Q116 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics
$e^{\int_0^{\pi / 2} \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x}=$
- A1
- B$2 \log 2$
- C$2 \log \sqrt{2}$
- Correct answerD2
Solution
$$
\begin{aligned}
& \text { Let } I=e^{\int_0^{\pi / 2} \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x}=e^{\int_0^{\pi / 2} \sqrt{\frac{(\cos x-\sin x)^2}{(\cos x+\sin x)^2}} d x} \\
& =e^{\int_0^{\pi / 2} \sqrt{\tan ^2\left(\frac{\pi}{4}-x\right) d x}}=e^{\int_0^{\pi / 2} \mid \tan (\pi / 4-x \mid d x} \\
& =e^{\left[\int_0^{\pi / 2} \tan (\pi / 4-x) d x-\int_{\pi / 4}^{\pi / 2} \tan (\pi / 4-x) d x\right]} \\
& =e^{\left[\left[-\log \left(\sec \left(\frac{\pi}{4}-x\right)\right]_0^{\pi / 4}-\left[-\log \mid\left(\sec \left(\frac{\pi}{4}-x\right) \mid\right]_{\pi / 4}^{\pi / 2}\right]\right.\right.} \\
& =e^{[(0+\log \sqrt{2})-(-\log \sqrt{2}+0)]} \\
& =e^{2 \log \sqrt{2}}=e^{\log 2}=2
\end{aligned}
$$
Hence, the correct answer is option: 'D'.