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Thread: Integration by substitution and part

  1. #1
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    Integration by substitution and part

    okay i got two question which i have real problems with. Any help would be great cheers guys

    a)intergrate this using the substitution u = x^2 + 1

    I = ∫2x(x^2 + 1)^3 dx

    b) intergrate this using parts

    J = ∫3xsinx dx

    any help would be brilliant cheers guys
    Last edited by mxmadman_44; Aug 17th 2007 at 08:38 AM. Reason: typo
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mxmadman_44 View Post
    okay i got two question which i have real problems with. Any help would be great cheers guys

    a)intergrate this using the substitution u = x^2 + 1

    I = ∫2x(x^2 + 1)^3 dx
    Well... $\displaystyle u = x^2 + 1 \implies du = 2xdx$

    So
    $\displaystyle I = \int (x^2 + 1)^3 \cdot 2x dx = \int u^3 du$

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mxmadman_44 View Post
    b) intergrate this using parts

    J = ∫3xsinx dx

    any help would be brilliant cheers guys
    $\displaystyle I = 3\int x sin(x) dx$

    Let
    $\displaystyle u = x \implies du = dx$
    $\displaystyle dv = sin(x) dx \implies v = -cos(x)$

    So
    $\displaystyle I = 3\int x sin(x) dx = 3 \left ( -xcos(x) + \int cos(x) dx \right ) $

    -Dan
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  4. #4
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    Quote Originally Posted by mxmadman_44 View Post
    b) intergrate this using parts

    J = ∫3xsinx dx
    Here's another solution (I'm not gonna take the constant):

    $\displaystyle
    \begin{aligned}
    \int {x\sin x\,dx} &= \int {\left( {\cos x - \cos x + x\sin x} \right)\,dx}\\
    &= \int {\left[ {\cos x - \left( {\cos x - x\sin x} \right)} \right]\,dx}\\
    &= \int {\left( {\sin x - x\cos x} \right)'\,dx}\\
    &= \sin x - x\cos x + k
    \end{aligned}
    $
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