Evaluate the integral in Exercise: ^ 2 _ 0 ^ 5 d

Question

Evaluate the integral in Exercise:
$\int\limits^{\frac{\pi}{2}}_{0}\sqrt{\sin\phi}\cos^{5}\phi\text{ d }\phi$

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Answer

$\text{Let}\ \text{I}=\int\limits^\frac{\pi}{2}_{0}\sqrt{\sin\phi}\cos^{5}\phi\ \text{d}\ \phi$
$\text{putting}\ \sin\phi=\text{t}\ \Rightarrow\ \cos\phi=\frac{\text{dt}}{\text{d}\phi}\ \Rightarrow\ \cos\phi\ \text{d}\ \phi=\text{dt}$
To change the limits of integration from $\phi \text{to}\ \text{t}$
$\text{when}\phi=0,\text{t}=\sin\phi=\sin0^{\text{o}}=0$
$\text{when}\phi=\frac{\pi}{2},\text{t}=\sin\phi=\sin\frac{\pi}{2}=1$
$\therefore$ From eq.(i),$\ \text{I}=\int\limits_{0}^\frac{\pi}{2}\sqrt{\sin}\phi\cos^{5}\phi\cos\phi\ \text{d}\ \phi=\int^\frac{\pi}{2}_{0}\sqrt{\sin\phi}(\cos^{2}\phi)^{2}\cos\phi\ \text{d}\ \phi$
$=\int\limits_{0}^\frac{\pi}{2}\sqrt{\sin}\phi(1-\sin^{2}\phi)^{2}\cos\phi\ \text{d}\ \phi=\int\limits_{0}^{1}\sqrt{\text{t}}(1-\text{t}^{2})^{2}\text{dt}=\int\limits_{0}^{1}\text{t}^\frac{1}{2}(1+\text{t}^{4}-2\text{t}^{2})\text{dt}$
$=\int\limits_{0}^{1}\bigg(\text{t}^\frac{1}{2}+\text{t}^{\frac{1}{2}+4}-2\text{t}^{\frac{1}{2}+2}\bigg)\text{dt}=\int\limits_{0}^{1}\bigg(\text{t}^\frac{1}{2}+\text{t}^\frac{9}{2}-2\text{t}^\frac{5}{2}\bigg)\text{dt}=\int\limits_{0}^{1}\text{t}^\frac{1}{2}\text{dt}+\int\limits_{0}^{1}\text{t}^\frac{9}{2}\text{dt}-2\int\limits_{0}^{1}\text{t}^\frac{5}{2}\text{dt}$
$=\frac{\bigg(\text{t}^\frac{3}{2}\bigg)^{1}_{0}}{\frac{3}{2}}+\frac{\bigg(\text{t}^\frac{11}{2}\bigg)^{1}_{0}}{\frac{11}{2}}+\frac{\bigg(\text{t}\frac{7}{2}\bigg)^{1}_{0}}{\frac{7}{2}}=\frac{2}{3}(1-0)+\frac{2}{11}(1-0)-\frac{4}{7}(1-0)=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$
$=\frac{154+42-132}{231}=\frac{64}{231}$

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