Question

Evaluate the integrals using substitution.
${\int }_0^{\frac{\pi }{2}}\sqrt{sin\varphi }co{s}^5\varphi d\varphi .$

Answer 1

${\int }_0^{\frac{\pi }{2}}\sqrt{sin\varphi }co{s}^5\varphi d\varphi .={\int }_0^{\frac{\pi }{2}}\sqrt{sin\varphi }(1-si{n}^2\varphi {)}^{2}cos\varphi d\varphi (\because co{s}^2\theta +si{n}^2\theta =1)$
Put sin $\varphi =t\Rightarrow cos\varphi =\frac{dt}{d\varphi }\Rightarrow d\varphi =\frac{dt}{cos\varphi }$
For limit when $\varphi =0\Rightarrow t=1\text{and when}\varphi =\frac{\pi }{2}\Rightarrow t=1$
$\therefore I={\int }_0^1\sqrt{t}(1-t^2{)}^{2}dt={\int }_0^1\sqrt{t}(t^4+1-2t^2)dt$

$={\int }_0^1(t^{\frac{9}{2}}+t^{\frac{1}{2}}-2t^{\frac{5}{2}})dt={[\frac{t^{\frac{9}{2}+1}}{(\frac{9}{2}+1)}+\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}-2\left(\frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right)]}_0^1=[\frac{2}{11}+\frac{2}{3}-\frac{4}{7}]-0=\frac{42+154-132}{11\times 3\times 7}=\frac{64}{231}$