Question
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi\text{ d}\phi$
✓
Answer
Let $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi\text{ d}\phi$$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^4\phi\cos\phi\text{ d}\phi$
Also, let $\sin\phi=\text{t}\Rightarrow\cos\phi\text{ d}\phi=\text{dt}$ When, $\phi=0,\text{t}=0$ and when $\phi=\frac{\pi}{2},\text{t}=1$$\therefore\ \text{I}=\int_{0}^\limits{1}\sqrt{\text{t}}\big(1-\text{t}^2\big)^2\text{dt}$
$=\int_{0}^\limits{1}\text{t}^{\frac{1}{2}}\big(1+\text{t}^4-2\text{t}^2\big)\text{dt}$
$=\int_{0}^\limits{1}\Big[\text{t}^{\frac{1}{2}}+\text{t}^{\frac{9}{2}}-2\text{t}^{\frac{5}{2}}\Big]\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{\text{t}^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2\text{t}^{\frac{7}{2}}}{\frac{7}{2}}\Bigg]^1_0$
$=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$
$=\frac{64}{231}$
Also, let $\sin\phi=\text{t}\Rightarrow\cos\phi\text{ d}\phi=\text{dt}$ When, $\phi=0,\text{t}=0$ and when $\phi=\frac{\pi}{2},\text{t}=1$$\therefore\ \text{I}=\int_{0}^\limits{1}\sqrt{\text{t}}\big(1-\text{t}^2\big)^2\text{dt}$
$=\int_{0}^\limits{1}\text{t}^{\frac{1}{2}}\big(1+\text{t}^4-2\text{t}^2\big)\text{dt}$
$=\int_{0}^\limits{1}\Big[\text{t}^{\frac{1}{2}}+\text{t}^{\frac{9}{2}}-2\text{t}^{\frac{5}{2}}\Big]\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{\text{t}^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2\text{t}^{\frac{7}{2}}}{\frac{7}{2}}\Bigg]^1_0$
$=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$
$=\frac{64}{231}$
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