer

2 sin x - ln ∣ sec x + tan x ∣ + C

Step-by-step solution

1For $x \in (0, π)$ : $sin^{- 1} (cos x) = \frac{π}{2} - x$

Final answer

\frac{π}{2} x - \frac{x ^{2}}{2} + C

Step-by-step solution

1Let $u = 1 - x^{2}$ , $x^{2} = 1 - u$ , $x d x = - \frac{d u}{2}$

Final answer

- \frac{1}{3} (1 - x^{2})^{3/2} + (1 - x^{2})^{1/2} + C

Step-by-step solution

1Let $u = cos^{- 1} x$ , $d u = - \frac{d x}{1 - x ^{2}}$

Final answer

- \frac{( cos ^{- 1} x ) ^{2}}{2} + C

Step-by-step solution

1For $x > 0$ : $tan^{- 1} (1/ x) = \frac{π}{2} - tan^{- 1} x$

Final answer

\frac{π}{2} x - x tan^{- 1} x + \frac{1}{2} ln (1 + x^{2}) + C

Step-by-step solution

1Let $u = 1 + x^{2}$ , $u^{2} = 1 + x^{2}$ , $u d u = x d x$
2Then $\int \frac{u}{x} \cdot \frac{u d u}{x} = \int \frac{u ^{2}}{u ^{2} - 1} d u$

Final answer

1 + x^{2} - ln \frac{1 + x ^{2} + 1}{x} + C

Step-by-step solution

1Simplify: $\frac{x ^{2} - 1}{x ( x ^{2} + 1 )} = \frac{x}{x ^{2} + 1} - \frac{1}{x ( x ^{2} + 1 )}$

Final answer

ln ∣ x ∣ - \frac{1}{2} ln (x^{2} + 1) + C

Step-by-step solution

1Let $u = 1 + sin x$ , $d u = cos x d x$
2Rewrite integrand as $\frac{s i n x}{c o s x 1 + s i n x}$ and use substitution.

Final answer

2 (\frac{1}{1 + sin x} - ln ∣ 1 + sin x + 1∣) + C

Step-by-step solution

1Let $u = 1 + cos x$ , $d u = - sin x d x$

Final answer

\frac{1}{1 + cos x} + C

Step-by-step solution

1Write $sin x = A (sin x + cos x) + B (cos x - sin x)$ to find constants.

Final answer

\frac{x}{2} - \frac{1}{4} ln ∣1 + sin 2 x ∣ + C

Step-by-step solution

1Let $u = sin^{- 1} (x^{2})$ , $d u = \frac{2 x}{1 - x ^{4}} d x$

Final answer

\frac{( sin ^{- 1} x ^{2} ) ^{2}}{4} + C

Exercise 7.3 — Integration Using Trigonometric Identities

11Exercise questions

Step-by-step solution

1Use $sin^{2} x = \frac{1 - c o s 2 x}{2}$

Final answer

\frac{x}{2} - \frac{sin 2 x}{4} + C

Step-by-step solution

1Use $cos^{2} x = \frac{1 + c o s 2 x}{2}$

Final answer

\frac{x}{2} + \frac{sin 2 x}{4} + C

Step-by-step solution

1Write $cos^{3} x = cos x (1 - sin^{2} x)$

Final answer

sin x - \frac{sin ^{3} x}{3} + C

Step-by-step solution

1Use $sin^{2} x = \frac{1 - c o s 2 x}{2}$ then square it.

Final answer

\frac{3 x}{8} - \frac{sin 2 x}{4} + \frac{sin 4 x}{32} + C

Step-by-step solution

1Use $sin^{2} x = 1 - cos^{2} x = (1 - cos x) (1 + cos x)$

Final answer

cos x + C

Step-by-step solution

1Use $tan^{2} x = sec^{2} x - 1$ and $sec^{2} x = 1 + tan^{2} x$

Final answer

\frac{tan ^{3} x}{3} - tan x + x + C

Step-by-step solution

1Split into $\frac{s i n x}{c o s ^{2} x} + \frac{c o s x}{s i n ^{2} x} = sec x tan x + csc x cot x$

