#### Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 32 maths

$\frac{\pi }{4}$

Hint:

To solve this equation we use $\int_{b}^{a} f(x) d x$   formula.

Given:

$\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x$

Solution:

Let

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x \ldots(i) \\\\ &\int_{0}^{\frac{\pi}{2}} f(x) d x=\int_{0}^{\frac{\pi}{2}} f\left(\frac{\pi}{2}-x\right) d x \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \ldots(i i) \end{aligned}

\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x+\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x} d x \\\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x+\cos x}[\cos x+\sin x] d x \end{aligned}

\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} 1 d x \\\\ &2 I=[x]_{0}^{\frac{\pi}{2}} \end{aligned}

\begin{aligned} &2 I=\left[\frac{\pi}{2}-0\right] \\\\ &=\frac{\pi}{2} \times \frac{1}{2} \\\\ &I=\frac{\pi}{4} \end{aligned}