Question

Evaluate : $\int\limits^{\pi/2}_0 { \sqrt{\sin\phi}\cdotp\cos^{5}\phi} \, d\phi$

Answer 1

➡️➡️Hope it will help you‼️‼️

Answer 2

$\underline{\textsf{Solution:}}$

$\displaystyle \mathsf{Now,\:\int_{0}^{\pi/2} \sqrt{sin\phi}\:cos^{5}\phi\:d\phi}$

$\displaystyle \mathsf{=\int_{0}^{\pi/2} (sin\phi)^{1/2}\:cos^{4}\phi\:cos\phi\:d\phi}$

$\displaystyle \mathsf{=\int_{0}^{\pi/2} (sin\phi)^{1/2}\:(1-sin^{2}\phi)^{2}\:cos\phi\:d\phi}$

$\displaystyle \mathsf{=\int_{0}^{\pi/2} (sin\phi)^{1/2}\:(1-2\:sin^{2}\phi+sin^{4}\phi)\:cos\phi\:d\phi}$

$\displaystyle \mathsf{=\int_{0}^{\pi/2} (sin\phi)^{1/2}\:cos\phi\:d\phi-2\int_{0}^{\pi/2} (sin\phi)^{5/2}\:cos\phi\:d\phi+\int_{0}^{\pi/2}(sin\phi)^{9/2}\:cos\phi\:d\phi}$

$\displaystyle \mathsf{=\int_{0}^{\pi/2}(sin\phi)^{1/2}\:d(sin\phi)-2\int_{0}^{\pi/2}(sin\phi)^{5/2}\:d(sin\phi)+\int_{0}^{\pi/2}(sin\phi)^{9/2}\:d(sin\phi)}$

$\displaystyle \mathsf{=\frac{1}{1/2+1}[(sin\phi)^{3/2}]_{0}^{\pi/2}-\frac{2}{5/2+1}[(sin\phi)^{7/2}]_{0}^{\pi/2}+\frac{1}{9/2+1}[(sin\phi)^{11/2}]_{0}^{\pi/2}}$

$\displaystyle \mathsf{=\frac{2}{3}(1-0)-\frac{4}{7}(1-0)+\frac{2}{11}(1-0)}$

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

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

$\displaystyle \mathsf{=\frac{64}{231}}$ ,

$\textsf{which is the required value.}$