Correct Answer - Option 1 :
\(\frac{\pi}{4}\)
Concept:
The property of integration is:
\(\rm \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx\)
Calculation:
Given: \(\rm \int_{0}^{\frac{\pi}{2}}\frac{sinx}{sinx+cosx}dx\)
Let \(\rm I= \int_{0}^{\frac{\pi}{2}}\frac{sinx}{sinx+cosx}\ dx ....(1)\)
As we know that, \(\rm \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx\)
\(\rm \Rightarrow I= \int_{0}^{\frac{\pi}{2}}\frac{sinx}{sinx+cosx}dx= \int_{0}^{\frac{\pi}{2}}\frac{cosx}{sinx+cosx}dx....(2)\)
By adding equation (1) and (2), we get
\(\rm \Rightarrow 2I= \int_{0}^{\frac{\pi}{2}}\frac{sinx}{sinx+cosx}dx+ \int_{0}^{\frac{\pi}{2}}\frac{cosx}{sinx+cosx}dx= \int_{0}^{\frac{\pi}{2}}\frac{sinx+cosx}{sinx+cosx}dx=\int_{0}^{\frac{\pi}{2}}dx\)
\(\rm \Rightarrow 2I=\int_{0 }^{\frac{\pi}{2}} dx=\left [ x \right ]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}\)
\(\rm \Rightarrow I=\frac{\pi}{4}\)
Hence, the correct option is 1.
