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Math Help - 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 \frac{\sin x\cos x}{\sqrt{\cos ^{2}2x+2\sin ^{2}x\cos ^{2}x}}.

    Now it suffices to put u=\cos2x\implies du=-4\sin x\cos x\,dx so that \sin^2x=\frac{1-u}2 and \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:

    <br />
I = \int {\frac{{dx}}<br />
{{\sqrt {\tan ^2 x + \cot ^2 x} }} = \int {\frac{{\sin x\cos x}}<br />
{{\sqrt {\sin ^4 x + \cos ^4 x} }}dx}  = } \frac{1}<br />
{2}\int {\frac{{2\sin x\cos x}}<br />
{{\sqrt {1 - 2\sin ^2 x\cos ^2 x} }}dx} <br />

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

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

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