Question

If ${\int }_0^{\pi /2}\frac{dx}{3\cos x+5}=k{\cot }^{-1}2$, then what is the value of k?

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

Answer: (B)

$\frac{1}{2}$

$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{3\cos x+5}$

$={\int }_0^{\frac{\pi }{2}}\frac{dx}{2\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)x+5}$

$={\int }_0^{\frac{\pi }{2}}\frac{(1+{\tan }^2\frac{x}{2})dx}{2{\tan }^2\frac{x}{2}+8}$

$={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\frac{x}{2}dx}{2{\tan }^2\frac{x}{2}+8}$

Put $\tan \frac{x}{2}=t$

$\therefore {\sec }^{2}xdx=2dt$

$\therefore I={\int }_0^1\frac{2dt}{2t^2+8}$

$=\frac{1}{2}{\int }_0^1\frac{2dt}{t^2+4}$

$=\frac{1}{2}{\left[ta{n}^{-1}\left(\frac{t}{2}\right)\right]}_0^1$

$=\frac{1}{2}[ta{n}^{-1}\left(\frac{1}{2}\right)-ta{n}^{-1}\left(\frac{0}{2}\right)]$

$=\frac{1}{2}co{t}^{-1}\left(2\right)$

$\therefore k=\frac{1}{2}$

Answer is option (B)