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Question Number 97892 by pranesh last updated on 10/Jun/20

Answered by john santu last updated on 10/Jun/20

$$\sin {}^2\mathrm{x}=\frac{1}{2}-\frac{1}{2}\cos 2\mathrm{x}$$$$\sin {}^4\mathrm{x}=\frac{1}{4}-\frac{1}{2}\cos 2\mathrm{x}+\frac{1}{8}+\frac{1}{8}\cos 4\mathrm{x}$$$$=\frac{3}{8}-\frac{1}{2}\cos 2\mathrm{x}+\frac{1}{8}\cos 4\mathrm{x}$$$$\int \sin {}^4\mathrm{x}\mathrm{dx}=\frac{3}{8}\mathrm{x}-\frac{1}{4}\sin 2\mathrm{x}+\frac{1}{32}\sin 4\mathrm{x}+\mathrm{c}$$

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