How do you find the antiderivative of #int sin^3xcosxdx#?

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Answer 1 · 2016-11-23T17:23:09

#" "intsin^3xcosxdx" "=1/4sin^4x+C #

no need for substitution here if you recognise that

#y=sin^nx=>(dy)/(dx)=nsin^(n-1)xcosx" " #using the chain rule

so#" "intsin^3xcosxdx" "#suggests a function of the type

#y=sin^4x#

lets check this by differentiating.

#u=sinx=>(du)/(dx)=cosx#

#y=u^4=>(dy)/(dx)=4u^3#

#(dy)/(dx)=4u^3cosx=4sin^3xcosx#

#" "intsin^3xcosxdx" "=1/4sin^4x+C #

Answer 2 · 2016-11-23T17:36:49

#sin^4x/4+C#.

Since you have a cosine terms hanging around some sine terms, it might be helpful to try the substitution #u=sinx#, #du=cosxdx#.

Using this substitution, #intsin^3xcosxdx=intu^3du#.

#intu^3du=u^4/4+C=sin^4x/4+C#.