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Math Help - integration problem no.11

  1. #1
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    integration problem no.11

    integration problem no.11
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  2. #2
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    \int\frac{1}{a^{2}cos^{2}{\theta}+b^{2}sin^{2}{\th  eta}}d{\theta}

    Make the sub u=tan{\theta}, \;\ tan^{-1}(u)={\theta}, \;\ d{\theta}=\frac{1}{u^{2}+1}du

    Then you get:

    \int\frac{1}{a^{2}+b^{2}u^{2}}du

    Now, let w=tan^{-1}(\frac{ub}{a}), \;\ \frac{atan(w)}{b}=u, \;\ du=\frac{a}{bcos^{2}(w)}dw

    Making the subs whittles it down to:

    \int\frac{1}{ab}dw=\frac{w}{ab}

    Now make the resubs:

    \frac{tan^{-1}(\frac{ub}{a})}{ab}

    \frac{tan^{-1}(\frac{btan({\theta})}{a})}{ab}
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  3. #3
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    Hello, afeasfaerw23231233!


    A slight variation of Galactus' excellent solution . . .


    22)\;\int^{\frac{\pi}{2}}_0\frac{d\theta}{a^2\cos^  2\!\theta + b^2\sin^2\!\theta}
    Divide top and bottom by \cos^2\!\theta\!:\;\;\int^{\frac{\pi}{2}}_0\frac{\  dfrac{d\theta}{\cos^2\theta}}{\dfrac{a^2\cos^2\the  ta}{\cos^2\theta} + \dfrac{b^2\sin^2\theta}{\cos^2\theta}}

    . . and we have: . \int^{\frac{\pi}{2}}_0\frac{\sec^2\theta\,d\theta}  {a^2 + b^2\tan^2\theta}


    Let: u \:=\:b\tan\theta\quad\Rightarrow\quad du \:=\:b\sec^2\theta\,d\theta


    Substitute: . \frac{1}{b}\int\frac{du}{a^2+u^2} \;=\;\frac{1}{b}\cdot\frac{1}{a}\arctan\left(\frac  {u}{a}\right) + C


    \text{Back-substitute: }\;\frac{1}{ab}\arctan\left(\frac{b\tan\theta}{a}\  right) + C \;=\;\frac{1}{ab}\arctan\left(\dfrac{b}{a}\!\cdot\  !\tan\theta\right) + C

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