Q115 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
If $\int(\sqrt{\operatorname{cosec} x+1}) d x=k \tan ^{-1}(f(x))+c$, then $\frac{1}{k} f\left(\frac{\pi}{6}\right)=$
- A$\frac{1}{2}$
- B$\frac{1}{4}$
- C$-\frac{1}{4}$
- Correct answerD$-\frac{1}{2}$
Solution
$\begin{aligned} & \mathrm{I}=\int \sqrt{\operatorname{cosec} x+1} d x \\ = & \int \sqrt{\frac{\tan ^2\left(\frac{x}{2}\right)+1}{2 \tan \left(\frac{x}{2}\right)}+1} d x\end{aligned}$
$\operatorname{Let} u=\tan \frac{x}{2} \Rightarrow d x=\frac{2}{u^2+1} d u$
$\begin{aligned} \mathbb{I} & =\sqrt{2} \int \frac{\sqrt{\frac{u^2+\mathbb{1}}{u}+2}}{u^2+1} d u \\ & =\sqrt{2} \int \frac{u+1}{\sqrt{u}\left(u^2+1\right)} d u\end{aligned}$
Let $v=\sqrt{u} \Rightarrow d v=\frac{1}{2 \sqrt{u}}$
$\begin{aligned} I & =2 \sqrt{2} \int \frac{v^2+1}{v^4+1} d v \\ & =2 \sqrt{2} \int \frac{v^2+1}{\left(v^2-\sqrt{2} v+1\right)\left(v^2+\sqrt{2} v+1\right)} d v \\ & =\sqrt{2} \int \frac{1}{v^2+\sqrt{2} v+1} d v+\sqrt{2} \int \frac{1}{v^2-\sqrt{2} v+1} d v \\ & =\sqrt{2} \int \frac{1}{\left(v+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} d v+\sqrt{2} \int \frac{1}{\left(v-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} d v\end{aligned}$
$\begin{aligned} & =2 \tan ^{-1}(\sqrt{2} v+1)+2 \tan ^{-1}(\sqrt{2} v-1) \\ & =2 \tan ^{-1}(\sqrt{2} u+1)+2 \tan ^{-1}(\sqrt{2} u-1)\end{aligned}$
$=2 \tan ^{-1}\left(\frac{\sqrt{2} u}{1-u}\right)=2 \tan ^{-1}\left(\frac{\sqrt{2 \tan \frac{x}{2}}}{1-\tan \frac{x}{2}}\right)$
$\Rightarrow k=2, f(x)=\frac{\sqrt{2 \tan \frac{x}{2}}}{1-\tan \frac{x}{2}} \quad\left[\because \tan \frac{\pi}{12}=2-\sqrt{3}\right]$
$\Rightarrow \frac{1}{k} f\left(\frac{\pi}{6}\right)=\frac{1}{2} \frac{\sqrt{4-2 \sqrt{3}}}{1-(2-\sqrt{3})}=\frac{1}{2} \frac{\sqrt{(1-\sqrt{3})^2}}{\sqrt{3}-1}=\frac{-1}{2}$
Hence, the correct answer is option: 'D'.