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Thread: Integral of tan^3(x) dx where did i go wrong?

  1. #1
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    Integral of tan^3(x) dx where did i go wrong?

    Hi

    I am trying to solve the integral of tan^3(x) dx but i seem to have made a slight error with the answer.

    This is the calculations i made, i can't see where my logic failed me:




    The answer seems to be natural log of cos x instead of sec x. But i am not quite getting how ?

    Hope some one can spot my error.
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  2. #2
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    Re: Integral of tan^3(x) dx where did i go wrong?

    $\begin{gathered}
    \int {{{\tan }^3}(x)dx} \hfill \\
    \int {{{\tan }^2}} (x)\tan (x)dx \hfill \\
    \int {({{\sec }^2}(x) - 1)\tan (x)dx} \hfill \\
    \int {(\tan (x){{\sec }^2}(x) - \tan (x))dx} \hfill \\
    \frac{1}{2}{\tan ^2}(x) + \log (\cos (x)) \hfill \\
    \end{gathered} $
    Last edited by Plato; Feb 1st 2018 at 02:34 PM.
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    Re: Integral of tan^3(x) dx where did i go wrong?

    Quote Originally Posted by thefollower View Post
    Hope some one can spot my error.
    There isn't one, I don't think.
    $$\ln|\sec x|=\ln \left|\frac1{\cos x}\right|=\ln|\cos x|^{-1}=-\ln|\cos x|$$
    Plato has a sign error in his integral of $\int \tan x \mathrm dx = \ln |\cos x|$
    Last edited by Archie; Feb 1st 2018 at 02:31 PM.
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    Re: Integral of tan^3(x) dx where did i go wrong?

    Quote Originally Posted by Archie View Post
    There isn't one, I don't think.
    $$\ln|\sec x|=\ln \left|\frac1{\cos x}\right|=\ln|\cos x|^{-1}=-\ln|\cos x|$$
    Plato has a sign error in his integral of $\int \tan x \mathrm dx = \ln |\cos x|$
    Oh really ? I got marked wrong which is kinda annoying. I am also confused how Plato got 1/2 tan^2(x) not 1/2sec^2(x) for the first term.
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    Re: Integral of tan^3(x) dx where did i go wrong?

    Quote Originally Posted by thefollower View Post
    Oh really ? I got marked wrong which is kinda annoying. I am also confused how Plato got 1/2 tan^2(x) not 1/2sec^2(x) for the first term.
    You do this and post the result:
    The derivative of $\dfrac{1}{2}\tan^2(x)$ is:
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    Re: Integral of tan^3(x) dx where did i go wrong?

    Quote Originally Posted by thefollower View Post
    I am also confused how Plato got 1/2 tan^2(x) not 1/2sec^2(x) for the first term.
    $$\frac{\mathrm d}{\mathrm dx}\tan x = \sec^2 x$$
    So, by writing $t=\tan x$ we get $\mathrm dt = \sec^2 x\, \mathrm d x$ and perform the substitution.
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    Re: Integral of tan^3(x) dx where did i go wrong?

    Quote Originally Posted by thefollower View Post
    Oh really ? I got marked wrong ...
    Other than that vertical bar next to the constant of integration (which
    looks like a typo), the derivative of your anti-derivative simplifies
    to the original expression that you were
    taking the integral of. The instructor should
    go back and give you credit for the problem.
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