Correct Answer - Option 3 :
\(\rm \frac{\pi}{4}\)
Concept:
Properties of definite integral:
\(\rm \int_{a}^{b}f(x)dx= \int_{a}^{b}f(a+b-x)dx\)
\(\rm \sin \left ( \frac{\pi}{2} -x\right )= \cos x\)
Calculation:
I = \(\rm \int_{0}^{\pi/2}\frac{ \sin x}{ \sin x + \cos x}dx\) ....(i)
Using identity , \(\rm \int_{a}^{b}f(x)dx= \int_{a}^{b}f(a+b-x)dx\)
I = \(\rm \int_{0}^{\pi/2}\frac{\sin \left ( \frac{\pi}{2} -x\right )}{\sin \left ( \frac{\pi}{2}-x \right )+\cos\left ( \frac{\pi}{2}-x \right )}\)dx
⇒ I = \(\rm \int_{0}^{\pi/2}\frac{ \cos x}{ \cos x + \sin x}dx\) ....(ii)
On adding eqn. (i) and (ii) ,we get
2I = \(\rm \int_{0}^{\pi/2}\frac{ \sin x+ \cos x}{ \sin x + \cos x}dx\)
⇒ 2I = \(\rm \int_{0}^{\pi/2}dx\)
⇒ 2I = \(\rm \frac{\pi}{2}\)
⇒ I = \(\rm \frac{\pi}{4}\) .
The correct option is 3.
