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Question Number 53051 by Abror last updated on 16/Jan/19

$$\int \sin \left(2x\right)\cos xd\left(x\right)=$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

$$\frac{1}{2}\int 2sin2xcosxdx$$$$=\frac{1}{2}\int (sin3x+sinx)dx$$$$=\frac{1}{2}[\frac{-cos3x}{3}+\frac{-cosx}{}]+c$$

Answered by kaivan.ahmadi last updated on 16/Jan/19

$$\int 2\mathrm{sinxcosx}.\mathrm{cosxdx}=2\int \mathrm{sinx}.{\cos }^2\mathrm{xdx}$$$$\mathrm{u}=\mathrm{cosx}\Rightarrow \mathrm{du}=-\mathrm{sinxdx}\Rightarrow$$$$-2\int {\mathrm{u}}^2\mathrm{du}=-2\frac{{\mathrm{u}}^3}{3}+\mathrm{C}=-\frac{2}{3}{\cos }^3\mathrm{x}+\mathrm{C}$$

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