Q109 / 160 · AP EAMCET 2023 (16 May Shift 2) · Mathematics
$\int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sqrt{x}\left(\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}\right)} d x=$
- A$\frac{2}{\pi}\left[\sin ^{-1} \sqrt{x}(2 x-1)+\sqrt{x(1-x)}\right]+x+C$
- Correct answerB$\frac{8}{\pi}\left(\left(\sqrt{x} \sin ^{-1} \sqrt{x}-\sqrt{1-x}\right)\right)-2 \sqrt{x}+C$
- C$\frac{2}{\pi}\left[(2 x-1) \sin ^{-1} \sqrt{x}-\sqrt{x(1-x)}\right]-x+C$
- D$\frac{2}{\pi}\left[(2 x-1) \sin ^{-1} \sqrt{x}-\sqrt{x(1-x)}\right]+x+C$
Solution
Let $I=\int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sqrt{x}\left(\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}\right)} d x$
$\begin{aligned}
& \text { Let } \sqrt{\mathrm{x}}=\mathrm{t} \Rightarrow \frac{1}{2 \sqrt{\mathrm{x}}} \mathrm{dx}=\mathrm{dt} \\
& \mathrm{I}=2 \int \frac{\sin ^{-1} \mathrm{t}-\cos ^{-1} \mathrm{t}}{\frac{\pi}{2}} \mathrm{dt}=\frac{4}{\pi} \int\left(2 \sin ^{-1} \mathrm{t}-\frac{\pi}{2}\right) \mathrm{dt} \\
& =\frac{8}{\pi} \int \sin -1 \mathrm{tdt}-2 \int \mathrm{dt} \\
& =\frac{8}{\pi}\left[\sqrt{\mathrm{x}} \sin ^{-1} \sqrt{\mathrm{x}}-\sqrt{1-\mathrm{x}}\right]-2 \sqrt{\mathrm{x}}+\mathrm{C}
\end{aligned}$
Hence, the correct answer is option: 'B'.