Integrals: 3 marks Questions (CBSE Class 12)

21 different 3 marks questions on Integrals from CBSE Class 12 Maths board exams 2026, newest first.

Q263 marksShort AnswerIntegrals CBSE 2026 · 65/1/1

Evaluate : $\int_{0}^{1} x tan^{- 1} x d x$ .

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Answer:

\frac{π}{4} - \frac{1}{2}

By parts: $\int_{0}^{1} x tan^{- 1} x d x = [\frac{x ^{2}}{2} tan^{- 1} x]_{0}^{1} - \int_{0}^{1} \frac{x ^{2}}{2 ( 1 + x ^{2} )} d x$ .
First term $= \frac{1}{2} \cdot \frac{π}{4} = \frac{π}{8}$ .
$\int_{0}^{1} \frac{x ^{2}}{1 + x ^{2}} d x = \int_{0}^{1} (1 - \frac{1}{1 + x ^{2}}) d x = 1 - \frac{π}{4}$ .
Value $= \frac{π}{8} - \frac{1}{2} (1 - \frac{π}{4}) = \frac{π}{4} - \frac{1}{2}$ .

Q273 marksShort AnswerIntegrals CBSE 2026 · 65/1/1

Find $\int \frac{x + 2}{x - 2} d x$

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Answer:

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

$\frac{x + 2}{x - 2} = \frac{x + 2}{x ^{2} - 4}$ .
$\int \frac{x}{x ^{2} - 4} d x = x^{2} - 4$ (put $t = x^{2} - 4$ ).
$\int \frac{2}{x ^{2} - 4} d x = 2 lo g x + x^{2} - 4$ .
So the integral $= x^{2} - 4 + 2 lo g x + x^{2} - 4 + C$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q27 (OR) (OR)3 marksShort AnswerIntegrals CBSE 2026 · 65/1/1

Find : $\int \frac{x ^{2}}{( x ^{2} + 9 ) ( x ^{2} + 16 )} d x$

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Answer:

- \frac{3}{7} tan^{- 1} \frac{x}{3} + \frac{4}{7} tan^{- 1} \frac{x}{4} + C

Put $x^{2} = t$ for the partial fractions: $\frac{t}{( t + 9 ) ( t + 16 )} = \frac{A}{t + 9} + \frac{B}{t + 16}$ .
$t = - 9$ : $A = \frac{- 9}{7}$ ; $t = - 16$ : $B = \frac{- 16}{- 7} = \frac{16}{7}$ .
So $\frac{x ^{2}}{( x ^{2} + 9 ) ( x ^{2} + 16 )} = - \frac{9}{7} \cdot \frac{1}{x ^{2} + 9} + \frac{16}{7} \cdot \frac{1}{x ^{2} + 16}$ .
Integral $= - \frac{9}{7} \cdot \frac{1}{3} tan^{- 1} \frac{x}{3} + \frac{16}{7} \cdot \frac{1}{4} tan^{- 1} \frac{x}{4} + C = - \frac{3}{7} tan^{- 1} \frac{x}{3} + \frac{4}{7} tan^{- 1} \frac{x}{4} + C$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q283 marksShort AnswerIntegrals CBSE 2026 · 65/1/1

If $I_{1} = \int_{- π /4}^{π /4} \frac{d x}{1 + c o s 2 x}$ and $I_{2} = \int_{- 1/2}^{1/2} ∣ x ∣ d x$ , then show that $I_{1} - 4 I_{2} = 0$ .

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Answer: Proved.

$I_{1} = \int_{- π /4}^{π /4} \frac{d x}{2 c o s ^{2} x} = \frac{1}{2} \int_{- π /4}^{π /4} sec^{2} x d x = \frac{1}{2} [tan x]_{- π /4}^{π /4} = \frac{1}{2} (1 + 1) = 1$ .
$∣ x ∣$ is even, so $I_{2} = 2 \int_{0}^{1/2} x d x = 2 \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{4}$ .
$I_{1} - 4 I_{2} = 1 - 4 \cdot \frac{1}{4} = 0$ . Hence proved.

