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Math Help - Evaluate the Integral

  1. #1
    Member FalconPUNCH!'s Avatar
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    Evaluate the Integral

    Is there a quick way to do this one?

     \int sin(8x)cos(5x)dx
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        \int\sin (c_1x)\cos (c_2x)\;dx = -\frac{\cos(c_1+c_2)x}{2(c_1+c_2)}-\frac{\cos(c_1-c_2)x}{2(c_1-c_2)} \qquad\mbox{(for }|c_1|\neq|c_2|\mbox{)}\,\!

    Thats the really quick way though. How to derive it, I don't know off hand, and don't have the time to work through it. Someone else here who is good with integrals probably can though.
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  3. #3
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by xifentoozlerix View Post
        \int\sin (c_1x)\cos (c_2x)\;dx = -\frac{\cos(c_1+c_2)x}{2(c_1+c_2)}-\frac{\cos(c_1-c_2)x}{2(c_1-c_2)} \qquad\mbox{(for }|c_1|\neq|c_2|\mbox{)}\,\!

    Thats the really quick way though. How to derive it, I don't know off hand, and don't have the time to work through it. Someone else here who is good with integrals probably can though.
    Interesting. I haven't actually seen this method before.
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  4. #4
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    From \sin(a+b)=\sin a\cos b+\cos a\sin b, combine this one with \sin(a-b) and you'll get a formula to split the original product into a sum.
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  5. #5
    Member FalconPUNCH!'s Avatar
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    I have another one that I tried but couldn't get the answer.

    \int_{0}^{\frac{pi}{3}} tan^{5}x sec^{6}x dx

    Here's my work

    u = sec(x) du = sec(x)tan(x)dx

    \int_{0}^{\frac{pi}{3}} (tan^{2}x)^{2} sec^{5}x  tan(x)sec(x)dx

    \int_{0}^{\frac{pi}{3}} (1-sec^{2}x)^{2} sec^{5}x  tan(x)sec(x)dx

    \int_{0}^{\frac{pi}{3}} (1- u^{2})^{2} u^{5}  du


    \int_{0}^{\frac{pi}{3}} u^{5} - 2u^{7} + u^{9} du


     \frac{sec^{6}x}{6} - \frac{sec^{8}x}{8} + \frac{sec^{10}x}{10}
    from 0 to \frac{Pi}{3}

    I plug those in and I get a huge number that isn't correct.
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  6. #6
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    Quote Originally Posted by FalconPUNCH! View Post
    I have another one that I tried but couldn't get the answer.

    \int_{0}^{\frac{pi}{3}} tan^{5}x sec^{6}x dx
    Do the make-up first:

    \tan ^5 x\sec ^6 x = \tan ^5 x\left( {\sec ^2 x} \right)^2 \sec ^2 x = \tan ^5 x\left( {1 + \tan ^2 x} \right)^2 \sec ^2 x.

    Now just make the obvious substitution. (Don't forget to change integration limits for u, or whatever.)
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  7. #7
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Krizalid View Post
    Do the make-up first:

    \tan ^5 x\sec ^6 x = \tan ^5 x\left( {\sec ^2 x} \right)^2 \sec ^2 x = \tan ^5 x\left( {1 + \tan ^2 x} \right)^2 \sec ^2 x.

    Now just make the obvious substitution. (Don't forget to change integration limits for u, or whatever.)
    To change the limits do I just plug them into whatever u equals?
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  8. #8
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    Yes.

    So by setting u=\tan x get the lower integration limit by makin' x=0 and the upper integration limit by makin' x=\frac\pi3.
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  9. #9
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Krizalid View Post
    Yes.

    So by setting u=\tan x get the lower integration limit by makin' x=0 and the upper integration limit by makin' x=\frac\pi3.
    Thanks if I have anymore questions I'll just post them here.
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  10. #10
    Member FalconPUNCH!'s Avatar
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    Does this look right?

    \frac{1}{6}tan^{6}(\frac{1}{\sqrt{3}}) + \frac{2}{8}tan^{8}(\frac{1}{\sqrt{3}}) + \frac{1}{10}tan^{10}(\frac{1}{\sqrt{3}})

    it just doesn't look right to me
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  11. #11
    Member FalconPUNCH!'s Avatar
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    I left it in terms of u so I ended up getting

    \frac{1}{6}\sqrt{3}^{6} - \frac{2}{8}\sqrt{3}^{8} + \frac{1}{10}\sqrt{3}^{10}

    I get \frac{171}{10}

    Can anyone check if it's right?
    Last edited by FalconPUNCH!; February 21st 2008 at 03:50 PM.
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  12. #12
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    Can't see your TeX, but I plugged the integral into my TI-89 and got \frac{981}{20}. So, uhh, yeah.
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  13. #13
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by xifentoozlerix View Post
    Can't see your TeX, but I plugged the integral into my TI-89 and got \frac{981}{20}. So, uhh, yeah.
    I edited it
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  14. #14
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    \frac{1}{6}\sqrt{3}^{6} - \frac{2}{8}\sqrt{3}^{8} + \frac{1}{10}\sqrt{3}^{10}
    There is a typo here. You should have \frac{1}{6}\sqrt{3}^{6} + \frac{2}{8}\sqrt{3}^{8} + \frac{1}{10}\sqrt{3}^{10}.
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  15. #15
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by badgerigar View Post
    There is a typo here. You should have \frac{1}{6}\sqrt{3}^{6} + \frac{2}{8}\sqrt{3}^{8} + \frac{1}{10}\sqrt{3}^{10}.
    hmm..I see I'll have to go look through my work to find out where I made the mistake. Thanks
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