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Thread: Relating a function and its derivative?

  1. #1
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    Relating a function and its derivative?

    I am trying to find the definite integral of tanx^3, with pi/6 at the top of the integral sign and -pi/6 on the bottom.

    The only problem I have is that I do not see how this fits the profile of:

    f(g(x))*g'(x)

    What would the g(x) be in tanx^3? Once I have that I have no problems finding the rest.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by redman223 View Post
    I am trying to find the definite integral of tanx^3, with pi/6 at the top of the integral sign and -pi/6 on the bottom.

    The only problem I have is that I do not see how this fits the profile of:

    f(g(x))*g'(x)

    What would the g(x) be in tanx^3? Once I have that I have no problems finding the rest.
    $\displaystyle \int\tan^3(x)dx=\int\tan(x)(\sec^2(x)-1)dx=$$\displaystyle \int\bigg[\tan(x)\sec^2(x)-\tan(x)\bigg]dx$

    The last one you should know how to do.

    If not, for the first part let $\displaystyle u=\tan(x)$ and for the second rewrite it as a quotient of sine and cosine and then let $\displaystyle u=\cos(x)$
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  3. #3
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    I don't understand what to do with u=tanx

    If I use tanx for u, then u' would be secx^2, but what would f(x) be?

    I only know how to solve these with the equation f(g(x))*g'(x), where g(x) is u, but I am having trouble setting it up.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by redman223 View Post
    I don't understand what to do with u=tanx

    If I use tanx for u, then u' would be secx^2, but what would f(x) be?

    I only know how to solve these with the equation f(g(x))*g'(x), where g(x) is u, but I am having trouble setting it up.
    $\displaystyle u=\tan(x)\Rightarrow{du=\sec^2(x)}$

    Or you can see that $\displaystyle f(x)=\left(\tan(x)\right)^1$ and $\displaystyle g(x)=\tan(x)\Rightarrow{g'(x)=\sec^2(x)}$
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  5. #5
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    So tanx*secx^2 = tanx^3?

    If that is true, I am having trouble seeing it.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by redman223 View Post
    So tanx*secx^2 = tanx^3?

    If that is true, I am having trouble seeing it.
    No...$\displaystyle \tan(x)\times\left[\sec^2(x)-1\right]=tan^3(x)$

    But we need to distribute it through...

    thus, we get $\displaystyle \tan(x)\sec^2(x)-\tan(x)$

    Integrating we have $\displaystyle \int \tan(x)\sec^2(x)\,dx-\int\tan(x)\,dx$

    Focusing on the first integral...
    $\displaystyle \int \tan(x)\sec^2(x)\,dx$

    Let $\displaystyle u=\tan(x)$ [note that the value for u is also our g(x)]. Thus, $\displaystyle \frac{du}{dx}=\sec^2(x)$ [again, this is the same as g'(x)].

    Thus, what we are left with then is $\displaystyle \int u\,du=\frac{1}{2}u^2=\frac{1}{2}\tan^2(x)$

    Now the second integral...

    $\displaystyle \int\tan(x)\,dx$

    Rewriting the integrand you have $\displaystyle \int\frac{\sin(x)}{\cos(x)}\,dx$

    Let $\displaystyle u=\cos(x)$ [this is our g(x)]. Thus, $\displaystyle \frac{du}{dx}=-\sin(x)$ [this is g'(x)].

    However, this doesn't quite match the g'(x) value in the integrand $\displaystyle \sin(x)$. To fix this, multiply by $\displaystyle \frac{-1}{-1}$

    Thus, we see the integral become $\displaystyle -\int\frac{\,du}{u}=-\ln|u|=-\ln|\cos(x)|$.

    Combining the two, we see $\displaystyle \int\tan^3(x)\,dx=\int\left[\tan(x)\sec^2(x)-\tan(x)\right]\,dx=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|$

    Now evaluate from $\displaystyle -\frac{\pi}{6}$ to $\displaystyle \frac{\pi}{6}$

    Does this make a little more sense?
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  7. #7
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    The only thing that I am having trouble understanding is why tan(x)sec(x)^2-tan(x) = tan(x)^3.
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  8. #8
    Lord of certain Rings
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    Hey you dont need all that!

    $\displaystyle I = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \tan^3 x \, dx $

    Substitute $\displaystyle x \to -x, dx \to -dx$

    $\displaystyle I = \int_{\frac{\pi}{6}}^{-\frac{\pi}{6}} \tan^3 (-x) \, (-dx) $

    $\displaystyle I = -\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} -\tan^3 (x) \, (-dx) $

    $\displaystyle I = -\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \tan^3 (x) \, dx $

    $\displaystyle I = -I \Rightarrow 2I = 0 \Rightarrow I = 0$

    In fact for any odd function f(x), and any real number a,

    $\displaystyle \int_{-a}^a f(x)\, dx = 0$
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