Also asked in: 2026 65/1/2, 2026 65/1/3

Q303 marksShort AnswerIntegrals CBSE 2026 · 65/1/2

Evaluate : $\int_{0}^{1} x sin^{- 1} x d x$

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Answer:

\frac{π}{8}

By parts: $\int_{0}^{1} x sin^{- 1} x d x = [\frac{x ^{2}}{2} sin^{- 1} x]_{0}^{1} - \frac{1}{2} \int_{0}^{1} \frac{x ^{2}}{1 - x ^{2}} d x$ .
First term $= \frac{1}{2} \cdot \frac{π}{2} = \frac{π}{4}$ .
Put $x = sin θ$ : $\int_{0}^{1} \frac{x ^{2}}{1 - x ^{2}} d x = \int_{0}^{π /2} sin^{2} θ d θ = \frac{π}{4}$ .
Value $= \frac{π}{4} - \frac{1}{2} \cdot \frac{π}{4} = \frac{π}{8}$ .

Q263 marksShort AnswerIntegrals CBSE 2026 · 65/1/3

Evaluate : $\int_{0}^{1} lo g (1 + x^{2}) d x$

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Answer:

lo g 2 - 2 + \frac{π}{2}

By parts: $\int_{0}^{1} lo g (1 + x^{2}) d x = [x lo g (1 + x^{2})]_{0}^{1} - \int_{0}^{1} \frac{2 x ^{2}}{1 + x ^{2}} d x$ .
First term $= lo g 2$ .
$\int_{0}^{1} \frac{2 x ^{2}}{1 + x ^{2}} d x = 2 \int_{0}^{1} (1 - \frac{1}{1 + x ^{2}}) d x = 2 (1 - \frac{π}{4}) = 2 - \frac{π}{2}$ .
Value $= lo g 2 - 2 + \frac{π}{2}$ .

Q263 marksShort AnswerIntegrals CBSE 2026 · 65/2/1

Find : $\int \frac{x + 2}{9 x - x ^{2}} d x$

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Answer:

- 9 x - x^{2} + \frac{13}{2} sin^{- 1} (\frac{2 x - 9}{9}) + C

Write $x + 2 = - \frac{1}{2} (9 - 2 x) + \frac{13}{2}$ , where $9 - 2 x = \frac{d}{d x} (9 x - x^{2})$ .
$\int \frac{- \frac{1}{2} ( 9 - 2 x )}{9 x - x ^{2}} d x = - 9 x - x^{2}$ .
$9 x - x^{2} = (\frac{9}{2})^{2} - (x - \frac{9}{2})^{2}$ , so $\int \frac{d x}{9 x - x ^{2}} = sin^{- 1} \frac{x - \frac{9}{2}}{\frac{9}{2}} = sin^{- 1} \frac{2 x - 9}{9}$ .
Answer: $- 9 x - x^{2} + \frac{13}{2} sin^{- 1} (\frac{2 x - 9}{9}) + C$ .

Q273 marksShort AnswerIntegrals CBSE 2026 · 65/2/1

Evaluate : $\int_{\frac{π}{12}}^{\frac{5 π}{12}} \frac{d x}{1 + c o t x}$

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Answer:

\frac{π}{6}

$I = \int_{π /12}^{5 π /12} \frac{s i n x}{s i n x + c o s x} d x$ .
Using $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ with $a + b = \frac{π}{2}$ : $I = \int_{π /12}^{5 π /12} \frac{c o s x}{c o s x + s i n x} d x$ .
Adding: $2 I = \int_{π /12}^{5 π /12} 1 d x = \frac{5 π}{12} - \frac{π}{12} = \frac{π}{3}$ .
$I = \frac{π}{6}$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q27 (OR) (OR)3 marksShort AnswerIntegrals CBSE 2026 · 65/2/1

Evaluate : $\int_{\frac{- π}{6}}^{\frac{π}{2}} (sin ∣ x ∣ + cos ∣ x ∣) d x$

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Answer:

