# Solve that equation

Question:

If $\int_{0}^{\frac{\pi}{3}} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} \mathrm{d} \theta=1-\frac{1}{\sqrt{2}},(\mathrm{k}>0)$, then the

value of $\mathrm{k}$ is :

1. 2

2. $\frac{1}{2}$

3. 4

4. 1

Correct Option: 1

Solution:

$\frac{1}{\sqrt{2 k}} \int_{0}^{\pi / 3} \frac{\tan \theta}{\sqrt{\sec \theta}} \mathrm{d} \theta=\frac{1}{\sqrt{2 \mathrm{k}}} \int_{0}^{\pi / 3} \frac{\sin \theta}{\sqrt{\cos \theta}} \mathrm{d} \theta$

$=-\left.\frac{1}{\sqrt{2 k}} 2 \sqrt{\cos \theta}\right|_{0} ^{\pi / 3}=-\frac{\sqrt{2}}{\sqrt{k}}\left(\frac{1}{\sqrt{2}}-1\right)$

given it is $1-\frac{1}{\sqrt{2}} \Rightarrow \mathrm{k}=2$