SATHEE: Indefinite Integration 1 Question 13

Indefinite Integration 1 Question 13

14. If $\int_{0}^{π / 3} \frac{\tan θ}{\sqrt{2 k \sec θ}} d θ = 1 - \frac{1}{\sqrt{2}}, (k > 0)$ , then the value of $k$ is

(a) 1

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

(d) 4

(2019 Main, 9 Jan II)

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Solution:

We have, $\int_{0}^{π / 3} \frac{\tan θ}{\sqrt{2 k \sec θ}} d θ = 1 - \frac{1}{\sqrt{2}}, (k > 0)$

$Let \begin{aligned} I = \int_{0}^{π / 3} & \frac{\tan θ}{\sqrt{2 k \sec θ}} d θ = \frac{1}{\sqrt{2 k}} \int_{0}^{π / 3} \frac{\tan θ}{\sqrt{\sec θ}} d θ \\ = \frac{1}{\sqrt{2 k}} \int_{0}^{π / 3} \frac{(\sin θ)}{(\cos θ) \sqrt{\frac{1}{\cos θ}}} d θ = \frac{1}{\sqrt{2 k}} \int_{0}^{π / 3} \frac{\sin θ}{\sqrt{\cos θ}} d θ \end{aligned}$

Let $\cos θ = t \Rightarrow - \sin θ d θ = d t \Rightarrow \sin θ d θ = - d t$

for lower limit, $θ = 0 \Rightarrow t = \cos 0 = 1$

for upper limit, $θ = \frac{π}{3} \Rightarrow t = \cos \frac{π}{3} = \frac{1}{2}$

$\begin{array}{ll} \Rightarrow I & = \frac{1}{\sqrt{2 k}} \int_{1}^{1 / 2} \frac{- d t}{\sqrt{t}} = \frac{- 1}{\sqrt{2 k}} \int_{I}^{1 / 2} t^{- \frac{1}{2}} d t \\ = - \frac{1}{\sqrt{2 k}} \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} = - \frac{1}{\sqrt{2 k}} [2 \sqrt{t}]_{1}^{\frac{1}{2}} \\ = - \frac{2}{\sqrt{2 k}} \sqrt{\frac{1}{2}} - \sqrt{1} = \frac{2}{\sqrt{2 k}} 1 - \frac{1}{\sqrt{2}} \\ ∵ & I = 1 - \frac{1}{\sqrt{2}} \\ ∴ \frac{2}{\sqrt{2 k}} 1 - \frac{1}{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}} \Rightarrow \frac{2}{\sqrt{2 k}} = 1 \\ \Rightarrow 2 = \sqrt{2 k} \Rightarrow 2 k = 4 \Rightarrow k = 2 \end{array}$

Indefinite Integration 1 Question 13

14. If $\int_{0}^{π / 3} \frac{\tan θ}{\sqrt{2 k \sec θ}} d θ = 1 - \frac{1}{\sqrt{2}}, (k > 0)$ , then the value of $k$ is

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Indefinite Integration 1 Question 13

14. If ∫0π/3tan⁡θ2ksec⁡θdθ=1−12,(k>0), then the value of k is

Solution:

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14. If $\int_{0}^{π / 3} \frac{\tan θ}{\sqrt{2 k \sec θ}} d θ = 1 - \frac{1}{\sqrt{2}}, (k > 0)$ , then the value of $k$ is