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Math Help - Help with integration by parts

  1. #1
    Junior Member TheMathTham's Avatar
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    Help with integration by parts

    So I have two integration by parts problems that I am having trouble with. For both of these problems, I have the answer but I cannot figure out how to get there.

    The first one is \int x^2 cos(x)dx. First off, I made u = x^2 and dv = cos(x)dx. Next step, I found that du = 2xdx and v = sin(x). So going off of u*v - \int v*du, I got x^2 sin(x) - \int sin(x)*2xdx. Simplifying that down further leaves me with x^2 sin(x) + 2cos(x). However, the answer is (x^2 - 2) sin(x) + 2cos(x). What did I mess up?

    And the second one is \int x^2 \sqrt{4x-1}. I made u = x^2 and dv = \sqrt{4x-1}. So that meant that du = 2xdx and v = 1/6(4x-1)^3/2. The next step left me with \frac{1}{6} x^2 (4x-1)^\frac{3}{2} - \int \frac{1}{6}(4x-1)^\frac{3}{2} * 2x . Simplifying that further left me with \frac{1}{6} x^2 (4x-1)^\frac{3}{2} - \frac{2}{15}x(4x-1)^\frac{5}{2} . But, the correct answer is \frac{x^2(4x-1)^\frac{3}{2}}{14} + \frac{x(4x-1)^\frac{3}{2}}{70} + \frac{(4x-1)^\frac{3}{2}}{420}. Again, where did I mess up so badly?
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  2. #2
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    pickslides's Avatar
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    For each question did you intergrate by parts once or twice?
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  3. #3
    Junior Member TheMathTham's Avatar
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    I did once for both of them. If it's twice? What do I separate in the equation? I'm left with a simple square root for one and a trig function for the other, both of which I thought, do not require integration by parts.
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  4. #4
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    Quote Originally Posted by TheMathTham View Post
    So I have two integration by parts problems that I am having trouble with. For both of these problems, I have the answer but I cannot figure out how to get there.

    The first one is \int x^2 cos(x)dx. First off, I made u = x^2 and dv = cos(x)dx. Next step, I found that du = 2xdx and v = sin(x). So going off of u*v - \int v*du, I got x^2 sin(x) - \int sin(x)*2xdx. Simplifying that down further leaves me with x^2 sin(x) + 2cos(x). However, the answer is (x^2 - 2) sin(x) + 2cos(x). What did I mess up?
    On the first one, you got this far correctly.

    \displaystyle \int x^2 \cos(x)dx=x^2 \sin(x) - 2\!\! \int x\sin(x)dx
    (I rearranged the integral on the right a little.)

    Now use integration by parts (again) to evaluate \displaystyle \int x\sin(x)dx .

    Use u=x and dv=\sin(x)dx.
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  5. #5
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    Quote Originally Posted by TheMathTham View Post
    I did once for both of them. If it's twice?
    You integrate twice so x^2 can be differentiated twice and become constant.
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