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Thread: An interesting integral

  1. #1
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    An interesting integral

    Integrate 1/sqrt[ (tan(x))^2 + (cot(x))^2 ] w.r.t. x

    Do you think this integral is very beautiful ??
    Try to solve and leave opinions for this , thank you
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  2. #2
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    Quote Originally Posted by simplependulum View Post
    Integrate 1/sqrt[ (tan(x))^2 + (cot(x))^2 ] w.r.t. x

    Do you think this integral is very beautiful ??
    Try to solve and leave opinions for this , thank you
    Not expressible in terms of elementary functions.
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  3. #3
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    The integrand is $\displaystyle \frac{\sin x\cos x}{\sqrt{\cos ^{2}2x+2\sin ^{2}x\cos ^{2}x}}.$

    Now it suffices to put $\displaystyle u=\cos2x\implies du=-4\sin x\cos x\,dx$ so that $\displaystyle \sin^2x=\frac{1-u}2$ and $\displaystyle \cos^2x=\frac{1+u}2$ and you'll get an easy integral to solve.
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  4. #4
    Member Nacho's Avatar
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    Other solution:

    $\displaystyle
    I = \int {\frac{{dx}}
    {{\sqrt {\tan ^2 x + \cot ^2 x} }} = \int {\frac{{\sin x\cos x}}
    {{\sqrt {\sin ^4 x + \cos ^4 x} }}dx} = } \frac{1}
    {2}\int {\frac{{2\sin x\cos x}}
    {{\sqrt {1 - 2\sin ^2 x\cos ^2 x} }}dx}
    $

    $\displaystyle
    \begin{gathered}
    I = \frac{1}
    {2}\int {\frac{{\sin \left( {2x} \right)}}
    {{\sqrt {1 - \frac{{\sin ^2 \left( {2x} \right)}}
    {2}} }}} dx = \frac{{\sqrt 2 }}
    {2}\int {\frac{{\sin \left( {2x} \right)}}
    {{\sqrt {2 - \left( {1 - \cos ^2 \left( {2x} \right)} \right)} }}dx} \hfill \\
    \hfill \\
    \tan z = \cos \left( {2x} \right) \Rightarrow \sec ^2 zdz = - 2\sin \left( {2x} \right)dx \hfill \\
    \end{gathered}
    $

    $\displaystyle
    I = \frac{{ - \sqrt 2 }}
    {4}\int {\frac{{\sec ^2 z}}
    {{\sqrt {1 + \tan ^2 z} }}dz} = \frac{{ - \sqrt 2 }}
    {4}\int {\sec zdz} = \frac{{ - \sqrt 2 }}
    {4}\ln \left| {\tan z + \sec z} \right| + C
    $

    If you comeback to initial variable, you obtain the primitive function
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