Evaluate the integral _ 0 ^ 2 ^ 5 ~ d using substitution.

Question

Evaluate the integral $\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi~ d \phi$ using substitution.

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Answer

Given: $\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi$
Let $\mathrm{I}=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi \mathrm{d} \phi=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{4} \phi \cos \phi \mathrm{d} \phi$
= $\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi}\left(\cos ^{2} \phi\right)^{2} \cos \phi d \phi$
$\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi}\left(1-\sin ^{2} \phi\right)^{2} \cos \phi d \phi$
Also, let $\sin \phi=t \Rightarrow \cos \phi d \phi=d t$
when $\phi=0, \mathrm{t}=0$ and when $\phi=\frac{\pi}{2}, \mathrm{t}=1$
So, $I=\int_{0}^{1} \sqrt{t}\left(1-t^{2}\right)^{2} d t$
= $\int_{0}^{1} t^{\frac{1}{2}}\left(1+t^{4}-2 t^{2}\right) d t$
= $\int_{0}^{1}\left(t^{\frac{1}{2}}+t^{\frac{9}{2}}-2 t^{\frac{5}{2}}\right) d t$
= $\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2 t^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{1}$
= $\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$
= $\frac{154+42-132}{231}=\frac{64}{231}$

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