Calculus

Integrals

Medium

core

Match List-I with List-II

List-I List-II
(A) $\displaystyle\int_{\pi/6}^{\pi/3} \cos x \, dx$ (I) $0$
(B) $\displaystyle\int_{-\pi}^{\pi} (\sin^{61}x + x^{123}) \, dx$ (II) $4$
(C) $\displaystyle\int_{0}^{\pi/2} \dfrac{1}{1+\tan^3 x} \, dx$ (III) $\dfrac{\sqrt{3}-1}{2}$
(D) $\displaystyle\int_{-2}^{2} (2 - \left\vert x \right\vert) \, dx$ (IV) $\dfrac{\pi}{4}$

Choose the correct answer from the options given below:

List-I	List-II
(A) $\displaystyle\int_{\pi/6}^{\pi/3} \cos x \, dx$	(I) $0$
(B) $\displaystyle\int_{-\pi}^{\pi} (\sin^{61}x + x^{123}) \, dx$	(II) $4$
(C) $\displaystyle\int_{0}^{\pi/2} \dfrac{1}{1+\tan^3 x} \, dx$	(III) $\dfrac{\sqrt{3}-1}{2}$
(D) $\displaystyle\int_{-2}^{2} (2 - \left\vert x \right\vert) \, dx$	(IV) $\dfrac{\pi}{4}$

(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

(A) - (III), (B) - (I), (C) - (IV), (D) - (II)

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

(A) - (III), (B) - (IV), (C) - (II), (D) - (I)

✅ Correct Option: 2

(A) $\sin\frac{\pi}{3}-\sin\frac{\pi}{6} = \frac{\sqrt{3}-1}{2}$ .

(B) the integrand is odd over a symmetric interval, so the value is 0.

(C) using the king property $\int_0^{a}f(x)dx = \int_0^a f(a-x)dx$ , the integral is half of $\frac{\pi}{2}$ , i.e. $\frac{\pi}{4}$ .

(D) $2\int_0^2(2-x)dx = 4$ .

2022: 10 Aug Shift 1

Match List I with List II

2023: 30 May Shift 3

$\int \left(x + \frac{1}{x}\right)^2 dx$ equals :

2025: 22 May Shift 1

$\int \frac{x}{(x-1)(x-2)} dx$ is equal to ( where $C$ is a constant of integration)

2026: 18 May Shift 1

If $\int \frac{1}{\tan x + \cot x}dx = k\cos 2x + C$ , where C is an arbitrary constant, then the value of $k$ is

2025: 13 May Shift 1

$\int \frac{dx}{9x^2 - 16}$ is equal to

2026: 19 May Shift 1

Match List-I with List-II

Choose the correct answer from the options given below:

Solution