integral of x sin ^2 (x) dx

[math]

integral of x sin ^2 (x) dx

attempt:

Let u = sin^2 (x) v' = x
u' = 2sinxcosx v = x^2 / 2

x^2/2 * sin^2 (x) - integral of x^2/2 * 2sinxcosx

 

[Count Iblis]

integral of x sin ^2 (x) dx

attempt:

Let u = sin^2 (x) v' = x
u' = 2sinxcosx v = x^2 / 2

x^2/2 * sin^2 (x) - integral of x^2/2 * 2sinxcosx

Use that sin^2(x) = 1/2 - 1/2 cos(2x)

You can then calculate te integral of x cos(2x) using partial integration, or using this trick: The integral of sin(px) is -1/p cos(px). If you differentiate this relation w.r.t. p, you find that the integral of x cos(px) is
1/p^2 cos(px) + x/p sin(px).

 

[galactus]

I have noticed you have been trying parts with these integrals. It's really not necessary with most of them.

For this one:

Using the identity, $\displaystyle sin^{2}(x)=\frac{1-cos(2x)}{2}$, rewrite as:

\(\displaystyle \L\\\int\left[\frac{1}{2}x-\frac{1}{2}xcos(2x)\right]dx=\int\frac{1}{2}xdx-\int\frac{1}{2}xcos(2x)dx\)