Q114 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
$\int_0^1 \sqrt{\frac{2+x}{2-x}} d x=$
- A$\pi+2$
- B$\frac{1}{2}(\pi+2)$
- C$\frac{\pi}{2}+2+\sqrt{3}$
- Correct answerD$\frac{\pi}{3}+2-\sqrt{3}$
Solution
$\begin{aligned}
& \mathrm{I}=\int_0^1 \sqrt{\frac{2+x}{2-x}} d x ; \text { Let } u=\sqrt{\frac{2-x}{2+x}} \\
& \Rightarrow d u=\left(-\frac{1}{2 \sqrt{2-x} \sqrt{x+2}}-\frac{\sqrt{2-x}}{2(x+2)^{\frac{3}{2}}}\right) d x \\
& \Rightarrow d u=\left(-\frac{1}{2 u(x+2)}-\frac{u}{2(x+2)}\right) d x=\frac{-\left(1+u^2\right)}{2 u(x+2)} d x \\
& \text {Now, } \mathrm{I}=-\int_1^{\frac{1}{\sqrt{3}}} \sqrt{\frac{2+x}{2-x}} \times \frac{2 u(x+2)}{\left(1+u^2\right)} d u=-8 \int_1^{\frac{1}{\sqrt{3}}}\frac{1}{\left(u^2+1\right)^2} d u
\end{aligned}$
Let $u=\tan v \Rightarrow d u=\sec ^2 v d v$
$\begin{aligned}
& \mathrm{I}=-8 \int_{\frac{\pi}{4}}^{\frac{\pi}{6}} \frac{\sec ^2 v}{\left(\tan ^2 v+1\right)^2} d v=-8 \int_{\frac{\pi}{4}}^{\frac{\pi}{6}} \cos ^2 v d v \\
& \mathrm{I}=-8\left[\frac{\sin 2 v}{4}+\frac{v}{2}\right]_{\frac{\pi}{4}}^{\frac{\pi}{6}}=\frac{\pi}{3}+2-\sqrt{3}
\end{aligned}$
Hence, the correct answer is option: 'D'.