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# Show that

Question:

$\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}$

Solution:

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

$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$

$\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_{0}^{\frac{\pi}{2}}$

$\Rightarrow 2 I=\frac{\pi}{2}$

$\Rightarrow I=\frac{\pi}{4}$