Question

The value of $\int\limits_0^π \frac {e^{cos⁡\ x} sin⁡x}{(1+cos^2⁡x)(e^{cos⁡\ x}+e^{−cos⁡\ x})}dx$ 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)

$\frac {\pi}{4}$

Answer 2

$\int\limits_0^π \frac {e^{cos⁡\ x} sin⁡x}{(1+cos^2⁡x)(e^{cos⁡\ x}+e^{−cos⁡\ x})}dx$
Let $cos\ x = t$
$sin\ x\ dx = dt$
Then _, _$\int\limits_{-1}^1 \frac {−e^tdt}{(1+t^2)(e^t+e^{−t})}$
Let $I =\int\limits_{-1}^1 \frac {−e^tdt}{(1+t^2)(e^t+e^{−t})}$ …..…(1)
$I =\int\limits_{-1}^1 \frac {−e^tdt}{(1+t^2)(e^{−t}+e^t)}$ ….…(2)
On adding eq(1) and eq(2)
$2I = \int\limits_{-1}^1 \frac {dt}{1+t^2}$

$2I = tan−t]_{−1}^1$

$2I = \frac {\pi}{4}−(−\frac {\pi}{4})$

$2I = \frac {π}{2}$

$I = \frac {π}{4}$

So, the correct option is (C): $\frac {π}{4}$