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Thread: Definite integration

  1. #1
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    Definite integration

    Using the substitution $\displaystyle t=\tan x$, or otherwise, find
    $\displaystyle \int^{\frac{\pi}{3}}_0\frac{1}{9-8\sin^2x}dx$

    $\displaystyle \int^{\frac{\pi}{3}}_0\frac{1}{5+4\cos2x}dx$
    when $\displaystyle x=\frac{\pi}{3}\rightarrow t=\sqrt{3}$ and $\displaystyle x=0\rightarrow t=0$
    $\displaystyle t=\tan x\rightarrow dt=sec^2xdx$
    $\displaystyle \int^{\frac{\pi}{3}}_0\frac{1}{5+4(\frac{1-t^2}{1+t^2})}dx$
    $\displaystyle \int^{\frac{\pi}{3}}_0\frac{1+t^2}{9+t^2}dx$
    $\displaystyle \int^{\sqrt{3}}_0\frac{1}{9+t^2}dt$
    $\displaystyle [\frac{1}{t}\ln(9+t^2)]^{\sqrt{3}}_0$
    now then the answer i would get would be
    $\displaystyle \frac{1}{\sqrt{3}}\ln(12)$
    but the answer is supposed to be $\displaystyle \frac{1}{18}\pi$ which cannot be arrived at from $\displaystyle [\frac{1}{t}\ln(9+t^2)]^{\sqrt{3}}_0$ as an inverse trig is needed
    thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arze View Post
    [snip]

    Using the substitution $\displaystyle t=\tan x$, or otherwise, find
    $\displaystyle \int^{\frac{\pi}{3}}_0\frac{1}{9-8\sin^2x}dx$

    $\displaystyle \int^{\frac{\pi}{3}}_0\frac{1}{5+4\cos2x}dx$
    when $\displaystyle x=\frac{\pi}{3}\rightarrow t=\sqrt{3}$ and $\displaystyle x=0\rightarrow t=0$
    $\displaystyle t=\tan x\rightarrow dt=sec^2xdx$
    $\displaystyle \int^{\frac{\pi}{3}}_0\frac{1}{5+4(\frac{1-t^2}{1+t^2})}dx$
    $\displaystyle \int^{\frac{\pi}{3}}_0\frac{1+t^2}{9+t^2}dx$
    $\displaystyle \int^{\sqrt{3}}_0\frac{1}{9+t^2}dt$
    $\displaystyle [\frac{1}{t}\ln(9+t^2)]^{\sqrt{3}}_0$

    [/snip]
    Everything seems ok, except you integrated the last integral incorrectly.

    You should have gotten $\displaystyle \left.\left[\frac{1}{3}\arctan\left(\frac{t}{3}\right)\right]\right|_0^{\sqrt{3}}$
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  3. #3
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    integration of (1/9+t^2) is not ln(9+t^2)....you need to use this formula for that purpose...

    integral of [1/ (a^2 + x^2) ] = (1/a) . arctan (x/a)

    hopefully that will help!

    ohh and arctan is the same as (inverse tan) ...

    Good luck!
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