Final answer

sec x - csc x + C

Step-by-step solution

1Rewrite as $\frac{c o s x}{s i n ^{2} x} = csc x cot x$

Final answer

- csc x + C

Step-by-step solution

1Use $sin^{2} x cos^{2} x = \frac{s i n ^{2} 2 x}{4} = \frac{1 - c o s 4 x}{8}$

Final answer

\frac{x}{8} - \frac{sin 4 x}{32} + C

Step-by-step solution

1Factor $cos 2 x = (cos x + sin x) (cos x - sin x)$

Final answer

\frac{sin x - cos x}{cos x + sin x} + C

Step-by-step solution

1Let $u = tan x$ , $d u = sec^{2} x d x$

Final answer

\frac{tan ^{5} x}{5} + \frac{tan ^{3} x}{3} + C

Exercise 7.4 — Integration of Particular Functions

11Exercise questions

Step-by-step solution

1Factor: $x^{2} + 3 x - 10 = (x + 5) (x - 2)$
2Partial fractions: $\frac{3 x + 5}{( x + 5 ) ( x - 2 )} = \frac{A}{x + 5} + \frac{B}{x - 2}$ , find $A = - 1, B = 2$

Final answer

- ln ∣ x + 5∣ + 2 ln ∣ x - 2∣ + C

Step-by-step solution

1Long division: $1 + \frac{- 4 x - 4}{( x + 1 ) ( x + 3 )}$
2Partial fractions on the remainder.

Final answer

x + \frac{2}{x + 1} - \frac{6}{x + 3} + C wait, partial fractions give: x - 2 ln ∣ x + 1∣ + ln ∣ x + 3∣ + C

Step-by-step solution

1Factor and use partial fractions: $\frac{x}{( x ^{2} + 1 ) ( x - 1 )} = \frac{A}{x - 1} + \frac{B x + C}{x ^{2} + 1}$ , $A = \frac{1}{2}, B = - \frac{1}{2}, C = \frac{1}{2}$

Final answer

\frac{1}{2} ln ∣ x - 1∣ - \frac{1}{4} ln (x^{2} + 1) + \frac{1}{2} tan^{- 1} x + C

Step-by-step solution

1Partial fractions: $\frac{1}{( x - 1 ) ( x ^{2} + 1 )} = \frac{1/2}{x - 1} + \frac{- x /2 + 1/2}{x ^{2} + 1}$

Final answer

\frac{1}{2} ln ∣ x - 1∣ - \frac{1}{4} ln (x^{2} + 1) + \frac{1}{2} tan^{- 1} x + C

Step-by-step solution

1Partial fractions: $\frac{x}{( x - 1 ) ( x - 2 ) ( x - 3 )} = \frac{1/2}{x - 1} - \frac{2}{x - 2} + \frac{3/2}{x - 3}$

Final answer

\frac{1}{2} ln ∣ x - 1∣ - 2 ln ∣ x - 2∣ + \frac{3}{2} ln ∣ x - 3∣ + C

Step-by-step solution

1Partial fractions: $\frac{2 x + 3}{( x + 1 ) ( x + 2 )} = \frac{1}{x + 1} + \frac{1}{x + 2}$

Final answer

ln ∣ x + 1∣ + ln ∣ x + 2∣ + C

Step-by-step solution

1Write $\frac{1 - x ^{2}}{x ( 1 + x ^{2} )} = \frac{1}{x} - \frac{2 x}{1 + x ^{2}}$

Final answer

ln ∣ x ∣ - ln (1 + x^{2}) + C

Step-by-step solution

1Long division: $x + \frac{- x - 1}{( x + 1 ) ( x + 2 )} = x + \frac{- 1}{x + 2}$

Final answer

\frac{x ^{2}}{2} - ln ∣ x + 2∣ + C

Step-by-step solution

1Long division: $5 + \frac{- 2 x - 4}{x ^{2} + x + 1}$
2Complete the square for the remainder integral.

