Question

By using properties of definite integrals, evaluate the integrals
${\int }_0^{\frac{\pi }{2}}\frac{si{n}^{\frac{3}{2}}}{si{n}^{\frac{3}{2}}x+co{s}^{\frac{3}{2}}x}dx.$

Answer 1

${\int }_0^{\frac{\pi }{2}}\frac{si{n}^{\frac{3}{2}}}{si{n}^{\frac{3}{2}}x+co{s}^{\frac{3}{2}}x}dx.\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{si{n}^{\frac{3}{2}}(\frac{\pi }{2}-x)}{si{n}^{\frac{3}{2}}(\frac{\pi }{2}-x)+co{s}^{\frac{3}{2}}(\frac{\pi }{2}-x)}dx(\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right)={\int }_0^{\frac{\pi }{2}}\frac{co{s}^{\frac{3}{2}}x}{co{s}^{\frac{3}{2}}x+si{n}^{\frac{3}{2}}X}[\because sin(\frac{\pi }{1}-X)=\text{cos x and cos}(\frac{pi}{2}-x)=sinx]$
on adding Eqs. (i) and (ii), we get
$2I={\int }_0^{\frac{\pi }{2}}\frac{si{n}^{\frac{3}{2}+}co{s}^{\frac{3}{2}}}{si{n}^{\frac{3}{2}}+co{s}^{\frac{3}{2}}x}dx={\int }_0^{\frac{\pi }{2}}1dx=[x{]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}-0\Rightarrow I=\frac{\pi }{4}$