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Q.
The value of the definite integral $\int\limits_0^{\frac{\pi}{2}} \frac{\sin 3 x}{\sin x+\cos x} d x$ is equal to
Integrals
Solution:
$I=\int\limits_0^{\frac{\pi}{2}} \frac{\sin 3 x-\sin x}{\sin x+\cos x} d x+\int\limits_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x$
$I = I _1+ I _2$
Now king and add in $I _2 \Rightarrow 2 I _2=\frac{\pi}{2} \Rightarrow I _2=\frac{\pi}{4}$
$I_1=\int\limits_0^{\frac{\pi}{2}} \frac{2(\cos 2 x) \sin x}{\sin x+\cos x} d x=2 \int\limits_0^{\frac{\pi}{2}} \sin x(\cos x-\sin x) d x=\int\limits_0^{\frac{\pi}{2}} \sin 2 x d x-2 \int\limits_0^{\frac{\pi}{2}} \sin ^2 x d x $
$\left.\therefore I_1=\frac{-1}{2} \cos 2 x\right]_0^{\frac{\pi}{2}}-2 \cdot \frac{\pi}{4}=1-\frac{\pi}{2}$
$\therefore I = I _1+ I _2=1-\frac{\pi}{2}+\frac{\pi}{4}$