Q41 / 75 · JEE Main 2025 (23 Jan Shift 2) · Mathematics
If $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} \mathrm{~d} x$, then $\int_0^{21} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} \mathrm{~d} x$ equals :
- A$\frac{\pi^2}{12}$
- B$\frac{\pi^2}{4}$
- Correct answerC$\frac{\pi^2}{16}$
- D$\frac{\pi^2}{8}$
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 $\tan ^2 x=t \Rightarrow 2 \tan x \sec ^2 x d x=d t$
$\begin{aligned}
& \Rightarrow I_0=\frac{\pi}{4} \int_0^{\infty} \frac{\frac{d t}{2}}{\left(1+t^2\right)}=\left.\frac{\pi}{8}\left(\tan ^{-1} t\right)\right|_0 ^{\infty}=\frac{\pi}{8}\left(\frac{\pi}{2}-0\right) \\
& \Rightarrow I_0=\frac{\pi^2}{16}
\end{aligned}$
Hence, the correct answer is option: 'C'.