Q114 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics
$\int_0^{\pi / 4} \frac{x^2}{(x \sin x+\cos x)^2} d x=$
- A$\frac{2-\pi}{2+\pi}$
- Correct answerB$\frac{4-\pi}{4+\pi}$
- C$\frac{6-\pi}{6+\pi}$
- D$\frac{8-\pi}{8+\pi}$
Solution
$\begin{aligned} & \text { } \quad I=\int \frac{x^2}{(x \sin x+\cos x)^2} d x \\ & =\int(x \sec x)\left(\frac{x \cos x}{(x \sin x+\cos x)^2}\right) d x \\ & =x \sec x \times \frac{(-1)}{(x \sin x+\cos x)} \\ & \quad-\int(\sec x+x \sec x \tan x) \frac{-1}{(x \sin x+\cos x)} d x\end{aligned}$
$\begin{aligned} & =\frac{-x \sec x}{(x \sin x+\cos x)}+\int \sec ^2 x d x \\ & =\frac{-x \sec x}{x \sin x+\cos x}+\tan x=\frac{-x+\sin ^2 x(x \sin x+\cos x)}{\cos x(x \sin x+\cos x)} \\ & =\frac{-x \cos ^2 x+\sin x \cos x}{\cos x(x \sin x+\cos x)}=\frac{\sin x-x \cos x}{x \sin x+\cos x} \\ & \int_0^{\pi / 4} \frac{x^2}{(x \sin x+\cos x)^2} d x=\left[\frac{\sin x-x \cos x}{x \sin x+\cos x}\right]_0^{\pi / 4}=\frac{4-\pi}{4+\pi} .\end{aligned}$
$\begin{aligned} & =\frac{-x \sec x}{(x \sin x+\cos x)}+\int \sec ^2 x d x \\ & =\frac{-x \sec x}{x \sin x+\cos x}+\tan x=\frac{-x+\sin ^2 x(x \sin x+\cos x)}{\cos x(x \sin x+\cos x)} \\ & =\frac{-x \cos ^2 x+\sin x \cos x}{\cos x(x \sin x+\cos x)}=\frac{\sin x-x \cos x}{x \sin x+\cos x} \\ & \int_0^{\pi / 4} \frac{x^2}{(x \sin x+\cos x)^2} d x=\left[\frac{\sin x-x \cos x}{x \sin x+\cos x}\right]_0^{\pi / 4}=\frac{4-\pi}{4+\pi} .\end{aligned}$
Hence, the correct answer is option: 'B'.