Question

How do you find the antiderivative of `((cosx)^2)*(sinx)`?

Answer 1

`int cos^2(x) sin(x) dx = -(cos^3(x))/(3) + C`

Explanation:

In order to evaluate this integral, we'd have to perform a u-substitution.

Let's write out our integral first, before we proceed:

`int cos^2(x) sin(x) dx`

When performing a `u`-substitution, the goal in mind is to find a factor which is the derivative of, namely `du`. In this case, we can let

`u = cos(x)`, so then `du = -sin(x) dx`, or simply `-du = sin(x) dx`

The reason why we've chosen `u = cos(x)` and not `u = cos^2(x)` is because when integrating, we can simply integrate `u` itself, whatever power it has, giving us

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

Since integration and derivation are related by the Fundamental Theorem of Calculus, we can even check our answer by taking a derivative.

Checking our answer:

`d/dx[-1/3 * cos^3(x)] =-cos^2(x) * -sin(x) = cos^2(x) sin(x)`