Question

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

Answer 1

`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)`.