$\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}$
Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$ ...(1)
$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)}{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)+\cos ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)} d x$ $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$
$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$ ...(2)
Adding (1) and (2), we obtain
$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$
$\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} 1 \cdot d x$
$\Rightarrow 2 I=[x]_{0}^{\frac{\pi}{2}}$
$\Rightarrow 2 I=\frac{\pi}{2}$
$\Rightarrow I=\frac{\pi}{4}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.