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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Definite Integration

NDA 2016 · Mathematics · Definite Integration

If ∫_0^ π / 2 dx 3 x+5 = ^ -1 ^ -1 2 , then what is the value of K ?

If ∫0π/2dx3cos⁡x+5=\mathrmk−1cot⁡−12\int_0^{\pi / 2} \frac{\mathrm{dx}}{3 \cos \mathrm{x}+5}=\mathrmk^{-1} \cot ^{-1} 2, then what is the value of K ?

Options

  1. A1/41 / 4
  2. B1/21 / 2
  3. C11
  4. D22

Correct answer

B. 1/21 / 2

Step-by-step solution

I=∫0π/2dx3cos⁡x+5I=∫0π/2dx3[1−tan⁡2x21+tan⁡2x2]+5\begin{aligned} & I=\int_0^{\pi / 2} \frac{d x}{3 \cos x+5} \\ & I=\int_0^{\pi / 2} \frac{d x}{3\left[\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right]+5}\end{aligned}I=∫0π/2(1+tan⁡2x2)dx3−3tan⁡2x2+5+5tan⁡2x2I=∫0π/2sec⁡2x2dx2tan⁡2x2+8I=12∫0π/2sec⁡2x2dxtan⁡2x2+22\begin{aligned} & I=\int_0^{\pi / 2} \frac{\left(1+\tan ^2 \frac{x}{2}\right) d x}{3-3 \tan ^2 \frac{x}{2}+5+5 \tan ^2 \frac{x}{2}} \\ & I=\int_0^{\pi / 2} \frac{\sec ^2 \frac{x}{2} d x}{2 \tan ^2 \frac{x}{2}+8} \\ & I=\frac{1}{2} \int_0^{\pi / 2} \frac{\sec ^2 \frac{x}{2} d x}{\tan ^2 \frac{x}{2}+2^2} \end{aligned}PutPut\tan \frac{x}{2}=y⇒12sec⁡2x2dx=dy⇒I=∫01dyy2+22⇒I=12tan⁡−1(y2)⇒I=12tan⁡−1(12)−0\begin{aligned} & \Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d y \\ & \Rightarrow I=\int_0^1 \frac{d y}{y^2+2^2} \\ & \Rightarrow I=\frac{1}{2} \tan ^{-1}\left(\frac{y}{2}\right) \\ & \Rightarrow I=\frac{1}{2} \tan ^{-1}\left(\frac{1}{2}\right)-0 \end{aligned}AlsoAlsoI=\frac{1}{2} \tan ^{-1} \frac{1}{2}=k \cot ^{-1}(2)(∵tan⁡−1(x)=cot⁡−1(1x))⇒12tan⁡−1(12)=ktan⁡−1(12)∴k=12\begin{aligned} & \quad\left(\because \tan ^{-1}(x)=\cot ^{-1}\left(\frac{1}{x}\right)\right) \\ & \Rightarrow \frac{1}{2} \tan ^{-1}\left(\frac{1}{2}\right)=k \tan ^{-1}\left(\frac{1}{2}\right) \\ & \therefore k=\frac{1}{2} \end{aligned}$
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