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

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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

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

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