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Question Number 12702 by mad last updated on 29/Apr/17
find ∫cos^2 2x dx
$${find}\:\int{cos}^{\mathrm{2}} \mathrm{2}{x}\:{dx} \\ $$
Answered by Joel577 last updated on 29/Apr/17
cos^2  2x = ((1 + cos 4x)/2)  ∫ cos^2  2x dx  = (1/2)∫ (1 + cos 4x) dx  = (1/2)(∫ 1 dx + ∫ cos 4x dx)  = (1/2)(x + ((sin 4x)/4)) + C  = (1/2)x + ((sin 4x)/8) + C
$$\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}\:=\:\frac{\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{2}} \\ $$$$\int\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}\:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\left(\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{4}{x}\right)\:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\int\:\mathrm{1}\:{dx}\:+\:\int\:\mathrm{cos}\:\mathrm{4}{x}\:{dx}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}\:+\:\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{4}}\right)\:+\:{C} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:+\:\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{8}}\:+\:{C} \\ $$

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