CBSE Maths › Class 12 PYQs › Integrals

Integrals: 2 marks Questions (CBSE Class 12)

14 different 2 marks questions on Integrals from CBSE Class 12 Maths board exams 2024–2026, newest first.

1 mark (58)2 marks (14)3 marks (53)5 marks (26)

Q222 marksVery Short AnswerIntegrals CBSE 2025 · 65/2/1

Evaluate : $\int_{0}^{\frac{π}{4}} 1 + sin 2 x d x$

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Answer: 1

$1 + sin 2 x = sin^{2} x + cos^{2} x + 2 sin x cos x = (sin x + cos x)^{2}$ .
On $[0, \frac{π}{4}]$ , $sin x + cos x > 0$ , so $1 + sin 2 x = sin x + cos x$ .
$\int_{0}^{\frac{π}{4}} (sin x + cos x) d x = [- cos x + sin x]_{0}^{\frac{π}{4}}$ .
$= (- \frac{1}{2} + \frac{1}{2}) - (- 1 + 0) = 1$ .

Q222 marksVery Short AnswerIntegrals CBSE 2025 · 65/2/2

Evaluate : $\int_{0}^{π} \frac{s i n 2 p x}{s i n x} d x$ , p $\in$ N.

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Answer: 0

Let $I = \int_{0}^{π} \frac{s i n 2 p x}{s i n x} d x$ .
Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ : $I = \int_{0}^{π} \frac{s i n ( 2 p π - 2 p x )}{s i n ( π - x )} d x$ .
Since $p \in N$ , $sin (2 p π - 2 p x) = - sin 2 p x$ , and $sin (π - x) = sin x$ .
So $I = - \int_{0}^{π} \frac{s i n 2 p x}{s i n x} d x = - I$ .
$2 I = 0$ , hence $I = 0$ .

Q222 marksVery Short AnswerIntegrals CBSE 2025 · 65/2/3

Find : $\int 2 x^{3} e^{x^{2}} d x$ .

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Answer:

e^{x^{2}} (x^{2} - 1) + C

Put $x^{2} = t$ , so $2 x d x = d t$ .
$\int 2 x^{3} e^{x^{2}} d x = \int x^{2} e^{x^{2}} \cdot 2 x d x = \int t e^{t} d t$ .
By parts: $\int t e^{t} d t = t e^{t} - \int e^{t} d t = t e^{t} - e^{t} + C$ .
$= e^{x^{2}} (x^{2} - 1) + C$ .

Q232 marksVery Short AnswerIntegrals CBSE 2024 · 65/1/1

Find :
$\int x 1 + 2 x d x$

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Answer:

\frac{1}{10} (1 + 2 x)^{\frac{5}{2}} - \frac{1}{6} (1 + 2 x)^{\frac{3}{2}} + C

Put $1 + 2 x = t$ , so $x = \frac{t - 1}{2}$ and $d x = \frac{d t}{2}$ .
$I = \int \frac{t - 1}{2} t \frac{d t}{2} = \frac{1}{4} \int (t^{\frac{3}{2}} - t^{\frac{1}{2}}) d t$ .
$= \frac{1}{4} (\frac{2}{5} t^{\frac{5}{2}} - \frac{2}{3} t^{\frac{3}{2}}) + C$ .
$= \frac{1}{10} (1 + 2 x)^{\frac{5}{2}} - \frac{1}{6} (1 + 2 x)^{\frac{3}{2}} + C$ .

Also asked in: 2024 65/1/2, 2024 65/1/3

Q23 (OR) (OR)2 marksVery Short AnswerIntegrals CBSE 2024 · 65/1/1

Evaluate :
$\int_{0}^{\frac{π ^{2}}{4}} \frac{s i n x}{x} d x$

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Answer: 2

Put $x = t$ , so $\frac{d x}{2 x} = d t$ ; when $x = 0$ , $t = 0$ and when $x = \frac{π ^{2}}{4}$ , $t = \frac{π}{2}$ .
$I = 2 \int_{0}^{\frac{π}{2}} sin t d t = 2 [- cos t]_{0}^{\frac{π}{2}}$ .
$= 2 (0 + 1) = 2$ .

Also asked in: 2024 65/1/2, 2024 65/1/3

Q242 marksVery Short AnswerIntegrals CBSE 2024 · 65/2/1

Find :
$\int \frac{e ^{4 x} - 1}{e ^{4 x} + 1} d x$

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Answer:

\frac{1}{2} lo g (e^{2 x} + e^{- 2 x}) + C

Divide numerator and denominator by $e^{2 x}$ : integrand $= \frac{e ^{2 x} - e ^{- 2 x}}{e ^{2 x} + e ^{- 2 x}}$ .
Put $t = e^{2 x} + e^{- 2 x}$ , $d t = 2 (e^{2 x} - e^{- 2 x}) d x$ .
Integral $= \frac{1}{2} \int \frac{d t}{t} = \frac{1}{2} lo g ∣ t ∣ + C = \frac{1}{2} lo g (e^{2 x} + e^{- 2 x}) + C$ .

Also asked in: 2024 65/2/3

Q222 marksVery Short AnswerIntegrals CBSE 2024 · 65/2/2

Evaluate :
$\int_{0}^{a^{3}} \frac{x ^{2}}{x ^{6} + a ^{6}} d x$

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Answer:

\frac{1}{3 a ^{3}} tan^{- 1} (a^{6})

(for a > 0)

Put $t = x^{3}$ , $d t = 3 x^{2} d x$ ; x = 0 gives t = 0 and $x = a^{3}$ gives $t = a^{9}$ .
$I = \frac{1}{3} \int_{0}^{a^{9}} \frac{d t}{t ^{2} + ( a ^{3} ) ^{2}} = \frac{1}{3} \cdot \frac{1}{a ^{3}} [tan^{- 1} \frac{t}{a ^{3}}]_{0}^{a^{9}}$ .
$I = \frac{1}{3 a ^{3}} tan^{- 1} (a^{6})$ .

