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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

$$3)\int {sin}^3\left(2x\right)cos\left(2x\right)dx$$$$t=sin2xdt=2cos2xdx$$$$=\frac{1}{2}\int t^{3}dt$$$$=\frac{1}{2}\times \frac{t^4}{4}+c$$$$=\frac{{\left(sin2x\right)}^4}{8}+c$$$$4)\int {cos}^5\left(2x\right){sin}^3\left(2x\right)dx$$$$t=cos2xdt=-2sin2xdx$$$$\int t^5\times (1-t^2)\times \frac{dt}{-2}$$$$=\frac{-1}{2}\int t^5-t^{7}dt$$$$=\frac{-1}{2}[\frac{t^6}{6}-\frac{t^8}{8}]+c$$$$=\frac{-1}{2}[\frac{{\left(cos2x\right)}^6}{6}-\frac{{\left(cos2x\right)}^8}{8}]+c$$

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