Calculus

Integrals

Medium

core

Match List-I with List-II

List-I List-II
(A) $\displaystyle\int_{-2}^{2}\cos^4x\sin^5x\,dx$ (I) $\pi$
(B) $\displaystyle\int_{-1}^{1}\lvert x \rvert dx$ (II) $\dfrac{\pi}{4}$
(C) $\displaystyle\int_0^{\pi/2}\cos^2x\,dx$ (III) 0
(D) $\displaystyle\int_{-\pi/2}^{\pi/2}(x^3+x\cos x+\tan^5x+1)dx$ (IV) 1

Choose the correct answer from the options given below:

List-I	List-II
(A) $\displaystyle\int_{-2}^{2}\cos^4x\sin^5x\,dx$	(I) $\pi$
(B) $\displaystyle\int_{-1}^{1}\lvert x \rvert dx$	(II) $\dfrac{\pi}{4}$
(C) $\displaystyle\int_0^{\pi/2}\cos^2x\,dx$	(III) 0
(D) $\displaystyle\int_{-\pi/2}^{\pi/2}(x^3+x\cos x+\tan^5x+1)dx$	(IV) 1

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

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

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

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

✅ Correct Option: 1

(A) $\cos^4x\sin^5x$ is odd, so the symmetric integral is $0$ .

(B) $|x|$ is even: $2\int_0^1 x\,dx = 1$ .

(D) The first three terms are odd and cancel, leaving $\int_{-\pi/2}^{\pi/2}1\,dx = \pi$ .

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$\int\left\{\dfrac{1}{\log_e x} - \dfrac{1}{(\log_e x)^2}\right\}dx$ is equal to : (where $C$ is an arbitrary constant)

2025: 16 May Shift 1

The value of $\int_0^1 [\log x - \log(1-x)] dx$ is

2023: 22 May Shift 3

$\int_1^2 \frac{x \, dx}{(x+1)(x+2)} =$

2026: 14 May Shift 2

$\int \frac{x^8}{(x^3+1)^{1/3}} dx$ is equal to

2025: 30 May Shift 2

Value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \log(\tan x)dx$ is

Solution