Q98 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics
$\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{\sin x+\cos x} d x=$
- A$\sqrt{2} \log (\sqrt{2}+1)$
- Correct answerB$\frac{1}{\sqrt{2}} \log (\sqrt{2}+1)$
- C$\log (\sqrt{2}+1)$
- D$\frac{1}{\sqrt{2}} \log (\sqrt{2}-1)$
Solution
$I=\int_0^{\pi / 2} \frac{\sin ^2 x}{\sin x+\cos x} d x ... (i)$
$I=\int_0^{\pi / 2} \frac{\sin ^2\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x$
$\begin{aligned} & I=\int_0^{\pi / 2} \frac{\cos ^2 x}{\sin x+\cos x} d x ... (ii) \\ & \text { (i) }+ \text { (ii) } \\ & 2 I=\int_0^{\pi / 2} \frac{\sin ^2 x+\cos ^2 x}{\sin x+\cos x} d x \\ & \Rightarrow \quad 2 I=\int_0^{\pi / 2} \frac{d x}{\sin x+\cos x} \\ & \Rightarrow \quad 2 I=\int_0^{\pi / 2} \frac{d x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right)} \\ & \Rightarrow I=\frac{1}{2 \sqrt{2}} \int_0^{\pi / 2} \frac{d x}{\sin \left(x+\frac{\pi}{4}\right)} \\ & \Rightarrow I=\frac{1}{2 \sqrt{2}} \int_0^{\pi / 2} \operatorname{cosec}\left(\frac{\pi}{4}+x\right) d x \\ & \Rightarrow I=\frac{1}{2 \sqrt{2}}\left[\log \left\{\operatorname{cosec}\left(\frac{x+\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right\}\right]_0^{\pi / 2} \\ & \Rightarrow I=\frac{1}{2 \sqrt{2}}\left[\log \left(\operatorname{cosec} \frac{3 \pi}{4}-\cot \frac{3 \pi}{4}\right)-\log \left(\operatorname{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right)\right] \\ & \Rightarrow I=\frac{1}{2 \sqrt{2}}[\log (1+\sqrt{2})-\log (\sqrt{2}-1)] \\ & \Rightarrow I=\frac{1}{2 \sqrt{2}} \log \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) \\ & \Rightarrow I(\sqrt{2}+1)^2 \\ & \Rightarrow I \log (\sqrt{2}+1) .\end{aligned}$
Hence, the correct answer is option: 'B'.