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Math Help - Intergal Question

  1. #1
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    Intergal Question

    \int \dfrac{7}{t^{3} + 5} dt

    \int 7(t^{3} + 5}^{-1} dt

    u = (t^{3} + 5)

    du = 3t^{2}

    \dfrac{7}{3t^{2}}du = 7 dt

    \dfrac{7}{3t^{2}} \int u^{-1}

    \dfrac{7}{3t^{2}}  \dfrac{u^{0}}{0} + C

    \dfrac{7}{3t^{2}} \ln|t^{3} + 5| + C Right?
    Last edited by Jason76; October 22nd 2013 at 11:22 PM.
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    Re: Intergal Question

    No firstly the integral of u^(-1) is log u. secondly you are treating t^2 as a constant which is not the case. In its present form i can't think of way, I hope you have copied the question correctly.
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    Re: Intergal Question

    Question is copied correctly
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    Re: Intergal Question

    This is not pretty, but here's what Wolfram gets. Sign in and you can see the step-by-step solution.
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    Re: Intergal Question

    A way through it is to let t^{3}+5=t^{3}+a^{3} where a is the cube root of 5.
    Factorize and split into two partial fractions.
    One integrates to a log and the other (after a routine bit of algebra) to a log and an inverse tangent.
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    Re: Intergal Question

    Quote Originally Posted by Jason76 View Post
    u = (t^{3} + 5)
    The point behind the substitution is to rewrite the integrand in terms of u only. So
    u = t^3 + 5 \implies t = (u - 5)^{1/3}

    and
    du = 3t^2 ~ dt = 3 \left ( (u - 5)^{1/3}\right ) ^2 \cdot dt

    which gives
    dt = \frac{du}{3 (u - 5)^{2/3}}

    Now you have a value for both t and dt to put into your integrand.

    This substitution doesn't help you do the problem, I just wanted to run you through how to set it up properly.

    -Dan
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    Re: Intergal Question

    Try t^3 = 5\tan^2\theta. Then t = \sqrt[3]{5}\tan^{2/3}\theta} and dt = \dfrac{2\sqrt[3]{5}}{3}\sec^{7/3}\theta\sin^{1/3}\theta d\theta

    \begin{align*}\int \dfrac{7}{t^3+5}dt & = \dfrac{14\sqrt[3]{5}}{3}\int \dfrac{\sec^{7/3}\theta\sin^{1/3}\theta d\theta}{5(\tan^2\theta + 1)} \\ & = \dfrac{14\sqrt[3]{5}}{15}\int \tan^{1/3}\theta d\theta\end{align*}

    That's an alternate possible place to begin.
    Last edited by SlipEternal; October 23rd 2013 at 11:53 AM.
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    Re: Intergal Question

    This is the method I outlined a couple of days ago.

    Let the integral equal I and let t^{3}+5 \equiv t^{3}+a^{3}, (to avoid various fractional powers of 5 complicating the calculation).

    Then I=\int \frac{7}{t^{3}+a^{3}} dt =7 \int \frac{1}{(t+a)(t^{2}-at+a^{2})} dt

    Split the integrand into two (partial) fractions.

    Let \frac{1}{(t+a)(t^{2}-at+a^{2})} \equiv \frac{A}{(t+a)}+\frac{Bt+C}{(t^{2}-at+a^{2})},

    then 1 \equiv A(t^{2}-at+a^{2})+(t+a)(Bt+C).

    Substituting t=-a and equating coefficients of t^{2} and constants across the identity leads to

    A=1/3a^{2}, \quad B=-1/3a^{2} \text{ and } C=2/3a.

    Substituting in and tidying up a little produces

    I = \frac{7}{3a^{2}}\int \frac{1}{(t+a)}- \frac{t-2a}{(t^{2}-at+a^{2})} dt

    Rewrite the top line of the second fraction as \frac{1}{2}(2t-4a) = \frac{1}{2}(2t-a-3a) and split this fraction into two, then

    I = \frac{7}{3a^{2}}\int \frac{1}{(t+a)}- \frac{1}{2}\frac{2t-a}{(t^{2}-at+a^{2})}+\frac{3a}{2}\frac{1}{(t^{2}-at+a^{2})} dt

    Complete the square in the denominator of the third fraction and introduce the factor a\sqrt{3}/2 top and bottom to arrive at

    I = \frac{7}{3a^{2}}\int \frac{1}{(t+a)}- \frac{1}{2}\frac{2t-a}{(t^{2}-at+a^{2})}+\sqrt{3}\frac{a\sqrt{3}/2}{(t-a/2)^{2}+(a\sqrt{3}/2)^{2}}. dt

    Finally, integrate to get

    I=\frac{7}{3a^{2}}\left( \ln(t+a)-\frac{1}{2}\ln(t^{2}-at+a^{2}) + \sqrt{3}\arctan\left[\frac{2(t-a/2)}{a\sqrt{3}}\right]\right)+D, D a constant of integration (and where a is the cube root of 5).
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