Question Number 44250 by pieroo last updated on 24/Sep/18
$$\mathrm{Evaliate}\:\int\mathrm{sin}^{\mathrm{4}} \mathrm{x}\:\mathrm{dx} \\ $$$$\mathrm{please}\:\mathrm{show}\:\mathrm{working}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Sep/18
![∫(((1−cos2x)/2))^2 dx (1/4)∫1−2cos2x+cos^2 2x dx (1/4)[∫dx−2∫cos2xdx+∫((1+cos4x)/2)dx] (1/4)[∫dx−2∫cos2xdx+(1/2)∫dx+(1/2)∫cos4x dx] (1/4)[x−2×((sin2x)/2)+(x/2)+(1/2)×((sin4x)/4)]+c (1/4)[((3x)/2)−sin2x+((sin4x)/8)]+c ((3x)/8)−((sin2x)/4)+((sin4x)/(32))+c](https://www.tinkutara.com/question/Q44252.png)
$$\int\left(\frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{2}}\right)^{\mathrm{2}} {dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{1}−\mathrm{2}{cos}\mathrm{2}{x}+{cos}^{\mathrm{2}} \mathrm{2}{x}\:{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left[\int{dx}−\mathrm{2}\int{cos}\mathrm{2}{xdx}+\int\frac{\mathrm{1}+{cos}\mathrm{4}{x}}{\mathrm{2}}{dx}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left[\int{dx}−\mathrm{2}\int{cos}\mathrm{2}{xdx}+\frac{\mathrm{1}}{\mathrm{2}}\int{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int{cos}\mathrm{4}{x}\:{dx}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left[{x}−\mathrm{2}×\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{{sin}\mathrm{4}{x}}{\mathrm{4}}\right]+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{3}{x}}{\mathrm{2}}−{sin}\mathrm{2}{x}+\frac{{sin}\mathrm{4}{x}}{\mathrm{8}}\right]+{c} \\ $$$$\frac{\mathrm{3}{x}}{\mathrm{8}}−\frac{{sin}\mathrm{2}{x}}{\mathrm{4}}+\frac{{sin}\mathrm{4}{x}}{\mathrm{32}}+{c} \\ $$
Commented by pieroo last updated on 24/Sep/18
$$\mathrm{Thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$