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Math Help - integration check!

  1. #1
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    integration check!

    \int (4sec^2(5x-6)+7sec(9-8x)tan(9-8x)) dx


    = \frac {-7}{8}sec(9 - 8x) - \frac {4}{5}tan(6 - 5x)

    is this correct? thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jvignacio View Post
    \int (4sec^2(5x-6)+7sec(9-8x)tan(9-8x)) dx


    = \frac {-7}{8}sec(9 - 8x) - \frac {4}{5}tan(6 - 5x)

    is this correct? thanks
    you are forgetting a +C, but other than that, it looks good.

    i will say that you have something weird here though. it seems counter-intuitive that you would integrate backwards here. and that you would change the angle of the tangent by a factor of -1 and take the negative of the function

    had i done the problem, it would look like \frac 45 \tan (5x - 6) - \frac 78 \sec (9 - 8x) + C
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    you are forgetting a +C, but other than that, it looks good.

    i will say that you have something weird here though. it seems counter-intuitive that you would integrate backwards here. and that you would change the angle of the tangent by a factor of -1 and take the negative of the function

    had i done the problem, it would look like \frac 45 \tan (5x - 6) - \frac 78 \sec (9 - 8x) + C
    very true. thank you
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