# An interesting integral

• Feb 17th 2009, 03:30 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
• Feb 17th 2009, 04:18 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.
• Feb 17th 2009, 07:09 AM
Krizalid
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.
• Feb 17th 2009, 07:42 AM
Nacho
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