integral of x sin^2x dx

and would you show me the proceedure please?

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- Feb 16th 2009, 03:00 PMyasu7help with this integral
integral of x sin^2x dx

and would you show me the proceedure please?

- Feb 16th 2009, 03:06 PMMush
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 $ - Feb 16th 2009, 03:28 PMyasu7
I've got the answer, and it is like this:

$\displaystyle 1/8(2x^2-2xsin2x-cos2x) + c$

could someone help me get to this? - Feb 16th 2009, 03:32 PMMush
$\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 $ - Feb 16th 2009, 03:42 PMyasu7
I don't get where does the $\displaystyle \frac{1}{2}\int\frac {sin(2x)}{2}$ comes from,

is that a rule or something? thanks in advance - Feb 16th 2009, 03:47 PMMush
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 $