\frac{7 - 3}{2}

For $x < 0$ : $sin ∣ x ∣ + cos ∣ x ∣ = - sin x + cos x$ ; for $x \geq 0$ : $sin x + cos x$ .
$\int_{- π /6}^{0} (cos x - sin x) d x = [sin x + cos x]_{- π /6}^{0} = 1 - (- \frac{1}{2} + \frac{3}{2}) = \frac{3 - 3}{2}$ .
$\int_{0}^{π /2} (sin x + cos x) d x = [- cos x + sin x]_{0}^{π /2} = 1 - (- 1) = 2$ .
Total $= \frac{3 - 3}{2} + 2 = \frac{7 - 3}{2}$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q283 marksShort AnswerIntegrals CBSE 2026 · 65/2/1

If $\frac{d}{d x} (F (x)) = \frac{1}{e ^{x} + 1}$ , then find $F (x)$ given that $F (0) = lo g \frac{1}{2}$ .

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Answer:

F (x) = x - lo g (e^{x} + 1)

, i.e.

lo g \frac{e ^{x}}{e ^{x} + 1}

$F (x) = \int \frac{d x}{e ^{x} + 1} = \int \frac{e ^{- x}}{1 + e ^{- x}} d x$ .
Put $1 + e^{- x} = t$ , $- e^{- x} d x = d t$ : $F (x) = - lo g (1 + e^{- x}) + C$ .
$F (0) = - lo g 2 + C = lo g \frac{1}{2} \Rightarrow C = 0$ .
$F (x) = - lo g (1 + e^{- x}) = lo g \frac{e ^{x}}{e ^{x} + 1} = x - lo g (e^{x} + 1)$ .

Also asked in: 2026 65/2/2, 2026 65/2/3

Q303 marksShort AnswerIntegrals CBSE 2026 · 65/2/2

Find : $\int \frac{2 x + 1}{6 x + x ^{2}} d x$

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Answer:

2 x^{2} + 6 x - 5 lo g x + 3 + x^{2} + 6 x + C

Write $2 x + 1 = (2 x + 6) - 5$ , where $2 x + 6 = \frac{d}{d x} (x^{2} + 6 x)$ .
$\int \frac{2 x + 6}{x ^{2} + 6 x} d x = 2 x^{2} + 6 x$ .
$x^{2} + 6 x = (x + 3)^{2} - 3^{2}$ , so $\int \frac{d x}{x ^{2} + 6 x} = lo g x + 3 + x^{2} + 6 x$ .
Answer: $2 x^{2} + 6 x - 5 lo g x + 3 + x^{2} + 6 x + C$ .

Q273 marksShort AnswerIntegrals CBSE 2026 · 65/2/3

Find : $\int \frac{3 x - 1}{x ^{2} - 4 x} d x$

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Answer:

3 x^{2} - 4 x + 5 lo g x - 2 + x^{2} - 4 x + C

Write $3 x - 1 = \frac{3}{2} (2 x - 4) + 5$ , where $2 x - 4 = \frac{d}{d x} (x^{2} - 4 x)$ .
$\frac{3}{2} \int \frac{2 x - 4}{x ^{2} - 4 x} d x = 3 x^{2} - 4 x$ .
$x^{2} - 4 x = (x - 2)^{2} - 2^{2}$ , so $\int \frac{d x}{x ^{2} - 4 x} = lo g x - 2 + x^{2} - 4 x$ .
Answer: $3 x^{2} - 4 x + 5 lo g x - 2 + x^{2} - 4 x + C$ .

