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Math Help - Integration

  1. #1
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    Integration

    1. Evaluate integral(e^(7x)cos(2x)dx)

    I let u =e^(7x) and so du = 7e^(7x)dx. dv is cos(2x) so V= 1/2sin(2x). I did the procedure, but i cant get the answer. Please show me some intermediate steps please!

    2. Evaluate I = integral(486cot(6x)/sin^2(6x)+18sin(6x)+81) dx

    No exactly sure how to go about this problem. I can factor the bottom into (sin(6x)+9)^2. I am using partial fraction decomposition, but the numerator seems to confuse me.

    Help is much appreciated!
    Thanks to all who contribute.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Belowzero78 View Post
    1. Evaluate integral(e^(7x)cos(2x)dx)

    I let u =e^(7x) and so du = 7e^(7x)dx. dv is cos(2x) so V= 1/2sin(2x). I did the procedure, but i cant get the answer. Please show me some intermediate steps please!
    Your work is good up to here.

    Now take note that -\tfrac{7}{2}\int e^{7x}\sin\!\left(2x\right)\,dx appears in your answer. Apply integration by parts again to get some constant multiple of \int e^{7x}\cos\!\left(2x\right)\,dx to reappear. At this stage, treat the integral as a variable, and solve the equation for the integral (and don't forget the +C at the end!).

    Does this make sense? Can you continue where you left off?

    2. Evaluate I = integral(486cot(6x)/sin^2(6x)+18sin(6x)+81) dx

    No exactly sure how to go about this problem. I can factor the bottom into (sin(6x)+9)^2. I am using partial fraction decomposition, but the numerator seems to confuse me.

    Help is much appreciated!
    Thanks to all who contribute.
    Note that \int\frac{486\cot\!\left(6x\right)}{\sin^2\!\left(  6x\right)+18\sin\!\left(6x\right)+81}\,dx=\int\fra  c{486\cos\!\left(6x\right)}{\sin\!\left(6x\right)\  left(\sin\!\left(6x\right)+9\right)^2}\,dx

    Now apply the substitution u=\sin\!\left(6x\right) and the reduced integral can be solved using partial fractions.

    Does this make sense? Can you try to take it from here?
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  3. #3
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    Quote Originally Posted by Belowzero78 View Post
    1. Evaluate integral(e^(7x)cos(2x)dx)

    I let u =e^(7x) and so du = 7e^(7x)dx. dv is cos(2x) so V= 1/2sin(2x). I did the procedure, but i cant get the answer. Please show me some intermediate steps please!

    2. Evaluate I = integral(486cot(6x)/sin^2(6x)+18sin(6x)+81) dx

    No exactly sure how to go about this problem. I can factor the bottom into (sin(6x)+9)^2. I am using partial fraction decomposition, but the numerator seems to confuse me.

    Help is much appreciated!
    Thanks to all who contribute.
    1. \int{e^{7x}\cos{2x}\,dx}

    Let u = e^{7x} so that du = 7e^{7x}.

    Let dv = \cos{2x} so that v = \frac{1}{2}\sin{2x}.


    So \int{e^{7x}\cos{2x}\,dx} = \frac{1}{2}e^{7x}\sin{2x} - \int{\frac{7}{2}e^{7x}\sin{2x}\,dx}

     = \frac{1}{2}e^{7x}\sin{2x} - \frac{7}{2}\int{e^{7x}\sin{2x}\,dx}.


    Now let u = e^{7x} so that du = 7e^{7x}

    Let dv = \sin{2x} so that v = -\frac{1}{2}\cos{2x}.


    So \int{e^{7x}\cos{2x}\,dx} = \frac{1}{2}e^{7x}\sin{2x} - \int{\frac{7}{2}e^{7x}\sin{2x}\,dx}

     = \frac{1}{2}e^{7x}\sin{2x} - \left[-\frac{1}{2}e^{7x}\cos{2x} - \int{-\frac{7}{2}e^{7x}\cos{2x}\,dx}\right]

     = \frac{1}{2}e^{7x}\sin{2x} + \frac{1}{2}e^{7x}\cos{2x} - \frac{7}{2}\int{e^{7x}\cos{2x}\,dx}.


    So \frac{9}{2}\int{e^{7x}\cos{2x}\,dx} = \frac{1}{2}e^{7x}\sin{2x} + \frac{1}{2}e^{7x}\cos{2x}

    \frac{9}{2}\int{e^{7x}\cos{2x}\,dx} = \frac{1}{2}e^{7x}\left(\sin{2x} + \cos{2x}\right)

    \int{e^{7x}\cos{2x}\,dx} = \frac{1}{9}e^{7x}\left(\sin{2x} + \cos{2x}\right) + C.
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  4. #4
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    Ok, that works for to the 2nd one, so i decomposed into fractions and i get A/u +b/(U+9) + C/(u+9)^2. How do i get them so i compare coefficient to solve for the constants? I also found the common denominator.
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  5. #5
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    prove It, the answer given is e^(7x)(7cos(2x)+2sin(2x))/53 +C
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  6. #6
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    Quote Originally Posted by Belowzero78 View Post
    prove It, the answer given is e^(7x)(7cos(2x)+2sin(2x))/53 +C
    Then go over my work and find my mistake.
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  7. #7
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    Im not understanding how they get the denominator as 53. The denominator im getting is 56, so my integral is = 2/7(1/2e^(7x)sin(2x)+7/4e^(7x)cos(2x)). How do i get a 53 from that?
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