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Question Number 40 by user3 last updated on 25/Jan/15

Evaluate   ∫x cos^2 x dx.

$$\mathrm{Evaluate}\:\:\:\int{x}\:\mathrm{cos}\:^{\mathrm{2}} {x}\:{dx}. \\ $$

Answered by user3 last updated on 03/Nov/14

∫x cos^2 x dx =∫x(((1+cos 2x)/2))dx  =(1/2)∫x dx+(1/2)∫x cos 2x dx  =(x^2 /4)+(1/2)∙[x∙∫cos 2x dx−∫{(d/dx)(x)∙∫cos 2x dx}dx]  =(x^2 /4)+(1/2)[((x sin 2x)/2) −∫ ((sin 2x )/2)dx]  =(x^2 /4) + ((x sin 2x)/4) − (1/4)×(((−cos 2x))/2)+C  =(x^2 /4) + ((x sin 2x)/4) + ((cos 2x)/8)+C

$$\int{x}\:\mathrm{cos}^{\mathrm{2}} {x}\:{dx}\:=\int{x}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{x}\:{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int{x}\:\mathrm{cos}\:\mathrm{2}{x}\:{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\centerdot\left[{x}\centerdot\int\mathrm{cos}\:\mathrm{2}{x}\:{dx}−\int\left\{\frac{{d}}{{dx}}\left({x}\right)\centerdot\int\mathrm{cos}\:\mathrm{2}{x}\:{dx}\right\}{dx}\right] \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{x}\:\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}\:−\int\:\frac{\mathrm{sin}\:\mathrm{2}{x}\:}{\mathrm{2}}{dx}\right] \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:+\:\frac{{x}\:\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{4}}×\frac{\left(−\mathrm{cos}\:\mathrm{2}{x}\right)}{\mathrm{2}}+{C} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:+\:\frac{{x}\:\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{4}}\:+\:\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{8}}+{C} \\ $$

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