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

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Answer 1 · 2015-05-25T18:07:30

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

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 $u= cos 2 pi = 1$

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

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