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Thread: help with this integral

  1. #1
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    help with this integral

    integral of x sin^2x dx

    and would you show me the proceedure please?
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  2. #2
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    Quote Originally Posted by yasu7 View Post
    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 $
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  3. #3
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    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?
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  4. #4
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    Quote Originally Posted by yasu7 View Post
    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?
    $\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 $
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  5. #5
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    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
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  6. #6
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    Quote Originally Posted by yasu7 View Post
    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
    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 $
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