Question:

Evaluate the integral: \(\int_{0}^{π/2} \sqrt{\sin \phi}\cos^5 \phi\, d\,\phi\)

Updated On: Oct 7, 2023
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Solution and Explanation

Let \(I\)=\(\int_{0}^{π/2} \sqrt{\sin \phi}\cos^5 \phi\, d\,\phi\)=\(\int_{0}^{π/2} \sqrt{\sin \phi}\cos^4 \phi\, d\,\phi\)

Also, let \(\sin \phi=t\Rightarrow \cos \phi d\phi=dt\)

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

∴ \(I=\int^1_0\sqrt{t}(1-t^2)dt\)

\(I=\int^1_0t^{\frac{1}{2}}(1+t^4-2t^2)dt\)

=\(I=\int^1_0\bigg[t^{\frac{1}{2}}+t^{\frac{9}{2}}-2t^{\frac{5}{2}}\bigg]dt\)

=\(\bigg[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2t^{\frac{7}{2}}}{\frac{7}{2}}\bigg]^1_0\)

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

=\(\frac{154+42-132}{231}\)

=\(\frac{64}{231}\)

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