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Question Number 21721 by Isse last updated on 02/Oct/17

$${\int }_0^{\pi /2}{sin}^2{xcos}^{3}xdx$$

Answered by sma3l2996 last updated on 02/Oct/17

$$={\int }_0^{\pi /2}{sin}^{2}x(1-{sin}^{2}x)cosxdx={\int }_0^{\pi /2}({cosxsin}^{2}x-{cosxsin}^{4}x)dx$$$$={[\frac{1}{3}{sin}^{3}x-\frac{1}{5}{sin}^{5}x]}_0^{\pi /2}=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}$$

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