If I=∫_0^ (π)/(2) ^ (3)/(2) x ^ (3)/(2) x+ ^ (3)/(2) x ~d x , then ∫_0^ 21 (x x x)/( ^4 x+ ^4 x) ~d x equals :

Correct answer: C. (π^2)/(16) aligned & I=∫_0^ (π)/(2) ( x)^ (3)/(2) d x ( x)^ (3)/(2) x+( x)^ (3)/(2) =∫_0^ (π)/(2) ^ (3)/(2) ((π)/(2)-x) d x ^ (3)/(2) ((π)/(2)-x)+ ^ (3)/(2) ((π)/(2)-x) & Adding 2 l=∫_0^ (π)/(2) ( x)^ (3)/(2) +( x)^ (3)/(2) ( x)^ (3)/(2) +( x)^ (3)/(2) d x=(π)/(2) & I_0=∫_0^ (π)/(2) (x x x)/( ^4 x+ ^4 x) d x=∫_0^ (π)/(2) (((π)/(2)-x) x x)/(( x)^4+( x)^4) d x aligned Adding, 2 I_0=∫_0^ (π)/(2) ((π)/(2)( x) x)/(( ^4 x+ ^4 x)) d x I_0=(π)/(4) ∫_0^ (π)/(2) ( x( ^2 x) d x)/(1+ ^4 x) put ^2 x=t 2 x ^2 x d x=d t aligned & I_0=(π)/(4) ∫_0^ ∞ ((d t)/(2))/((1+t^2))=.(π)/(8)( ^ -1 t)|_0 ^ ∞ =(π)/(8)((π)/(2)-0) & I_0=(π^2)/(16) aligned

JEE Main 202523 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Definite Integration

If I=∫_0^ (π)/(2) ^ (3)/(2) x ^ (3)/(2) x+ ^ (3)/(2) x ~d x , then...

\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} \mathrm{~d} x

, then

\int_0^{21} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} \mathrm{~d} x

equals :

Options

A $\frac{\pi^2}{12}$
B $\frac{\pi^2}{4}$
C $\frac{\pi^2}{16}$
D $\frac{\pi^2}{8}$

Correct answer

C. $\frac{\pi^2}{16}$

Step-by-step solution

\begin{aligned} & I=\int_0^{\frac{\pi}{2}} \frac{(\sin x)^{\frac{3}{2}} d x}{(\sin x)^{\frac{3}{2}} x+(\cos x)^{\frac{3}{2}}}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right) d x}{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)+\cos ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)} \\ & \Rightarrow \text{ Adding } 2 l=\int_0^{\frac{\pi}{2}} \frac{(\sin x)^{\frac{3}{2}}+(\cos x)^{\frac{3}{2}}}{(\sin x)^{\frac{3}{2}}+(\cos x)^{\frac{3}{2}}} d x=\frac{\pi}{2} \\ & I_0=\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x=\int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin x \cos x}{(\sin x)^4+(\cos x)^4} d x \end{aligned}

Adding,

2 I_0=\int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2}(\sin x) \cos x}{\left(\sin ^4 x+\cos ^4 x\right)} d x \Rightarrow I_0=\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\tan x\left(\sec ^2 x\right) d x}{1+\tan ^4 x}

put

\tan ^2 x=t \Rightarrow 2 \tan x \sec ^2 x d x=d t

\begin{aligned} & \Rightarrow I_0=\frac{\pi}{4} \int_0^{\infty} \frac{\frac{d t}{2}}{\left(1+t^2\right)}=\left.\frac{\pi}{8}\left(\tan ^{-1} t\right)\right|_0 ^{\infty}=\frac{\pi}{8}\left(\frac{\pi}{2}-0\right) \\ & \Rightarrow I_0=\frac{\pi^2}{16} \end{aligned}

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