Q111 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics
$\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=$
- A$\sqrt{2} \sin ^{-1}(\sin x+\cos x)+c$
- B$\sqrt{2} \cos ^{-1}(\sin x+\cos x)+c$
- C$\sqrt{2} \cos ^{-1}(\sin x-\cos x)+c$
- Correct answerD$\sqrt{2} \sin ^{-1}(\sin x-\cos x)+c$
Solution
We have,
$$
\begin{aligned}
& \int(\sqrt{\tan x}+\sqrt{\cot x}) d x \\
& =\int\left(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}\right) d x=\int \frac{\sin x+\cos x}{\sqrt{\sin x \cos x}}
\end{aligned}
$$
multiply and divide by $\sqrt{2}$,
$$
\begin{aligned}
& \quad \sqrt{2} \int \frac{\sin x+\cos x}{\sqrt{2 \sin x \cos x}} d x=\sqrt{2} \int \frac{\sin x+\cos x}{\sqrt{1-1+2 \sin x \cos x}} d x \\
& =\sqrt{2} \int \frac{\sin x+\cos x}{1-(1-2 \sin x \cos x)} d x \\
& =\sqrt{2} \int \frac{\sin x+\cos x}{1-\left(\sin ^2 x+\cos ^2 x-2 \sin x \cos x\right)} d x \\
& =\sqrt{2} \int \frac{\sin x+\cos x}{1-(\sin x-\cos x)^2} d x \\
& \text { Put }(\sin x-\cos x)=t \Rightarrow \quad(\cos x+\sin x) d x=d t \\
& =\sqrt{2} \int \frac{d t}{1-t^2}=\sqrt{2} \sin ^{-1} t=\sqrt{2} \sin ^{-1}(\sin x-\cos x)+c
\end{aligned}
$$
Hence, the correct answer is option: 'D'.