Q263 marksShort AnswerIntegrals CBSE 2026 · 65/3/1

Find :
$\int \frac{d x}{x ^{1/2} + x ^{1/3}}$

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Answer:

2 x - 3 x^{1/3} + 6 x^{1/6} - 6 lo g ∣1 + x^{1/6} ∣ + C

Put $x = t^{6}$ , $d x = 6 t^{5} d t$ ; then $x^{1/2} = t^{3}$ , $x^{1/3} = t^{2}$ .
$I = \int \frac{6 t ^{5}}{t ^{3} + t ^{2}} d t = 6 \int \frac{t ^{3}}{t + 1} d t$ .
$\frac{t ^{3}}{t + 1} = t^{2} - t + 1 - \frac{1}{t + 1}$ .
$I = 6 (\frac{t ^{3}}{3} - \frac{t ^{2}}{2} + t - lo g ∣ t + 1∣) + C = 2 t^{3} - 3 t^{2} + 6 t - 6 lo g ∣ t + 1∣ + C$ .
With $t = x^{1/6}$ : $I = 2 x - 3 x^{1/3} + 6 x^{1/6} - 6 lo g ∣1 + x^{1/6} ∣ + C$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q26 (OR) (OR)3 marksShort AnswerIntegrals CBSE 2026 · 65/3/1

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

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Answer:

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

, i.e.

\frac{π x}{4} - x tan^{- 1} x + \frac{1}{2} lo g (1 + x^{2}) + C

Let $u = tan^{- 1} (\frac{1 - x}{1 + x})$ . By the chain rule, $\frac{d u}{d x} = \frac{1}{1 + \frac{( 1 - x ) ^{2}}{( 1 + x ) ^{2}}} \cdot \frac{- 2}{( 1 + x ) ^{2}} = \frac{- 2}{2 + 2 x ^{2}} = - \frac{1}{1 + x ^{2}}$ .
By parts: $I = x u - \int x (- \frac{1}{1 + x ^{2}}) d x = x tan^{- 1} (\frac{1 - x}{1 + x}) + \frac{1}{2} lo g (1 + x^{2}) + C$ .
(For $x > - 1$ , $tan^{- 1} \frac{1 - x}{1 + x} = \frac{π}{4} - tan^{- 1} x$ , so this equals $\frac{π x}{4} - x tan^{- 1} x + \frac{1}{2} lo g (1 + x^{2}) + C$ .)

Also asked in: 2026 65/3/2, 2026 65/3/3

Q273 marksShort AnswerIntegrals CBSE 2026 · 65/3/1

Evaluate :
$\int_{0}^{π} \frac{s i n ^{2026} x}{s i n ^{2026} x + c o s ^{2026} x} d x$

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Answer:

\frac{π}{2}

Let $f (x) = \frac{s i n ^{2026} x}{s i n ^{2026} x + c o s ^{2026} x}$ . Since 2026 is even, $f (π - x) = f (x)$ .
So $I = 2 \int_{0}^{π /2} f (x) d x = 2 J$ .
In J, replace x by $\frac{π}{2} - x$ : $J = \int_{0}^{π /2} \frac{c o s ^{2026} x}{c o s ^{2026} x + s i n ^{2026} x} d x$ .
Adding the two forms: $2 J = \int_{0}^{π /2} 1 d x = \frac{π}{2}$ , so $J = \frac{π}{4}$ .
$I = 2 J = \frac{π}{2}$ .

Q283 marksShort AnswerIntegrals CBSE 2026 · 65/3/1

Find :
$\int \frac{c o s x}{( 2 + s i n x ) ( 4 + s i n x )} d x$

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Answer:

\frac{1}{2} lo g \frac{2 + s i n x}{4 + s i n x} + C

Put $t = sin x$ , $d t = cos x d x$ : $I = \int \frac{d t}{( 2 + t ) ( 4 + t )}$ .
$\frac{1}{( 2 + t ) ( 4 + t )} = \frac{1}{2} (\frac{1}{2 + t} - \frac{1}{4 + t})$ .
$I = \frac{1}{2} lo g ∣2 + t ∣ - \frac{1}{2} lo g ∣4 + t ∣ + C = \frac{1}{2} lo g \frac{2 + s i n x}{4 + s i n x} + C$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q28 (OR) (OR)3 marksShort AnswerIntegrals CBSE 2026 · 65/3/1

Find :
$\int \frac{x + 3}{x ^{2} + 4 x + 5} d x$

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Answer:

