Question #103230
Integrate sin³x dx using reduction method
1
Expert's answer
2020-02-20T08:34:27-0500

sin3xdx=sin2xsinxdx=replacecosx=td(cosx)=dtsinxdx=dtsin2x=1cos2x=(1t2)dt=(tt33+C)==(cosxcos3x3+C)\int sin^3x dx=\int sin^2 x\cdot sinx dx=\\ replace\\ cos x=t\\ d(cosx)=dt\\ -sinxdx=dt\\ sin^2x=1-cos^2x\\ =-\int(1-t^2)dt=-(t-\frac{t^3}{3}+C)=\\ =-(cosx-\frac{cos^3x}{3}+C)


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