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I want to solve the following integral:

$$\int_{-\infty}^{\infty} (1+x^2)^{-3/2}$$

I thought maybe it's possible with $\sinh$ or $\cosh$ or something similar, but I can't figure it out. Thanks in advance for the help

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  • $\begingroup$ Try with $\sinh$, using the identity $\cosh^2=\sinh^2+1$. $\endgroup$ – Yves Daoust May 23 '15 at 8:21
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Write this as \begin{equation*} 2\int^{\infty}_0 \frac{1}{(x^2+1)^{3/2}}dx. \end{equation*} Using the substitution $x=\tan(u)$ and the identity $1+\tan^2(u)=\sec^2(u)$ to get \begin{equation*} 2\int^{\pi/2}_0 \frac{1}{\sqrt{\sec^2(u)}}du=2\int^{\pi/2}_{0}\cos(u)du=2 \end{equation*}

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  • $\begingroup$ ahh, thank you very much mate :D $\endgroup$ – DeltaChief May 23 '15 at 8:26
  • $\begingroup$ Not a problem :) $\endgroup$ – user230715 May 23 '15 at 8:26
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For fun I tried Yves' hyperbolic suggestion. $$ I = \int_{-\infty}^\infty (1+x^2)^{-3/2}dx $$ substitute $x=\sinh t$ so that $dx = \cosh t\,dt$ and $$ I = \int_{-\infty}^\infty (\cosh^2 t)^{-3/2} \cosh t\,dt= \int_{-\infty}^{\infty} (\cosh t)^{-2} dt = \tanh t \big|_{-\infty}^\infty = 2 $$

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