Match each definite integral in List-I…

Match each definite integral in List-I with its value in List-II. List-I (P) ∫0^π/2(sin ^5x)/(sin ^5x+cos ^5x) dx (Q) ∫0^π(xsin x)/(1+cos ^2x) dx (R)…

Single correct option

Match each definite integral in List-I with its value in List-II.

List-I
(P)

\displaystyle\int_0^{\pi/2}\frac{\sin^{5}x}{\sin^{5}x+\cos^{5}x}\,dx

(Q)

\displaystyle\int_0^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,dx

(R)

\displaystyle\int_0^{\pi/4}\ln(1+\tan x)\,dx

(S)

\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^{4}x}{1+e^{x}}\,dx

List-II
(1)

\dfrac{\pi}{4}

(2)

\dfrac{\pi^{2}}{4}

(3)

\dfrac{\pi}{8}\ln 2

(4)

\dfrac{3\pi}{16}

(5)

\dfrac{\pi}{2}

A. P→1, Q→2, R→3, S→4
B. P→5, Q→2, R→3, S→1
C. P→1, Q→5, R→3, S→4
D. P→1, Q→2, R→4, S→3

Show answer and solution

Answer: (A) P→1, Q→2, R→3, S→4

(P) Using

x\to\frac\pi2-x

and adding,

2I=\int_0^{\pi/2}1\,dx

, so

I=\frac\pi4

: (1).
(Q) Using

x\to\pi-x

I=\frac\pi2\int_0^\pi\frac{\sin x}{1+\cos^2x}dx=\frac\pi2[-\tan^{-1}\cos x]_0^\pi=\frac\pi2\cdot\frac\pi2=\frac{\pi^2}{4}

: (2).
(R) Using

x\to\frac\pi4-x

\ln(1+\tan x)+\ln\frac{2}{1+\tan x}=\ln2

, so

I=\frac12\cdot\frac\pi4\ln2=\frac\pi8\ln2

: (3).
(S) Adding

x\to-x

gives

\frac{\cos^4x}{1+e^x}+\frac{e^x\cos^4x}{1+e^x}=\cos^4x

, so

I=\frac12\int_{-\pi/2}^{\pi/2}\cos^4x\,dx=\frac12\cdot\frac{3\pi}{8}=\frac{3\pi}{16}

: (4).
All four values were also confirmed numerically by quadrature. The common theme is the reflection property: replace the variable by a mirror image, add the two forms of the integral and the integrand collapses. The correct code is P→1, Q→2, R→3, S→4.

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Original question written in the JEE Advanced 2023 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake