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

Question

5082 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-23T00:06:30

#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#

Answer 2 · 2017-01-23T01:49:21

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

then

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