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

The Correct Option is C

\(\int\limits_0^π \frac {e^{cos⁡\ x} sin⁡x}{(1+cos^2⁡x)(e^{cos⁡\ x}+e^{−cos⁡\ x})}dx\)
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): \(\frac {π}{4}\)