Question

The value of $\underset{0}{\overset{\pi /3}{\int }}\frac{\tan \theta }{\sqrt{2k\sec \theta }}d\theta$ such that $k>0$ is

• A. $\frac{\sqrt{2}-1}{\sqrt{k}}$
• B. $\frac{\sqrt{2}-1}{k}$
• C. $\frac{1-\sqrt{2}}{\sqrt{k}}$
• D. $\frac{1}{\sqrt{k}}$

Answer 1

The correct option is A $\frac{\sqrt{2}-1}{\sqrt{k}}$

Let $I=\underset{0}{\overset{\pi /3}{\int }}\frac{\tan \theta }{\sqrt{2k\sec \theta }}d\theta$
$\Rightarrow I=\underset{0}{\overset{\pi /3}{\int }}\frac{\sin \theta }{\sqrt{2k\cos \theta }}d\theta$
Substitute $\cos \theta =t\Rightarrow \sin \theta d\theta =-dt$
$\Rightarrow I=\frac{-1}{\sqrt{2k}}\underset{1}{\overset{1/2}{\int }}{t}^{-1/2}dt$
$\Rightarrow I=-\sqrt{\frac{2}{k}}{\sqrt{t}|}_1^{1/2}$
$\therefore I=\frac{\sqrt{2}-1}{\sqrt{k}}$