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Question Number 18281 by aplus last updated on 17/Jul/17

Answered by Tinkutara last updated on 18/Jul/17

$$\frac{1}{2}\int 2\sin x\cos xdx=\frac{1}{2}\int \sin 2xdx$$$$=\frac{1}{2}\left(\frac{-\cos 2x}{2}\right)=\frac{-\cos 2x}{4}+C$$

Answered by mrW1 last updated on 18/Jul/17

$$=\int \sin \mathrm{x}\mathrm{d}\left(\sin \mathrm{x}\right)=\frac{1}{2}{\sin }^2\mathrm{x}+\mathrm{C}$$

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