JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 14)

$$\int_\limits0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \text { is equal to }$$
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Explanation

$$\begin{aligned} & \int_\limits0^{\pi / 4} \frac{\cos ^2 x \cdot \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \\ & =\int_\limits0^{\pi / 4} \frac{\tan ^2 x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x \end{aligned}$$

Let $$\tan x=t$$

$$\int_\limits0^1 \frac{t^2 d t}{\left(1+t^3\right)^2}$$

Let $$1+t^3=\mathrm{z}$$

$$\begin{gathered} 3 t^2 d t=d z \\ \frac{1}{3} \int_\limits1^2 \frac{d z}{z^2}=\left.\frac{1}{3}\left(-\frac{1}{z}\right)\right|_1 ^2 \\ =-\frac{1}{3}\left(\frac{1}{2}-1\right)=\frac{1}{6}\end{gathered}$$

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