Verified Solution • 2025–26 Session moderate

QJEEM25Ma8ae2: JEE Main 2025 Mathematics - Class 12 JEEM25Ma8ae2 Solution: If , then equals :

If I=∫0π2sin⁡32xsin⁡32x+cos⁡32x 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} x , then ∫021xsin⁡xcos⁡xsin⁡4x+cos⁡4x dx\int_0^{21} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} \mathrm{~d} x equals :

Multiple Choice Options

  • A π212\frac{\pi^2}{12}
  • B π24\frac{\pi^2}{4}
  • C π216\frac{\pi^2}{16}
  • D π28\frac{\pi^2}{8}

Step-by-Step Verified Solution

For I

Apply king ( P−5\mathrm{P}-5 ) and add

amp;2I=∫0π/2dx=π2⇒I=π4amp;I2=∫0π/2xsin⁡xcos⁡xsin⁡4x+cos⁡4xdx\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}

Apply king and add

amp;I2=π4∫0π/2tan⁡xsec2xdxtan⁡4x+1amp; put tan⁡2x=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}

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