Final answer

5 x - ln (x^{2} + x + 1) + \frac{4}{3} tan^{- 1} \frac{2 x + 1}{3} + C

Step-by-step solution

1Partial fractions: $\frac{1}{x ( x ^{2} + 1 )} = \frac{1}{x} - \frac{x}{x ^{2} + 1}$

Final answer

ln ∣ x ∣ - \frac{1}{2} ln (x^{2} + 1) + C

Step-by-step solution

1Partial fractions: $\frac{1}{( x ^{2} + 1 ) ( x ^{2} + 4 )} = \frac{1}{3} (\frac{1}{x ^{2} + 1} - \frac{1}{x ^{2} + 4})$

Final answer

\frac{1}{3} tan^{- 1} x - \frac{1}{6} tan^{- 1} \frac{x}{2} + C

Exercise 7.5 — Integration by Partial Fractions

13Exercise questions

Step-by-step solution

1Partial fractions: $\frac{x}{( x + 1 ) ( x + 2 )} = \frac{- 1}{x + 1} + \frac{2}{x + 2}$

Final answer

- ln ∣ x + 1∣ + 2 ln ∣ x + 2∣ + C

Step-by-step solution

1Factor: $x^{2} - 9 = (x - 3) (x + 3)$
2 $\frac{1}{( x - 3 ) ( x + 3 )} = \frac{1}{6} (\frac{1}{x - 3} - \frac{1}{x + 3})$

Final answer

\frac{1}{6} ln \frac{x - 3}{x + 3} + C

Step-by-step solution

1Partial fractions: $\frac{3 x - 1}{( x - 1 ) ( x - 2 ) ( x - 3 )} = \frac{1}{x - 1} - \frac{5}{x - 2} + \frac{4}{x - 3}$

Final answer

ln ∣ x - 1∣ - 5 ln ∣ x - 2∣ + 4 ln ∣ x - 3∣ + C

Step-by-step solution

1Partial fractions: $\frac{x ^{2} + 1}{( x - 1 ) ^{2} ( x + 3 )} = \frac{1}{x + 3} + \frac{3}{( x - 1 ) ^{2}}$

Final answer

ln ∣ x + 3∣ - \frac{3}{x - 1} + C

Step-by-step solution

1Partial fractions: $\frac{1 - x ^{2}}{x ( 1 + x ^{2} )} = \frac{1}{x} - \frac{2 x}{1 + x ^{2}}$

Final answer

ln ∣ x ∣ - ln (1 + x^{2}) + C

Step-by-step solution

1Partial fractions: $\frac{x}{( x - 1 ) ( x - 2 ) ( x - 3 )} = \frac{1/2}{x - 1} - \frac{2}{x - 2} + \frac{3/2}{x - 3}$

Final answer

\frac{1}{2} ln ∣ x - 1∣ - 2 ln ∣ x - 2∣ + \frac{3}{2} ln ∣ x - 3∣ + C

Step-by-step solution

1Partial fractions: $\frac{x}{( x + 1 ) ( x ^{2} + 1 )} = \frac{- 1/2}{x + 1} + \frac{x /2 + 1/2}{x ^{2} + 1}$

Final answer

- \frac{1}{2} ln ∣ x + 1∣ + \frac{1}{4} ln (x^{2} + 1) + \frac{1}{2} tan^{- 1} x + C

Step-by-step solution

1Partial fractions: $\frac{x ^{2} - 3 x + 2}{x ( x - 1 ) ^{2}} = \frac{2}{x} - \frac{1}{x - 1} - \frac{1}{( x - 1 ) ^{2}}$

Final answer

2 ln ∣ x ∣ - ln ∣ x - 1∣ + \frac{1}{x - 1} + C

Step-by-step solution

1Long division: $1 + \frac{5 x - 8}{x ^{2} + 4}$
2Split: $\frac{5 x}{x ^{2} + 4} - \frac{8}{x ^{2} + 4}$

Final answer

x + \frac{5}{2} ln (x^{2} + 4) - 4 tan^{- 1} \frac{x}{2} + C

Step-by-step solution

1Factor: $x^{4} - 1 = (x^{2} - 1) (x^{2} + 1) = (x - 1) (x + 1) (x^{2} + 1)$
2Partial fractions.

