Question Number 48703 by cesar.marval.larez@gmail.com last updated on 27/Nov/18
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18
![3)∫sin^3 (2x)cos(2x)dx t=sin2x dt=2cos2xdx =(1/2)∫t^3 dt =(1/2)×(t^4 /4)+c =(((sin2x)^4 )/8)+c 4)∫cos^5 (2x)sin^3 (2x)dx t=cos2x dt=−2sin2xdx ∫t^5 ×(1−t^2 )×(dt/(−2)) =((−1)/2)∫t^5 −t^7 dt =((−1)/2)[(t^6 /6)−(t^8 /8)]+c =((−1)/2)[(((cos2x)^6 )/6)−(((cos2x)^8 )/8)]+c](Q48704.png)
$$\left.\mathrm{3}\right)\int{sin}^{\mathrm{3}} \left(\mathrm{2}{x}\right){cos}\left(\mathrm{2}{x}\right){dx} \\ $$$${t}={sin}\mathrm{2}{x}\:\:\:{dt}=\mathrm{2}{cos}\mathrm{2}{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{t}^{\mathrm{3}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{t}^{\mathrm{4}} }{\mathrm{4}}+{c} \\ $$$$=\frac{\left({sin}\mathrm{2}{x}\right)^{\mathrm{4}} }{\mathrm{8}}+{c} \\ $$$$\left.\mathrm{4}\right)\int{cos}^{\mathrm{5}} \left(\mathrm{2}{x}\right){sin}^{\mathrm{3}} \left(\mathrm{2}{x}\right){dx} \\ $$$${t}={cos}\mathrm{2}{x}\:\:\:{dt}=−\mathrm{2}{sin}\mathrm{2}{xdx} \\ $$$$\int{t}^{\mathrm{5}} ×\left(\mathrm{1}−{t}^{\mathrm{2}} \right)×\frac{{dt}}{−\mathrm{2}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\int{t}^{\mathrm{5}} −{t}^{\mathrm{7}} \:\:\:{dt} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\left[\frac{{t}^{\mathrm{6}} }{\mathrm{6}}−\frac{{t}^{\mathrm{8}} }{\mathrm{8}}\right]+{c} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\left[\frac{\left({cos}\mathrm{2}{x}\right)^{\mathrm{6}} }{\mathrm{6}}−\frac{\left({cos}\mathrm{2}{x}\right)^{\mathrm{8}} }{\mathrm{8}}\right]+{c} \\ $$