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 :
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 $1-\frac{1}{\sqrt{2}} \Rightarrow \mathrm{k}=2$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.