Question:
If $\int_{0}^{\pi / 3} \frac{\tan \theta}{\sqrt{2 \mathrm{k} \sec \theta}} \mathrm{d} \theta=1-\frac{1}{\sqrt{2}},(\mathrm{k}>0)$ then the value of $\mathrm{k}$
is:
Correct Option: , 4
Solution:
Let, $I=\int_{0}^{\pi / 3} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta$
$=\frac{1}{\sqrt{2 k}} \int_{0}^{\pi / 3} \frac{\sin \theta}{\sqrt{\cos \theta}} d \theta$
Let $\cos \theta=t^{2}$
$\therefore \quad \sin \theta d \theta=-2 t d t$
Hence, integral becomes,
$I=\frac{1}{\sqrt{2 k}} \int_{1}^{\sqrt{\frac{1}{2}}} \frac{-2 t d t}{t}$
$=\sqrt{\frac{2}{k}} \int_{\frac{1}{\sqrt{2}}}^{1} d t$
$=\sqrt{\frac{2}{k}}\left(1-\frac{1}{\sqrt{2}}\right)$
$=\frac{\sqrt{2}-1}{\sqrt{k}}$
$=1-\frac{1}{\sqrt{2}}$ (given)
$\therefore \quad k=2$