# Show that

Question:

$\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi$

Solution:

Let $I=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{4} \phi \cos \phi d \phi$

Also, let $\sin \phi=t \Rightarrow \cos \phi d \phi=d t$

When $\phi=0, t=0$ and when $\phi=\frac{\pi}{2}, t=1$

$\therefore I=\int_{0}^{t} \sqrt{t}\left(1-t^{2}\right)^{2} d t$

$=\int_{0}^{1} t^{\frac{1}{2}}\left(1+t^{4}-2 t^{2}\right) d t$

$=\int_{0}^{1}\left[t^{\frac{1}{2}}+t^{\frac{9}{2}}-2 t^{\frac{5}{2}}\right] d t$

$=\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2 t^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{1}$

$=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$

$=\frac{154+42-132}{231}$

$=\frac{64}{231}$