How do you find the integral of sin^3(x) cos^5(x) dx?

1 Answer
May 15, 2016

cos^8(x)/8-cos^6(x)/6+C

Explanation:

Recall that through the Pythagorean Identity sin^2(x)=1-cos^2(x).

Thus, sin^3(x)=sin(x)sin^2(x)=sin(x)(1-cos^2(x)). Substituting this into the integral we see:

intsin^3(x)cos^5(x)dx=intsin(x)(1-cos^2(x))cos^5(x)dx

Distributing just the cosines, this becomes

=int(cos^5(x)-cos^7(x))sin(x)dx

Now use the substitution: u=cos(x)" "=>" "du=-sin(x)dx

Noting that sin(x)dx=-du, the integral becomes:

=-int(u^5-u^7)du

Integrating, this becomes

=-(u^6/6-u^8/8)+C

Reordering and back-substituting with u=cos(x):

=cos^8(x)/8-cos^6(x)/6+C

Note that this integration could have also been done my modifying the cosines like:

cos^5(x)=cos(x)(cos^2(x))^2=cos(x)(1-sin^2(x))^2

And then proceeding by expanding and letting u=sin(x).