Q249 / 270 · NDA 2016 (Phase 2) · Mathematics
If $\int_0^{\pi / 2} \frac{\mathrm{dx}}{3 \cos \mathrm{x}+5}=\mathrm{k}^{-1} \cot ^{-1} 2$, then what is the value of K ?
- A$1 / 4$
- Correct answerB$1 / 2$
- C$1$
- D$2$
Solution
$\begin{aligned} & I=\int_0^{\pi / 2} \frac{d x}{3 \cos x+5} \\ & I=\int_0^{\pi / 2} \frac{d x}{3\left[\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right]+5}\end{aligned}$
$\begin{aligned}
& I=\int_0^{\pi / 2} \frac{\left(1+\tan ^2 \frac{x}{2}\right) d x}{3-3 \tan ^2 \frac{x}{2}+5+5 \tan ^2 \frac{x}{2}} \\
& I=\int_0^{\pi / 2} \frac{\sec ^2 \frac{x}{2} d x}{2 \tan ^2 \frac{x}{2}+8} \\
& I=\frac{1}{2} \int_0^{\pi / 2} \frac{\sec ^2 \frac{x}{2} d x}{\tan ^2 \frac{x}{2}+2^2}
\end{aligned}$
Put $\tan \frac{x}{2}=y$
$\begin{aligned}
& \Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d y \\
& \Rightarrow I=\int_0^1 \frac{d y}{y^2+2^2} \\
& \Rightarrow I=\frac{1}{2} \tan ^{-1}\left(\frac{y}{2}\right) \\
& \Rightarrow I=\frac{1}{2} \tan ^{-1}\left(\frac{1}{2}\right)-0
\end{aligned}$
Also $I=\frac{1}{2} \tan ^{-1} \frac{1}{2}=k \cot ^{-1}(2)$
$\begin{aligned}
& \quad\left(\because \tan ^{-1}(x)=\cot ^{-1}\left(\frac{1}{x}\right)\right) \\
& \Rightarrow \frac{1}{2} \tan ^{-1}\left(\frac{1}{2}\right)=k \tan ^{-1}\left(\frac{1}{2}\right) \\
& \therefore k=\frac{1}{2}
\end{aligned}$
Hence, the correct answer is option: 'B'.