Question

What is the integral of `cos^2[x] sin[x]`?

Answer 1

`-cos^3(x)/3+C`

Explanation:

We want to find:

`intcos^2(x)sin(x)dx`

Remember that the derivative of sine and cosine are basically one another. Since we only have one sine function, it will serve very well as the derivative of the cosine function when we substitute.

Let `u=cos(x)`. This implies that `du=-sin(x)dx`. Note that `sin(x)dx=-du`, which is what we have here. Also note that `cos^2(x)=u^2`.

Thus:

`intunderbrace(cos^2(x))_(u^2)overbrace(sin(x)dx)^(-du)=-intu^2du`

Then using the common rule: `intu^ndu=u^(n+1)/(n+1)+C`, where `n!=-1`, we see that:

`-intu^2du=-u^3/3+C=-cos^3(x)/3+C`