Question
By using the properties of definite integrals, evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}$

Answer

Given integral is: $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$ 
Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$ .....(i)
as $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right.$ 
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-\mathrm{x}\right)}{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-\mathrm{x}\right)+\cos ^{\frac{3}{2}}\left(\frac{\pi}{2}-\mathrm{x}\right)} \mathrm{dx}$ 
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{\frac{3}{2}} \mathrm{x}}{\cos ^{\frac{3}{2}} \mathrm{x}+\sin ^{\frac{3}{2}} \mathrm{x}} \mathrm{dx}$ ......(ii)
Adding (i) and (ii), we get
$2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$ 
$\Rightarrow 2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[1] \mathrm{d} \mathrm{x}$ 
$\Rightarrow 2 \mathrm{I}=[\mathrm{x}]_{0}^{\frac{\pi}{2}}$ 
$\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}-0$ 
$\Rightarrow 2 \mathrm{I}=\frac{\pi}{2}$ 
$\Rightarrow \mathrm{I}=\frac{\pi}{4}$

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