# An interesting integral

• February 17th 2009, 04:30 AM
simplependulum
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
• February 17th 2009, 05:18 AM
Mush
Quote:

Originally Posted by simplependulum
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.
• February 17th 2009, 08:09 AM
Krizalid
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.
• February 17th 2009, 08:42 AM
Nacho
Other solution:

$
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}
$

$
\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}
$

$
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