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How do you integrate #int cos^3xsinxdx#?

Answer 1

Emilia Aguilar

#int cos^3x sinx dx = - cos^4x/4+C#

Note that:

#-sinx dx = d(cosx)#

so if you substitute #u= cosx# you have:

#int cos^3x sinx dx = - int u^3 du = - u^4/4 +C = - cos^4x/4+C#

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Answer 2

Ezra Adams

Since #d/dx(cos(x))=-sin(x)#

then

#intcos^3(x)sin(x)dx=-intcos^3(x)dcos(x)=-(cos^(4)(x))/4+C#

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Answer 3

Adam Adkins

To integrate ( \int \cos^3(x) \sin(x) , dx ), you can use the substitution method.

Let ( u = \cos(x) ), then ( du = -\sin(x) , dx ).

This means ( \sin(x) , dx = -du ).

Substituting ( u ) and ( du ) into the integral:

[ \int \cos^3(x) \sin(x) , dx = \int -u^3 , du ]

Now, integrate ( -u^3 ) with respect to ( u ):

[ \int -u^3 , du = -\frac{1}{4}u^4 + C ]

Finally, resubstitute ( \cos(x) ) for ( u ):

[ -\frac{1}{4}\cos^4(x) + C ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.