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Question Number 20390 by tammi last updated on 26/Aug/17

$$\int \sin {}^{5}xdx$$

Answered by ajfour last updated on 26/Aug/17

$$\int {(1-\cos {}^{2}x)}^2\sin xdx$$$$let\cos x=t\Rightarrow -\sin xdx=dt$$$$\int \sin {}^{5}xdx=-\int {(1-t^2)}^{2}dt$$$$=-\int (1-2t^2+t^4)dt$$$$=-t+\frac{2t^3}{3}-\frac{t^5}{5}+C$$$$=-\cos x+\frac{2}{3}\cos {}^{3}x-\frac{1}{5}\cos {}^{5}x+C.$$

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