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

1 Answer
Jul 13, 2016

#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)#