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Math Help - integration by parts

  1. #1
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    integration by parts

    Evaluate the integral.

    i dont understand how this problem will end up finishing itself using integration by parts i guess i just dont understand it!
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  2. #2
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    Quote Originally Posted by ahawk1 View Post
    Evaluate the integral.

    i dont understand how this problem will end up finishing itself using integration by parts i guess i just dont understand it!
    u = sin 2x v = (1/5) e^5x
    du = 2 cos 2x dx dv = e^5x dx

    {sin 2x e^5x dx = (1/5) e^5x (sin 2x) - 2/5 {e^5x cos 2x dx

    Important note: whenever you're stuck with another "integration by parts" expression in the end, always try to use similar u, du, v, and dv identities. In this case, the u/du side appears to be trigonometric and the v/dv sides appears to be e-related. So...

    u = cos 2x v = (1/5)e^5x
    du = -2 sin 2x dv = e^5x

    {cos 2x e^5x dx = (1/5)e^5x cos 2x + 2/5{e^5x sin 2x dx

    So, returning to the original integration by parts expression...

    {sin 2x e^5x dx = (1/5) e^5x (sin 2x) - 2/5 {e^5x cos 2x dx

    Substitution gives us...

    {sin 2x e^5x dx = (1/5) e^5x (sin 2x) - 2/5 [(1/5)e^5x cos 2x + 2/5 {e^5x sin 2x dx]

    {sin 2x e^5x dx = (1/5) e^5x (sin 2x) - (2/25)e^5x cos 2x - 4/25{e^5x sin 2x dx

    Adding the same integral to both sides....

    (29/25){sin 2x e^5x dx = (1/5) e^5x (sin 2x) - (2/25)e^5x cos 2x

    {sin 2x e^5x dx = (25/145)e^5x (sin2x) - (50/725)e^5x cos 2x

    e^5x [(5/29)sin 2x - (2/29)cos 2x] + C
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  3. #3
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    Quote Originally Posted by ahawk1 View Post
    Evaluate the integral.

    i dont understand how this problem will end up finishing itself using integration by parts i guess i just dont understand it!
    Strong hint:

    One way is to apply integration by parts twice. The first time you end up with an integral involving e^{5 \theta} \cos(2 \theta).

    Apply integration by parts to this second integral and you now have an integral involving e^{5 \theta} \sin(2 \theta).

    It looks like you are in an infinite loop. But no! You have an equation involving \int e^{5 \theta} \sin(2 \theta) which you can solve for the value of the integral.
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