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Question-18281




Question Number 18281 by aplus last updated on 17/Jul/17
Answered by Tinkutara last updated on 18/Jul/17
(1/2)∫2 sin x cos x dx = (1/2)∫sin 2x dx  = (1/2)(((−cos 2x)/2)) = ((−cos 2x)/4) + C
$$\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{2}\:\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:\mathrm{2}{x}\:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}\right)\:=\:\frac{−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{4}}\:+\:{C} \\ $$
Answered by mrW1 last updated on 18/Jul/17
=∫sin x d(sin x)=(1/2)sin^2  x+C
$$=\int\mathrm{sin}\:\mathrm{x}\:\mathrm{d}\left(\mathrm{sin}\:\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \:\mathrm{x}+\mathrm{C} \\ $$

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