Hard MCQ +4 / -1 PYQ · JEE Mains 2025

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 $\int_0^{2I} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} \mathrm{~d} x$ equals :

  1. A $\frac{\pi^2}{12}$
  2. B $\frac{\pi^2}{4}$
  3. C $\frac{\pi^2}{16}$ Correct answer
  4. D $\frac{\pi^2}{8}$

Solution

<p>For I</p> <p>Apply king ($\mathrm{P}-5$) and add</p> <p>$$\begin{aligned} & 2 I=\int_0^{\pi / 2} d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4} \\ & I_2=\int_0^{\pi / 2} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \end{aligned}$$</p> <p>Apply king and add</p> <p>$$\begin{aligned} & \mathrm{I}_2=\frac{\pi}{4} \int_0^{\pi / 2} \frac{\tan \mathrm{xsec}{ }^2 \mathrm{xdx}}{\tan ^4 \mathrm{x}+1} \\ & \text { put } \tan ^2 \mathrm{x}=\mathrm{t} \\ & \frac{\pi}{8} \int_0^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^2+1} \\ & =\frac{\pi}{8} \cdot \frac{\pi}{2}=\frac{\pi^2}{16} \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals

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