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, $intsin^3xcosxdx=intu^3du$ .

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

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