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Math Help - Another integration by parts problem

  1. #1
    Member Jones's Avatar
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    Another integration by parts problem

    Howdy,

    Trying to integrate sin7x~\times~cos8x

    \text{let}~~u = sin7x, du = 7cos7x

    2 ~sin7x~ \times~cos8x-\int 7cos7x *\frac{sin8x}{8}

    Again..
    \text{Let}~~u = \frac{sin8x}{8}~~ \text{then}~~du=cos8x

    7cos7\times\frac{sin8x}{8}-\int sin7x*cos8x

    Substitute back into 2

    sin7x~ \times~cos8x-\bigg[ 7cos7\times\frac{sin8x}{8}-\int sin7x*cos8x \bigg]

    2\int sin7x~\times~cos8x = sin7x~ \times~cos8x-\bigg[ 7cos7\times\frac{sin8x}{8}\bigg]

    \int sin7x~\times~cos8x = \frac{sin7x~ \times~cos8x-\bigg[ 7cos7\times\frac{sin8x}{8}\bigg]}{2}

    But this is wrong
    ¿Porque?¿
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  2. #2
    MHF Contributor red_dog's Avatar
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    This is the general form of the integral:

    I=\int\sin\alpha x\cos\beta xdx=-\int\left(\frac{\cos\alpha x}{\alpha}\right)'\cos\beta xdx=

    =-\frac{1}{\alpha}\cos\alpha x\cos\beta x-\frac{\beta}{\alpha}\int\cos\alpha x\sin\beta xdx=

    =-\frac{1}{\alpha}\cos\alpha x\cos\beta x-\frac{\beta}{\alpha}\int\left(\frac{\sin\alpha x}{\alpha}\right)'\sin\beta xdx=

    =-\frac{1}{\alpha}\cos\alpha x\cos\beta x-\frac{\beta}{\alpha^2}\sin\alpha x\sin\beta x+\frac{\beta^2}{\alpha^2}I

    =\frac{\alpha^2-\beta^2}{\alpha^2}I=-\frac{1}{\alpha}\cos\alpha x\cos\beta x-\frac{\beta}{\alpha^2}\sin\alpha x\sin\beta x\Rightarrow

    \Rightarrow I=-\frac{\alpha}{\alpha^2-\beta^2}\cos\alpha x\cos\beta x-\frac{\beta}{\alpha^2-\beta^2}\sin\alpha x\sin\beta x+C

    Here is another method, without using integration by parts:

    \int\sin\alpha x\cos\beta xdx=\frac{1}{2}\int[\sin(\alpha+\beta)x+\sin(\alpha-\beta)x]dx=

    -\frac{1}{2(\alpha+\beta)}\cos(\alpha+\beta)x-\frac{1}{2(\alpha-\beta)}\cos(\alpha-\beta)x+C
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  3. #3
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    parts is too much work ...

    \sin(15x) = \sin(7x+8x) = \sin(7x)\cos(8x) + \cos(7x)\sin(8x)<br />

    \sin(-x) = \sin(7x-8x) = \sin(7x)\cos(8x) - \cos(7x)\sin(8x)<br />
    ----------------------------------------------------------

    add up the terms of both equations ...

    \sin(15x) + \sin(-x) = 2\sin(7x)\cos(8x)<br />

    \sin(7x)\cos(8x) = \frac{1}{2}[\sin(15x) - \sin(x)]<br />

    integrate the RHS
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  4. #4
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    Hello, Jones!

    skeeter is right . . . "by parts" is a lot of work.
    Here it is . . . in case you ever get that desperate.


    Integrate: . I \;=\;\int \sin7x\cos8x\,dx

    By parts: . \begin{array}{cccccccc}u &=&\sin7x & & dv &=& \cos8x\,dx \\ du &=& 7\cos7x\,dx & & v&=&\frac{1}{8}sin8x \end{array}

    . . We have: . I \;=\;\tfrac{1}{8}\sin7x\sin8x - \tfrac{7}{8}\!\!\int\sin8x\cos7x\,dx


    By parts: . \begin{array}{cccccccc}u &=& \cos7x && dv &=& \sin8x\,dx \\ <br />
du &=& \text{-}7\sin7x\,dx & & v &=& \text{-}\frac{1}{8}\cos8x \end{array}

    . . We have: . I \;=\;\tfrac{1}{8}\sin7x\sin9x - \tfrac{7}{8}\bigg[\text{-}\tfrac{1}{8}\cos7x\cos8x - \tfrac{7}{8}\int\!\!\sin7x\cos8x\,dx\bigg]

    \text{Hence: }\;I \;=\;\tfrac{1}{8}\sin7x\sin8x + \tfrac{7}{64}\cos7x\cos8x + \tfrac{49}{64}\!\!\underbrace{\int\sin7x\cos8x\,dx  }_{\text{This is }I}+C


    We have: . I \;=\;\tfrac{1}{8}\sin7x\sin8x + \tfrac{7}{64}\cos7x\cos8x + \tfrac{49}{64}I + C

    . . Then: . \tfrac{15}{64}I \;=\;\tfrac{1}{8}\sin7x\sin8x + \tfrac{7}{64}cos7x\cos8x + C

    . . Multiply by \tfrac{64}{15}\!:\quad I \;=\;\tfrac{8}{15}\sin7x\sin8x + \tfrac{7}{15}\cos7x\cos8x + C


    Therefore: . \int\sin7x\cos8x\,dx \;=\;\frac{8}{15}\sin7x\sin8x + \frac{7}{15}\cos7x\cos8x + C

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  5. #5
    Member Jones's Avatar
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    Thanks,

    How do you know when to use integration by parts and when not to use it?

    I thought you always had to use it if you had a product of two functions =/
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  6. #6
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    Quote Originally Posted by Jones View Post
    Thanks,

    How do you know when to use integration by parts and when not to use it?

    I thought you always had to use it if you had a product of two functions =/
    You use integration by parts if two situations hold...

    1. One of the functions in the product should be easy to differentiate and one should be easy to integrate.

    2. By integrating and differentiating the correct functions, the new integral is reduced to something easier to integrate.
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