Integrals: 4 marks Questions (CBSE Class 12)

8 different 4 marks questions on Integrals from CBSE Class 12 Maths board exams 2022–2026, newest first.

1 mark (78)2 marks (23)3 marks (104)4 marks (8)5 marks (25)

Q134 marksLong AnswerIntegrals CBSE 2022 · 65/1/1

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

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Answer:

π

Let $I = \int_{0}^{π} \frac{x}{1 + s i n x} d x$ . Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ : $I = \int_{0}^{π} \frac{π - x}{1 + s i n x} d x$ .
Adding: $2 I = π \int_{0}^{π} \frac{d x}{1 + s i n x}$ .
$\frac{1}{1 + s i n x} = \frac{1 - s i n x}{c o s ^{2} x} = sec^{2} x - sec x tan x$ , with antiderivative $tan x - sec x$ .
$[tan x - sec x]_{0}^{π} = (0 + 1) - (0 - 1) = 2$ (the integrand is continuous on $[0, π]$ , and $tan x - sec x = \frac{s i n x - 1}{c o s x}$ extends continuously at $\frac{π}{2}$ ).
So $2 I = 2 π$ , giving $I = π$ .

Also asked in: 2023 65/5/2, 2022 65/1/2, 2022 65/1/3

Q124 marksLong AnswerIntegrals CBSE 2022 · 65/2/1

Evaluate :
$\int_{0}^{1} x (1 - x)^{n} d x$

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Answer:

\frac{1}{( n + 1 ) ( n + 2 )}

Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ : $\int_{0}^{1} x (1 - x)^{n} d x = \int_{0}^{1} (1 - x) x^{n} d x$ .
$= \int_{0}^{1} (x^{n} - x^{n + 1}) d x = [\frac{x ^{n + 1}}{n + 1} - \frac{x ^{n + 2}}{n + 2}]_{0}^{1}$ .
$= \frac{1}{n + 1} - \frac{1}{n + 2} = \frac{1}{( n + 1 ) ( n + 2 )}$ .

Also asked in: 2022 65/2/3

Q134 marksLong AnswerIntegrals CBSE 2022 · 65/2/2

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

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Answer:

\frac{π}{2 2} lo g (2 + 1)

Let $I = \int_{0}^{π /2} \frac{x}{s i n x + c o s x} d x$ . Replacing $x$ by $\frac{π}{2} - x$ : $I = \int_{0}^{π /2} \frac{\frac{π}{2} - x}{c o s x + s i n x} d x$ .
Adding: $2 I = \frac{π}{2} \int_{0}^{π /2} \frac{d x}{s i n x + c o s x}$ .
$sin x + cos x = 2 sin (x + \frac{π}{4})$ , so $\int_{0}^{π /2} \frac{d x}{s i n x + c o s x} = \frac{1}{2} \int_{π /4}^{3 π /4} cosec u d u$ .
$= \frac{1}{2} [lo g ∣ cosec u - cot u ∣]_{π /4}^{3 π /4} = \frac{1}{2} (lo g (2 + 1) - lo g (2 - 1)) = \frac{1}{2} lo g (2 + 1)^{2} = 2 lo g (2 + 1)$ .
$2 I = \frac{π}{2} 2 lo g (2 + 1)$ , so $I = \frac{π}{2 2} lo g (2 + 1)$ .

Also asked in: 2025 65/6/1, 2025 65/6/2, 2025 65/6/3

Q124 marksLong AnswerIntegrals CBSE 2022 · 65/3/1

Find : $\int \frac{x ^{2}}{( x ^{2} + 1 ) ( 3 x ^{2} + 4 )} d x$

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Answer:

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

Put $x^{2} = t$ : $\frac{t}{( t + 1 ) ( 3 t + 4 )} = \frac{A}{t + 1} + \frac{B}{3 t + 4}$
$t = A (3 t + 4) + B (t + 1)$ ; $t = - 1$ : $A = - 1$ ; $t = - \frac{4}{3}$ : $B = 4$
So integrand $= - \frac{1}{x ^{2} + 1} + \frac{4}{3 x ^{2} + 4}$
$\int \frac{4}{3 x ^{2} + 4} d x = \frac{4}{3} \int \frac{d x}{x ^{2} + ( \frac{2}{3} ) ^{2}} = \frac{4}{3} \cdot \frac{3}{2} tan^{- 1} \frac{3 x}{2} = \frac{2}{3} tan^{- 1} \frac{3 x}{2}$
Answer: $- tan^{- 1} x + \frac{2}{3} tan^{- 1} (\frac{3 x}{2}) + C$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q12 (OR) (OR)4 marksLong AnswerIntegrals CBSE 2022 · 65/3/1

