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Question Number 20389 by tammi last updated on 26/Aug/17

$$\int \sin {}^{4}xdx$$

Answered by Joel577 last updated on 26/Aug/17

$$I=\int {\left(\frac{1-\cos 2x}{2}\right)}^{2}dx$$$$=\frac{1}{4}\int (1-2\cos 2x+{\cos }^{2}2x)dx$$$$=\frac{1}{4}\int (1-2\cos 2x+\frac{1+\cos 4x}{2})dx$$$$=\frac{1}{4}\int (\frac{3}{2}-2\cos 2x+\frac{1}{2}\cos 4x)dx$$$$=\frac{1}{4}(\frac{3}{2}x-\sin 2x+\frac{1}{8}\sin 4x)+C$$

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