Final answer

\frac{1}{4} ln \frac{x - 1}{x + 1} - \frac{1}{2} tan^{- 1} x + C

Step-by-step solution

1Partial fractions: $\frac{2 x - 1}{( x - 1 ) ( x - 2 ) ( x - 3 )} = \frac{1/2}{x - 1} - \frac{3}{x - 2} + \frac{5/2}{x - 3}$

Final answer

\frac{1}{2} ln ∣ x - 1∣ - 3 ln ∣ x - 2∣ + \frac{5}{2} ln ∣ x - 3∣ + C

Step-by-step solution

1Partial fractions in $u = x^{2}$ : $\frac{u}{( u + 1 ) ( u + 4 )} = \frac{- 1/3}{u + 1} + \frac{4/3}{u + 4}$

Final answer

- \frac{1}{3} tan^{- 1} x + \frac{4}{3} \cdot \frac{1}{2} tan^{- 1} \frac{x}{2} + C = - \frac{1}{3} tan^{- 1} x + \frac{2}{3} tan^{- 1} \frac{x}{2} + C

Step-by-step solution

1Long division: $1 + \frac{2}{x ^{2} - 1}$
2Partial fractions: $\frac{2}{( x - 1 ) ( x + 1 )} = \frac{1}{x - 1} - \frac{1}{x + 1}$

Final answer

x + ln \frac{x - 1}{x + 1} + C

Exercise 7.6 — Integration by Parts

12Exercise questions

Step-by-step solution

1By parts: $u = x, d v = sin x d x$ → $d u = d x, v = - cos x$

Final answer

- x cos x + sin x + C

Step-by-step solution

1By parts: $u = x, d v = sin 3 x d x$

Final answer

- \frac{x cos 3 x}{3} + \frac{sin 3 x}{9} + C

Step-by-step solution

1By parts twice.

Final answer

x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + C

Step-by-step solution

1By parts: $u = x, d v = e^{x} d x$

Final answer

x e^{x} - e^{x} + C

Step-by-step solution

1By parts: $u = lo g x, d v = x d x$

Final answer

\frac{x ^{2}}{2} lo g x - \frac{x ^{2}}{4} + C

Step-by-step solution

1By parts: $u = lo g x, d v = x^{2} d x$

Final answer

\frac{x ^{3}}{3} lo g x - \frac{x ^{3}}{9} + C

Step-by-step solution

1By parts: $u = sin^{- 1} x, d v = x d x$

Final answer

\frac{x ^{2}}{2} sin^{- 1} x - \frac{1}{4} (x 1 - x^{2} + sin^{- 1} x) + C

Step-by-step solution

1By parts: $u = cos^{- 1} x, d v = x d x$

Final answer

\frac{x ^{2}}{2} cos^{- 1} x + \frac{1}{4} (x 1 - x^{2} - cos^{- 1} x) + C

Step-by-step solution

1Recognise $\frac{d}{d x} (\frac{e ^{x}}{x}) = \frac{e ^{x}}{x} - \frac{e ^{x}}{x ^{2}}$

Final answer

\frac{e ^{x}}{x} + C

Step-by-step solution

1Recognise $\frac{d}{d x} (\frac{e ^{x}}{x}) = \frac{x e ^{x} - e ^{x}}{x ^{2}}$

Final answer

- \frac{e ^{x}}{x} + C

Step-by-step solution

1Rewrite as $\int e^{x} (\frac{x - 1}{x + 1} + \frac{2}{( x + 1 ) ^{2}}) d x$ or recognise the standard form.

Final answer

\frac{e ^{x} ( x - 1 )}{x + 1} + C

Step-by-step solution

1Simplify the rational function, then use the formula $\int e^{x} (f (x) + f^{'} (x)) d x = e^{x} f (x) + C$

Final answer

e^{x} (x - 2) + C (after simplification)

Exercise 7.7 — Definite Integrals

8Exercise questions

Step-by-step solution

1Use $cos^{2} x = \frac{1 + c o s 2 x}{2}$ :
2 $[\frac{x}{2} + \frac{s i n 2 x}{4}]_{0}^{π /2} = \frac{π}{4}$

Final answer

\frac{π}{4}

Step-by-step solution

1By symmetry with $cos^{2} x$ , the answer is the same.

