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Math Help - Primitive function

  1. #1
    Senior Member Twig's Avatar
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    Primitive function

    Hi!

    Problem: Find the primitive function to

    \int \frac{1}{sin^{2}(x)+2cos^{2}(x)} \; dx

    Solution attempt:

    I tried the substitution  t = tan\left(\frac{x}{2}\right) , which gives us sin(x)=\frac{2t}{1+t^{2}} \; , \; cos(x)=\frac{1-t^{2}}{1+t^{2}} \mbox{ and } dx = \frac{2}{1+t^{2}}dt

    Substituting into the integral I got (if I havenīt done any algebraic mistake)

    \int \frac{1+t^{2}}{1+t^{4}}\; dt


    And this is where I am stuck.

    Thx
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  2. #2
    Super Member malaygoel's Avatar
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    <br />
\int \frac{1}{sin^{2}(x)+2cos^{2}(x)} \; dx<br />

    divide every term by cos^2x

    <br />
\int \frac{sec^2(x)}{tan^{2}(x)+2} \; dx<br />

    and then substitute tanx=t

    <br />
\int \frac{1}{t^2+2} \; dx<br />
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  3. #3
    Senior Member Twig's Avatar
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    Hi

    Thank you! This solution is like a million times easier.
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  4. #4
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    Hello, Twig!

    Another approach . . . but malaygoel's is the best!


    Find the primitive function to: . \int \frac{1}{\sin^2\!x+2\cos^2\!x}\,dx

    The denominator is: . \sin^2\!x + 2\cos^2\!x \;=\;\frac{1-\cos2x}{2} + 2\!\cdot\!\frac{1+\cos2x}{2} \;=\;\frac{3+\cos2x}{2}

    The integral becomes: . 2\int\frac{dx}{3+\cos2x}


    Let t \:=\:\tan x \quad\Rightarrow\quad dx \:=\:\frac{dt}{1+t^2}\quad\Rightarrow\quad \cos 2x \:=\:\frac{1-t^2}{1+t^2}

    Substitute: . 2\int\frac{\frac{dt}{1+t^2}}{3 + \frac{1-t^2}{1+t^2}} \quad\Rightarrow\quad \int\frac{dt}{2+t^2} \quad\Rightarrow\quad \frac{1}{\sqrt{2}}\arctan\left(\frac{t}{\sqrt{2}}\  right) + C


    Back-substitute: . \frac{1}{\sqrt{2}}\arctan\left(<br />
\frac{\tan x}{\sqrt{2}}\right) + C

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