Question

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

Answer 1

`int sin^{3}(x)\ dx=1/3 cos^{3}(x)-cos(x)+C`

Explanation:

Use the identity `sin^{2}(x)=1-cos^{2}(x)` and then let `u=cos(x)` so that `du = -sin(x)\ dx` and

`int sin^{3}(x)\ dx=int(1-cos^{2}(x))sin(x)\ dx=int (u^2-1)\ du`

`=u^3/3-u+C=1/3 cos^{3}(x)-cos(x)+C`