Q232 marksVery Short AnswerIntegrals CBSE 2024 · 65/3/1

Evaluate :
$\int_{0}^{π /2} sin 2 x cos 3 x d x$

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Answer:

- \frac{2}{5}

$sin 2 x cos 3 x = \frac{1}{2} [sin 5 x + sin (- x)] = \frac{1}{2} [sin 5 x - sin x]$ .
$\int_{0}^{π /2} = \frac{1}{2} [- \frac{c o s 5 x}{5} + cos x]_{0}^{π /2}$ .
$= \frac{1}{2} [(0 + 0) - (- \frac{1}{5} + 1)] = \frac{1}{2} (- \frac{4}{5}) = - \frac{2}{5}$ .

Also asked in: 2024 65/3/2, 2024 65/3/3

Q23 (OR) (OR)2 marksVery Short AnswerIntegrals CBSE 2024 · 65/3/1

Given $\frac{d}{d x} F (x) = \frac{1}{2 x - x ^{2}}$ and F(1) = 0, find F(x).

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Answer:

F (x) = sin^{- 1} (x - 1)

$2 x - x^{2} = 1 - (x - 1)^{2}$ .
$F (x) = \int \frac{d x}{1 - ( x - 1 ) ^{2}} = sin^{- 1} (x - 1) + C$ .
$F (1) = sin^{- 1} 0 + C = 0$ , so $C = 0$ .
$F (x) = sin^{- 1} (x - 1)$ .

Also asked in: 2024 65/3/2, 2024 65/3/3

Q252 marksVery Short AnswerIntegrals CBSE 2024 · 65/4/1

Find : $\int \frac{1}{x ( x ^{2} - 1 )} d x$ .

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Answer:

\frac{1}{2} lo g \frac{x ^{2} - 1}{x ^{2}} + C

$\frac{1}{x ( x - 1 ) ( x + 1 )} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$ .
$x = 0$ : $A = - 1$ ; $x = 1$ : $B = \frac{1}{2}$ ; $x = - 1$ : $C = \frac{1}{2}$ .
Integral $= - lo g ∣ x ∣ + \frac{1}{2} lo g ∣ x - 1∣ + \frac{1}{2} lo g ∣ x + 1∣ + C$
$= \frac{1}{2} lo g ∣ x^{2} - 1∣ - lo g ∣ x ∣ + C = \frac{1}{2} lo g \frac{x ^{2} - 1}{x ^{2}} + C$ .

Q252 marksVery Short AnswerIntegrals CBSE 2024 · 65/4/2

Evaluate : $\int_{\frac{- 1}{2}}^{\frac{1}{2}} cos x \cdot lo g (\frac{1 + x}{1 - x}) d x$

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Answer: 0

Let $f (x) = cos x \cdot lo g (\frac{1 + x}{1 - x})$ .
$f (- x) = cos x \cdot lo g (\frac{1 - x}{1 + x}) = - cos x \cdot lo g (\frac{1 + x}{1 - x}) = - f (x)$ , so f is odd.
The integral of an odd function over $[- \frac{1}{2}, \frac{1}{2}]$ is 0.

Q252 marksVery Short AnswerIntegrals CBSE 2024 · 65/4/3

Find : $\int \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} - 4 )} d x$ .

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Answer:

\frac{1}{5} lo g \frac{x ^{2} - 4}{x ^{2} + 1} + C

Put $x^{2} = t$ , $2 x d x = d t$ : integral $= \int \frac{d t}{( t + 1 ) ( t - 4 )}$ .
$\frac{1}{( t + 1 ) ( t - 4 )} = \frac{1}{5} (\frac{1}{t - 4} - \frac{1}{t + 1})$ .
Integral $= \frac{1}{5} lo g ∣ t - 4∣ - \frac{1}{5} lo g ∣ t + 1∣ + C = \frac{1}{5} lo g \frac{x ^{2} - 4}{x ^{2} + 1} + C$ .

Q242 marksVery Short AnswerIntegrals CBSE 2024 · 65/5/1

Find : $\int cos^{3} x e^{l o g s i n x} d x$

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Answer:

- \frac{c o s ^{4} x}{4} + C

$e^{l o g s i n x} = sin x$ , so the integral is $\int cos^{3} x sin x d x$ .
Put $t = cos x$ , $d t = - sin x d x$ : $- \int t^{3} d t = - \frac{t ^{4}}{4} + C$ .
$= - \frac{c o s ^{4} x}{4} + C$ .

Also asked in: 2024 65/5/2, 2024 65/5/3

Q24 (OR) (OR)2 marksVery Short AnswerIntegrals CBSE 2024 · 65/5/1

Find : $\int \frac{1}{5 + 4 x - x ^{2}} d x$

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Answer:

\frac{1}{6} lo g \frac{1 + x}{5 - x} + C

$5 + 4 x - x^{2} = 9 - (x - 2)^{2} = 3^{2} - (x - 2)^{2}$ .
$\int \frac{d x}{3 ^{2} - ( x - 2 ) ^{2}} = \frac{1}{2 \cdot 3} lo g \frac{3 + ( x - 2 )}{3 - ( x - 2 )} + C$ .
$= \frac{1}{6} lo g \frac{1 + x}{5 - x} + C$ .

Also asked in: 2024 65/5/2, 2024 65/5/3

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