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Thread: [SOLVED] Definite Integral, parts, with trig substution

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

    Okay I have: $\displaystyle \int \frac {sec^{14}{20x}}{cot(20x)}$
    I know I should break up $\displaystyle sec{x}$ but I am not sure what to do for the trig substitutions.
    Thanks,
    Matt
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by matt3D View Post
    Okay I have: $\displaystyle \int \frac {sec^{14}{20x}}{cot(20x)}$
    I know I should break up $\displaystyle sec{x}$ 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 $\displaystyle \sec(20x)$

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

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

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

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

    I hope this makes sense!

    --Chris
    Last edited by Chris L T521; Sep 9th 2008 at 08:42 PM.
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    Talking

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

    RonL
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