If

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:

  1. (1) 4

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

  3. (3) 1

  4. (4) 2


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)

$\therefore \quad k=2$

 

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