Indefinite Integration 1 Question 13
14. If $\int _0^{\pi / 3} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta=1-\frac{1}{\sqrt{2}},(k>0)$, then the value of $k$ is
(a) 1
(b) $\frac{1}{2}$
(c) 2
(d) 4
(2019 Main, 9 Jan II)
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Solution:
- We have, $\int _0^{\pi / 3} \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta=1-\frac{1}{\sqrt{2}},(k>0)$
$$ \text { Let } \begin{aligned} I=\int _0^{\pi / 3} & \frac{\tan \theta}{\sqrt{2 k \sec \theta}} d \theta=\frac{1}{\sqrt{2 k}} \int _0^{\pi / 3} \frac{\tan \theta}{\sqrt{\sec \theta}} d \theta \\ & =\frac{1}{\sqrt{2 k}} \int _0^{\pi / 3} \frac{(\sin \theta)}{(\cos \theta) \sqrt{\frac{1}{\cos \theta}}} d \theta=\frac{1}{\sqrt{2 k}} \int _0^{\pi / 3} \frac{\sin \theta}{\sqrt{\cos \theta}} d \theta \end{aligned} $$
Let $\cos \theta=t \Rightarrow-\sin \theta d \theta=d t \Rightarrow \sin \theta d \theta=-d t$
for lower limit, $\theta=0 \Rightarrow t=\cos 0=1$
for upper limit, $\theta=\frac{\pi}{3} \Rightarrow t=\cos \frac{\pi}{3}=\frac{1}{2}$
$$ \begin{array}{ll} \Rightarrow \quad I & =\frac{1}{\sqrt{2 k}} \int _1^{1 / 2} \frac{-d t}{\sqrt{t}}=\frac{-1}{\sqrt{2 k}} \int _I^{1 / 2} t^{-\frac{1}{2}} d t \\ & =-\frac{1}{\sqrt{2 k}} \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=-\frac{1}{\sqrt{2 k}}[2 \sqrt{t}] _1^{\frac{1}{2}} \\ & =-\frac{2}{\sqrt{2 k}} \sqrt{\frac{1}{2}}-\sqrt{1}=\frac{2}{\sqrt{2 k}} \quad 1-\frac{1}{\sqrt{2}} \\ \because \quad & I=1-\frac{1}{\sqrt{2}} \\ \therefore \quad \frac{2}{\sqrt{2 k}} \quad 1-\frac{1}{\sqrt{2}}=1-\frac{1}{\sqrt{2}} \Rightarrow \frac{2}{\sqrt{2 k}}=1 \\ \Rightarrow \quad 2=\sqrt{2 k} \Rightarrow 2 k=4 \Rightarrow k=2 \end{array} $$
