CUET PG 2026 — Mathematics PYQ

Solution

Let $I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{1+\sin x\cos x}dx\dots (1)$

Using property ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$:

$I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^2(\frac{\pi }{2}-x)}{1+\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)}dx$

$I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{1+\cos x\sin x}dx\dots (2)$

Adding $(1)$ and $(2)$:

$2I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x+{\cos }^{2}x}{1+\sin x\cos x}dx$

$2I={\int }_0^{\frac{\pi }{2}}\frac{1}{1+\sin x\cos x}dx$

Divide numerator and denominator by ${\cos }^{2}x$:

$2I={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{{\sec }^{2}x+\tan x}dx$

$2I={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{1+{\tan }^{2}x+\tan x}dx$

Let $\tan x=t⟹{\sec }^{2}xdx=dt$

Limits: $x=0\to t=0$, $x=\frac{\pi }{2}\to t=\infty$

$2I={\int }_0^{\infty }\frac{dt}{t^2+t+1}$

$2I={\int }_0^{\infty }\frac{dt}{(t+\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}$

$2I={\left[\frac{1}{\frac{\sqrt{3}}{2}}{\tan }^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]}_0^{\infty }$

$2I=\frac{2}{\sqrt{3}}\left[{\tan }^{-1}(\infty )-{\tan }^{-1}(\frac{1}{\sqrt{3}})\right]$

$2I=\frac{2}{\sqrt{3}}\left[\frac{\pi }{2}-\frac{\pi }{6}\right]$

$2I=\frac{2}{\sqrt{3}}\left[\frac{\pi }{3}\right]$

$I=\frac{\pi }{3\sqrt{3}}$

Correct Option: (a)