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Math Help - [SOLVED] Definite Integral, parts, with trig substution

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    Question [SOLVED] Definite Integral, parts, with trig substution

    Okay I have:  \int \frac {sec^{14}{20x}}{cot(20x)}
    I know I should break up sec{x} but I am not sure what to do for the trig substitutions.
    Thanks,
    Matt
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by matt3D View Post
    Okay I have:  \int \frac {sec^{14}{20x}}{cot(20x)}
    I know I should break up sec{x} but I am not sure what to do for the trig substitutions.
    Thanks,
    Matt
    You're correct, you first need to split off a factor of \sec(20x)

    \int\frac{\sec^{14}(20x)}{\cot(20x)}\,dx=\int\frac  {\sec^{13}(20x)\cdot\sec(20x)}{\cot(20x)}\,dx

    Now note that \frac{1}{\cot(20x)}=\tan(20x)

    So we now see that \int\frac{\sec^{13}(20x)\cdot\sec(20x)}{\cot(20x)}  \,dx=\int \sec^{13}(20x)\cdot\sec(20x)\tan(20x)\,dx

    Now it should be clear what the proper substitution should be.

    I hope this makes sense!

    --Chris
    Last edited by Chris L T521; September 9th 2008 at 09:42 PM.
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    Talking

    Thanks Chris! I forgot to mention that it had limits from 0 to pi/60 after I did u substitution I got \frac {16383}{280}
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by matt3D View Post
    Okay I have:  \int \frac {sec^{14}{20x}}{cot(20x)}
    I know I should break up sec{x} but I am not sure what to do for the trig substitutions.
    Thanks,
    Matt
     \int_0^{\pi/60} \frac{(\sec(20x))^{14}}{\cot(20x)} \ dx= \int_0^{\pi/60} \frac{\sin(20x)}{(\cos(20x))^{15}} \ dx= \int_0^{\pi/60} \frac{1}{20 \times 14}\frac{d}{dx} (\cos(20x))^{-14} \ dx

    RonL
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