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

Write $x + 3 = \frac{1}{2} (2 x + 4) + 1$ .
$I = \frac{1}{2} \int \frac{2 x + 4}{x ^{2} + 4 x + 5} d x + \int \frac{d x}{( x + 2 ) ^{2} + 1}$ .
$I = \frac{1}{2} lo g (x^{2} + 4 x + 5) + tan^{- 1} (x + 2) + C$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q313 marksShort AnswerIntegrals CBSE 2026 · 65/3/2

Evaluate :
$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{c o s ^{2} x}{2 ^{x} + 1} d x$

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Answer:

\frac{π}{4}

Let $I = \int_{- π /2}^{π /2} \frac{c o s ^{2} x}{2 ^{x} + 1} d x$ . Using $\int_{- a}^{a} f (x) d x = \int_{- a}^{a} f (- x) d x$ :
$I = \int_{- π /2}^{π /2} \frac{c o s ^{2} x}{2 ^{- x} + 1} d x = \int_{- π /2}^{π /2} \frac{2 ^{x} c o s ^{2} x}{1 + 2 ^{x}} d x$ .
Adding: $2 I = \int_{- π /2}^{π /2} cos^{2} x d x = 2 \int_{0}^{π /2} cos^{2} x d x = 2 \cdot \frac{π}{4} = \frac{π}{2}$ .
$I = \frac{π}{4}$ .

Q283 marksShort AnswerIntegrals CBSE 2026 · 65/3/3

Evaluate :
$\int_{0}^{\frac{π}{2}} \frac{s i n ^{2} x}{1 + s i n 2 x} d x$

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Answer:

\frac{1}{2}

Let $I = \int_{0}^{π /2} \frac{s i n ^{2} x}{1 + s i n 2 x} d x$ . Replacing x by $\frac{π}{2} - x$ : $I = \int_{0}^{π /2} \frac{c o s ^{2} x}{1 + s i n 2 x} d x$ .
Adding: $2 I = \int_{0}^{π /2} \frac{d x}{1 + s i n 2 x} = \int_{0}^{π /2} \frac{d x}{( s i n x + c o s x ) ^{2}} = \int_{0}^{π /2} \frac{s e c ^{2} x}{( 1 + t a n x ) ^{2}} d x$ .
Put $t = tan x$ : $2 I = \int_{0}^{\infty} \frac{d t}{( 1 + t ) ^{2}} = [- \frac{1}{1 + t}]_{0}^{\infty} = 1$ .
$I = \frac{1}{2}$ .

Q273 marksShort AnswerIntegrals CBSE 2026 · 65/5/1

Find :
$\int \frac{x - s i n x}{1 - c o s x} d x$

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Answer:

- x cot \frac{x}{2} + C

$1 - cos x = 2 sin^{2} \frac{x}{2}$ and $sin x = 2 sin \frac{x}{2} cos \frac{x}{2}$ .
Integrand $= \frac{x}{2} cosec^{2} \frac{x}{2} - cot \frac{x}{2}$ .
By parts: $\int \frac{x}{2} cosec^{2} \frac{x}{2} d x = - x cot \frac{x}{2} + \int cot \frac{x}{2} d x$ .
So $I = - x cot \frac{x}{2} + \int cot \frac{x}{2} d x - \int cot \frac{x}{2} d x = - x cot \frac{x}{2} + C$ .

Also asked in: 2026 65/5/2, 2026 65/5/3

Q27 (OR) (OR)3 marksShort AnswerIntegrals CBSE 2026 · 65/5/1

Evaluate :
$\int_{0}^{2} \frac{1}{x ^{2} + 2 x + 3} d x$

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Answer:

lo g (\frac{3 + 11}{1 + 3})

$x^{2} + 2 x + 3 = (x + 1)^{2} + (2)^{2}$ .
$\int \frac{d x}{( x + 1 ) ^{2} + 2} = lo g x + 1 + x^{2} + 2 x + 3$ .
Value $= lo g (3 + 11) - lo g (1 + 3) = lo g (\frac{3 + 11}{1 + 3})$ .

Also asked in: 2026 65/5/2, 2026 65/5/3

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