Final answer

\frac{π}{4}

Step-by-step solution

1Write $sin^{3} x = sin x (1 - cos^{2} x)$ , let $u = cos x$

Final answer

\frac{2}{3}

Step-by-step solution

1By symmetry with $sin^{3} x$

Final answer

\frac{2}{3}

Step-by-step solution

1Use power-reduction formulas twice.

Final answer

\frac{3 π}{8}

Step-by-step solution

1Use power-reduction: $sin^{4} x cos^{2} x = \frac{1 - c o s 2 x}{2} \cdot \frac{1 - c o s 2 x}{2} \cdot \frac{1 + c o s 2 x}{2}$

Final answer

\frac{π}{32}

Step-by-step solution

1Use $sin^{4} x = (\frac{1 - c o s 2 x}{2})^{2}$

Final answer

\frac{3 π}{4}

Step-by-step solution

1Reduction: $sin^{5} x = sin x (1 - cos^{2} x)^{2}$ , let $u = cos x$

Final answer

\frac{8}{15}

Exercise 7.8 — Evaluation of Definite Integrals by Substitution

8Exercise questions

Step-by-step solution

1Let $u = x + 1$ :
2 $\int_{1}^{2} \frac{u - 1}{u} d u = [u - ln u]_{1}^{2} = 1 - ln 2$

Final answer

1 - ln 2

Step-by-step solution

1Let $u = x^{2} + 1$ :
2 $\frac{1}{2} \int_{1}^{3} u^{- 1/2} d u = [u]_{1}^{3} = 3 - 1$

Final answer

3 - 1

Step-by-step solution

1Long division: $1 - \frac{4}{x ^{2} + 4}$

Final answer

2 - 2 tan^{- 1} 1 = 2 - \frac{π}{2}

Step-by-step solution

1Let $u = tan^{- 1} x$ , $d u = \frac{d x}{1 + x ^{2}}$

Final answer

\frac{π ^{2}}{32}

Step-by-step solution

1Standard integral: $\int \frac{d x}{1 - x ^{2}} = sin^{- 1} x$

Final answer

\frac{π}{2}

Step-by-step solution

1Antiderivative: $tan^{- 1} x$

Final answer

\frac{π}{4}

Step-by-step solution

1Let $u = x^{3}$ , $d u = 3 x^{2} d x$

Final answer

\frac{1}{3} tan^{- 1} 1 = \frac{π}{12}

Step-by-step solution

1Simplify: $\frac{x}{2 x} = \frac{1}{2}$

Final answer

\frac{1}{2}

Exercise 7.9 — Some Properties of Definite Integrals

7Exercise questions

Step-by-step solution

1Let $t = a + b - x$ : $d x = - d t$
2When $x = a, t = b$ ; when $x = b, t = a$
3 $\int_{b}^{a} f (t) (- d t) = \int_{a}^{b} f (t) d t$

Final answer

Proved.

Step-by-step solution

1Apply the property with $a = 0, b = π /2$ :
2 $I = \int_{0}^{π /2} \frac{c o s ^{n} x}{c o s ^{n} x + s i n ^{n} x} d x$ ; add: $2 I = \int_{0}^{π /2} 1 d x = \frac{π}{2}$

Final answer

\frac{π}{4}

Step-by-step solution

1Use the property with $a = 0, b = π$

Final answer

\frac{π ^{2}}{4}

Step-by-step solution

1Let $u = 1 - x$ and use the property.

Final answer

- 1

Step-by-step solution

1The integrand has period $π$ ; over a full period of an odd power of cosine, the integral is zero.

Final answer

0

Step-by-step solution

1Let $u = cos x$ :
2 $\int_{0}^{1} u^{3} (1 - u^{2}) d u = \int_{0}^{1} (u^{3} - u^{5}) d u$

Final answer

\frac{1}{4} - \frac{1}{6} = \frac{1}{12}

Step-by-step solution

1By the same property as the $sin^{n} x$ problem, the answer is always $π /4$

Final answer

\frac{π}{4}

Exercise 7.10 — More Definite Integral Problems

7Exercise questions

Step-by-step solution

1Use $sin^{4} x = (\frac{1 - c o s 2 x}{2})^{2}$

Final answer

\frac{3 π}{16}

Step-by-step solution

1By symmetry: same as $\int_{0}^{π /2} sin^{4} x d x$

Final answer

\frac{3 π}{16}

Step-by-step solution

1Use reduction formula.

