Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 21.

Answer: $\frac{\pi}{2}$

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: $\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sin x +\cos x}dx$

Solution: $\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sin x +\cos x}dx$

\begin{aligned} &\text { Put } \sin x=k(\sin x+\cos x)+L \frac{d}{d x}(\sin x+\cos x) \; \; \; \; \; \; \; \; \; \quad\left[\frac{d}{d x} \sin x=\cos x, \frac{d}{d x} \cos x=-\sin x\right] \\ &\sin x=k \sin x+k \cos x+L(\cos x-\sin x) \\ &\sin x=k \sin x+k \cos x+L \cos x-L \sin x \end{aligned}

$\sin x(k-L)\sin x+(k+L)\cos x$

Equating coefficient of sinx  and cosx  respectively, we get

$k-L=1............(i)$

$k+L=1............(ii)$

Solving equation (i)  and (ii)

k-L=1

k+L=0

2k=1

k=12

ii=k+L=0

12+L=0

L=-12

\begin{aligned} &\therefore \sin x=\frac{1}{2}(\sin x+\cos x)-\frac{1}{2}(\cos x-\sin x) \\ &I=\int_{0}^{\pi} \frac{\sin x}{\sin x+\cos x} d x \\ &=\int_{0}^{\pi \frac{1}{2}(\sin x+\cos x)-\frac{1}{2}(\cos x-\sin x)}{\sin x+\cos x} d x \\ &=\int_{0}^{\pi \frac{1}{2}(\sin x+\cos x)-\frac{1}{2}(\cos x-\sin x)}{\sin x+\cos x} d x \\ &=\frac{1}{2} \int_{0}^{\pi}\left[\frac{(\sin x+\cos x)}{(\sin x+\cos x)}-\frac{(\cos x-\sin x)}{(\sin x+\cos x)}\right] d x \\ &=\frac{1}{2} \int_{0}^{\pi} 1 d x-\frac{1}{2} \int_{0}^{\pi} \frac{(\cos x-\sin x)}{(\sin x+\cos x)} d x \end{aligned}

Put $\sin x+\cos x=t$

\begin{aligned} &(\cos x-\sin x) d x=d t \\ &I=\int_{0}^{\pi} \frac{1}{2} d x-\frac{1}{2} \int_{0}^{\pi} \frac{d t}{t} \\ &=\frac{1}{2}\left[\frac{x^{0+1}}{0+1}\right]_{0}^{\pi}-\frac{1}{2}[\log |t|]_{0}^{\pi} \\ &=\frac{1}{2}[x]_{0}^{\pi}-\frac{1}{2}[\log |\sin x+\cos x|]_{0}^{\pi} \end{aligned}

\begin{aligned} &=\frac{1}{2}[\pi-0]-\frac{1}{2}[\log |\sin \pi+\cos \pi|-\log |\sin 0+\cos 0|] \\ &=\frac{1}{2} \pi-\frac{1}{2}[\log |0+(-1)|-\log |0+1|] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\cos \pi=-1, \cos 0=1, \sin \pi=\sin 0=0] \end{aligned}

\begin{aligned} &=\frac{1}{2} \pi-\frac{1}{2}[\log |-1|-\log |1|] \\ &=\frac{1}{2} \pi-\frac{1}{2}[\log |1|-\log |1|] \quad[\log 1=0] \\ &=\frac{1}{2} \pi-0 \\ &=\frac{\pi}{2} \end{aligned}