Question

How do you evaluate the integral of `(cos^2xsinx dx)` from 0 to 2pi?

Answer 1

`int _0^(2pi) cos^2x sinx dx`

First remember that `cos^2x = (cosx)^2`, so we have:

`int _0^(2pi) (cosx)^2 sinx dx`. Does that help?

I have something squared times, almost (but not quite) the derivative of that something.

Use substitution:

Let `u = cosx`, so that `du = - sinx dx`.

We have a choice now. We can integrate by `u` substituion and go back to `x`'x before putting in the limits of integration )`0` and `2 pi`),

or we can change the limits of integrations ahen we substitute and just do the new integral:

When `x = 0`, we get `u = cos 0 =1`

When `x=2 pi`, we get `u= cos 2 pi = 1`

`int _0^(2pi) cos^2x sinx dx = -int _1^1 u^2 du =0`