Q87 / 200 · Manipal (MET) 2012 · Mathematics
Evaluate $\int(\sqrt{\tan x}+\sqrt{\cot x}) d x$
- A$\sqrt{2} \tan ^{-1}(\sqrt{\tan x}-\sqrt{\cot x})+C$
- B$\frac{2}{\sqrt{2}} \tan ^{-1} x+C$
- Correct answerC$\sqrt{2} \tan ^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)+C$
- DNone of the above
Solution
$\text { Let } \begin{aligned}
I & =\int(\sqrt{\tan x}+\sqrt{\cot x}) d x \\
& =\int \frac{\tan x+1}{\sqrt{\tan x}} d x
\end{aligned}$
Put $\tan x=t^2$
$\begin{aligned} & \Rightarrow \quad \sec ^2 x d x=2 t d t \\ & \Rightarrow \quad d x=\frac{2 t d t}{1+\tan ^2 t}=\frac{2 t}{1+t^4} d t \\ & \begin{aligned} \therefore \quad I & =\int \frac{t^2+1}{\sqrt{t^2}} \cdot \frac{2 t}{t^4+1} d t \\ & =2 \int \frac{t^2+1}{t^4+1} d t\end{aligned}\end{aligned}$
$\begin{aligned} &=2 \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}-2+2} d t \\ &=2 \int \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+(\sqrt{2})^2} d t \\ &=2 \int \frac{d u}{u^2+(\sqrt{2})^2} \\ & \text { where, } \quad \begin{aligned} u & =t-\frac{1}{t} \Rightarrow d u=\left(1+\frac{1}{t^2}\right) d t \\ \Rightarrow \quad I & =\frac{2}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+C \\ & =\sqrt{2} \tan ^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)+C\end{aligned} \$\end{aligned}$
Hence, the correct answer is option: 'C'.