Verified Solution • 2025–26 Session moderate QJEEM25Ma8ae2: JEE Main 2025 Mathematics - Class 12 JEEM25Ma8ae2 Solution: If , then equals : If I=∫0π2sin32xsin32x+cos32x dx\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} xI=∫02πsin23x+cos23xsin23x dx , then ∫021xsinxcosxsin4x+cos4x dx\int_0^{21} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} \mathrm{~d} x∫021sin4x+cos4xxsinxcosx dx equals : Multiple Choice Options A π212\frac{\pi^2}{12}12π2 B π24\frac{\pi^2}{4}4π2 C π216\frac{\pi^2}{16}16π2 D π28\frac{\pi^2}{8}8π2 Step-by-Step Verified Solution For I Apply king ( P−5\mathrm{P}-5P−5 ) and add amp;2I=∫0π/2dx=π2⇒I=π4amp;I2=∫0π/2xsinxcosxsin4x+cos4xdx\begin{aligned}& 2 I=\int_0^{\pi / 2} d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4} \\& I_2=\int_0^{\pi / 2} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x\end{aligned}amp;2I=∫0π/2dx=2π⇒I=4πamp;I2=∫0π/2sin4x+cos4xxsinxcosxdx Apply king and add amp;I2=π4∫0π/2tanxsec2xdxtan4x+1amp; put tan2x=tamp;π8∫0∞dtt2+1amp;=π8⋅π2=π216\begin{aligned}& \mathrm{I}_2=\frac{\pi}{4} \int_0^{\pi / 2} \frac{\tan \mathrm{xsec}{ }^2 \mathrm{xdx}}{\tan ^4 \mathrm{x}+1} \\& \text { put } \tan ^2 \mathrm{x}=\mathrm{t} \\& \frac{\pi}{8} \int_0^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^2+1} \\& =\frac{\pi}{8} \cdot \frac{\pi}{2}=\frac{\pi^2}{16}\end{aligned}amp;I2=4π∫0π/2tan4x+1tanxsec2xdxamp; put tan2x=tamp;8π∫0∞t2+1dtamp;=8π⋅2π=16π2 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring