integral of x sin^2x dx
and would you show me the proceedure please?
Parts!
$\displaystyle \sin^2(x) = \frac{1}{2} - \frac{1}{2}\cos(2x) $
Hence
$\displaystyle \int x \sin^2(x) dx = \int x\bigg( \frac{1}{2} - \frac{1}{2}\cos(2x) \bigg)dx = \frac{1}{2}\int x - x \cos(2x) dx $
The first term is easily integrated, and the 2nd term is integrated by parts.
$\displaystyle \int x \cos(2x) dx = \bigg(\frac{x \sin(2x)}{2}\bigg) - \int \frac{\sin(2x)}{2}dx $
$\displaystyle \sin^2(x) = \frac{1}{2} - \frac{1}{2}\cos(2x) $
Hence the integral becomes:
$\displaystyle \int x \sin^2(x) dx = \int x\bigg( \frac{1}{2} - \frac{1}{2}\cos(2x) \bigg)dx = \frac{1}{2}\int x - x \cos(2x) dx $.
This can be split into two integrals as such:
$\displaystyle = \frac{1}{2}\int x dx - \frac{1}{2}\int x \cos(2x) dx $
The first integral can be calculated easily using the rule that $\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1} +C $. The second integral must be evaluated by parts:
First Integral
$\displaystyle = \frac{1}{2}\bigg(\frac{x^2}{2}\bigg)$
Second Integral
$\displaystyle \frac{1}{2}\bigg(x \frac{\sin(2x)}{2}\bigg) - \frac{1}{2} \int \frac{\sin(2x)}{2}dx$
$\displaystyle = x \frac{\sin(2x)}{4} +\bigg(\frac{1}{2} \frac{\cos(2x)}{4}\bigg)dx$
$\displaystyle = x \frac{\sin(2x)}{4} +\frac{\cos(2x)}{8} +C$
Total Integral = First Integral - Second Integral
$\displaystyle = \frac{1}{2}\bigg(\frac{x^2}{2}\bigg) - \bigg(x \frac{\sin(2x)}{4} +\frac{\cos(2x)}{8} + C\bigg) $
$\displaystyle = \frac{x^2}{4} - x \frac{\sin(2x)}{4} - \frac{\cos(2x)}{8} + C $
Take out a factor of $\displaystyle \frac{1}{8} $
$\displaystyle = \frac{1}{8} \bigg(2x^2 - 2x\sin(2x) -cos(2x)\bigg) + C $
Yes, it's called integration by parts.
$\displaystyle \int f(x)g'(x) = g(x)f(x) - \int g(x)f'(x) dx $
In your case $\displaystyle f(x) = x $ and $\displaystyle g'(x) = \cos(2x) $
So $\displaystyle f'(x) = 1 $ and $\displaystyle g(x) = \int g'(x)dx = \int(\cos(2x))dx = \frac{\sin(2x)}{2} $
Put it all together and you get:
$\displaystyle \int x \times \cos(2x) dx = \bigg(\frac{\sin(2x)}{2} \times x \bigg) - \int \bigg(\frac{\sin(2x)}{2} \times 1\bigg) dx $