Question

The value of $\underset{0}{\overset{\pi }{\int }}\frac{e^{\cos x}\sin x}{\left(1+{\cos }^{2}x\right)\left(e^{\cos x}+e^{-\cos x}\right)}dx$ is equal to

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

Answer 1

Answer: (C)

$\underset{0}{\overset{\pi }{\int }}\frac{e^{\cos x}\sin x}{\left(1+{\cos }^{2}x\right)\left(e^{\cos x}+e^{-\cos x}\right)}dx....$(1)
Use King's property
$I=\underset{0}{\overset{\pi }{\int }}\frac{e^{-\cos x}\sin x}{\left(1+{\cos }^{2}x\right)\left(e^{-\cos x}+e^{\cos x}\right)}dx.....$(2)
On adding equation (1) and (2), we get
$2I=\underset{0}{\overset{\pi }{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx=2\underset{0}{\overset{\pi /2}{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx$
On putting $\cos x=t$, we get
$I=\underset{0}{\overset{1}{\int }}\frac{dt}{1+t^2}={\left({\tan }^{-1}t\right)}_0^1=\frac{\pi }{4}$