Final answer

\frac{5 π}{32}

Step-by-step solution

1Use $sin^{3} x = sin x (1 - cos^{2} x)$

Final answer

\frac{4}{3}

Step-by-step solution

1Odd power of sine over full period: the integral is zero by symmetry.

Final answer

0

Step-by-step solution

1Odd function integrated over symmetric interval: zero.

Final answer

0

Step-by-step solution

1Use the property: add with $\frac{s i n ^{3} x}{s i n ^{3} x + c o s ^{3} x}$

Final answer

\frac{π}{4}

Exercise 7.11 — Properties of Definite Integrals

8Exercise questions

Step-by-step solution

1Use property: $\int_{0}^{π /2} lo g sin x d x = \int_{0}^{π /2} lo g cos x d x$
2Add: $2 I = \int_{0}^{π /2} lo g (\frac{s i n 2 x}{2}) d x$
3Substitute $u = 2 x$ and simplify.

Final answer

- \frac{π}{2} ln 2

Step-by-step solution

1By the property just proved, same as the above.

Final answer

- \frac{π}{2} ln 2

Step-by-step solution

1Use the substitution property: add the integral with $cos x$ in numerator.

Final answer

\frac{π}{4}

Step-by-step solution

1Recognise: $\frac{d}{d x} (\frac{e ^{2 x}}{2 x}) = \frac{e ^{2 x}}{x} - \frac{e ^{2 x}}{2 x ^{2}}$

Final answer

\frac{e ^{4}}{4} - \frac{e ^{2}}{2}

Step-by-step solution

1Use the property $\int_{0}^{π} f (cos x) d x = \int_{0}^{π} f (- cos x) d x$ and add.

Final answer

- π ln 2

Step-by-step solution

1Periodicity and symmetry: double the integral over $[0, π]$

Final answer

\frac{π}{8}

Step-by-step solution

1Use property with $f (x) = \frac{t a n x}{s e c x + t a n x}$

Final answer

\frac{π}{2} (\frac{π}{2} - ln 2)

Step-by-step solution

1Let $I = \int_{0}^{π /2} \frac{c o s ^{α} x}{c o s ^{α} x + s i n ^{α} x} d x$
2Apply property $I = \int_{0}^{π /2} \frac{s i n ^{α} x}{s i n ^{α} x + c o s ^{α} x} d x$
3Add: $2 I = \frac{π}{2}$

Final answer

I = \frac{π}{4}

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Indefinite integral

\int f (x) d x = F (x) + C

Definite integral

\int_{a}^{b} f (x) d x = F (b) - F (a)

Substitution

u = g (x) ⟹ d u = g^{'} (x) d x

By parts

\int u d v = uv - \int v d u

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Always attach + C to an indefinite integral — a definite integral evaluated by the fundamental theorem needs no constant.
In by parts choose u as the function that simplifies under differentiation and dv as the part you can integrate at sight.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Maths Chapter 7 (Integrals)?

There are 11 exercise questions in this chapter, numbered Ex 7.1 to Ex 7.11. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Integrals Class 12 Maths?

The formulas this chapter's questions actually turn on are: Indefinite integral, Definite integral, Substitution, By parts. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Integrals important for JEE Main?

Very important — substitution, by parts and partial fractions are the entire integration toolkit for boards and JEE Main, and the largest exercise set in the book.

Next Chapters

Continue learning

Previous: Application of Derivatives Next: Application of Integrals Integrals revision notes All Class 12 Maths NCERT Solutions

Same solutions, live doubt-clearing help

Reading a solution is step one — getting a doubt resolved in real time is what clears it. ClassApna runs small-batch CBSE, JEE & NEET coaching with daily doubt sessions and mock tests.

Book a Free Demo Class Explore Plans & Batches

Small batches · 1-on-1 personal mentorship · Live online & offline centre