Evaluate : $\int_{- 2}^{1} 5 - 4 x - x^{2} d x$

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Answer:

\frac{9 π}{4}

$5 - 4 x - x^{2} = 9 - (x + 2)^{2}$
$\int 9 - (x + 2)^{2} d x = \frac{x + 2}{2} 5 - 4 x - x^{2} + \frac{9}{2} sin^{- 1} \frac{x + 2}{3}$
At $x = 1$ : $\frac{3}{2} \cdot 0 + \frac{9}{2} \cdot \frac{π}{2} = \frac{9 π}{4}$
At $x = - 2$ : $0 + 0 = 0$
Value $= \frac{9 π}{4}$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q134 marksLong AnswerIntegrals CBSE 2022 · 65/4/1

Evaluate :
$\int_{0}^{π /2} (2 lo g cos x - lo g sin 2 x) d x$

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Answer:

- \frac{π}{2} lo g 2

(i.e.

\frac{π}{2} lo g \frac{1}{2}

)

$lo g sin 2 x = lo g 2 + lo g sin x + lo g cos x$ .
So the integrand is $lo g cos x - lo g sin x - lo g 2$ .
By $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ , $\int_{0}^{π /2} lo g cos x d x = \int_{0}^{π /2} lo g sin x d x$ , so those two terms cancel.
Integral $= - \int_{0}^{π /2} lo g 2 d x = - \frac{π}{2} lo g 2$ .

Q124 marksLong AnswerIntegrals CBSE 2022 · 65/4/2

Evaluate :
$\int_{0}^{π} \frac{x}{9 s i n ^{2} x + 16 c o s ^{2} x} d x$

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Answer:

\frac{π ^{2}}{24}

Let $I = \int_{0}^{π} \frac{x}{9 s i n ^{2} x + 16 c o s ^{2} x} d x$ . Using $x \to π - x$ : $I = \int_{0}^{π} \frac{π - x}{9 s i n ^{2} x + 16 c o s ^{2} x} d x$ .
Adding: $2 I = π \int_{0}^{π} \frac{d x}{9 s i n ^{2} x + 16 c o s ^{2} x} = 2 π \int_{0}^{π /2} \frac{d x}{9 s i n ^{2} x + 16 c o s ^{2} x}$ (the integrand is unchanged by $x \to π - x$ ).
So $I = π \int_{0}^{π /2} \frac{s e c ^{2} x d x}{9 t a n ^{2} x + 16}$ . Put $t = tan x$ : $I = π \int_{0}^{\infty} \frac{d t}{9 t ^{2} + 16}$ .
$\int_{0}^{\infty} \frac{d t}{9 t ^{2} + 16} = \frac{1}{9} \cdot \frac{3}{4} [tan^{- 1} \frac{3 t}{4}]_{0}^{\infty} = \frac{1}{12} \cdot \frac{π}{2} = \frac{π}{24}$ .
Hence $I = \frac{π ^{2}}{24}$ .

Q114 marksLong AnswerIntegrals CBSE 2022 · 65/4/3

Evaluate :
$\int_{- 3}^{3} \frac{x ^{4}}{1 + e ^{x}} d x$

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Answer:

\frac{243}{5}

Let $I = \int_{- 3}^{3} \frac{x ^{4}}{1 + e ^{x}} d x$ . Using $x \to - x$ : $I = \int_{- 3}^{3} \frac{x ^{4}}{1 + e ^{- x}} d x = \int_{- 3}^{3} \frac{x ^{4} e ^{x}}{1 + e ^{x}} d x$ .
Adding: $2 I = \int_{- 3}^{3} x^{4} d x = 2 \int_{0}^{3} x^{4} d x$ .
$I = \int_{0}^{3} x^{4} d x = [\frac{x ^{5}}{5}]_{0}^{3} = \frac{243}{5}$ .

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