Question

If $I={\int }_0^{\pi /2}\frac{dx}{5+3\sin x}=\lambda {\tan }^{-1}\left(\frac{1}{2}\right)$ then
value of $\lambda$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{1}{2}$

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

Substitute $\tan \left(x/2\right)=t,$ so that

$dx=\frac{2dt}{1+t^2}$ and $5+3\cos x=\frac{8+2t^2}{1+t^2}$

$\therefore I={\int }_0^1\frac{dt}{4+t^2}=\frac{1}{2}{\tan }^{-1}{\left(\frac{t}{2}\right)]}_0^1$

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

Thus, $\lambda =1/2$