Question

By using the properties of definite integrals, evaluate the integral ${\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}xdx}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}xdx}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx$ .................(1)
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}(\frac{\pi }{2}-x)}{{\sin }^{\frac{3}{2}}(\frac{\pi }{2}-x)+{\cos }^{\frac{3}{2}}(\frac{\pi }{2}-x)}dx,(\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx$ ...........(2)
Adding (1) and (2), we obtain
$2I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx$
$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}1\cdot dx$
$\Rightarrow 2I=[x{]}_0^{\frac{\pi }{2}}$
$\Rightarrow 2I=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}$