# Indefinite integration

• May 3rd 2012, 12:55 AM
kyou
Indefinite integration
INTEGRATE (x sec x)^2 dx
• May 3rd 2012, 01:45 AM
FernandoRevilla
Re: Indefinite integration
Express $\int (x\sec x)^2\;dx=\int x^2\sec^2 x\;dx$ and use integration by parts with $u=x^2,\;dv=\sec^2 x\;dx.$
• May 3rd 2012, 01:51 AM
Prove It
Re: Indefinite integration
Quote:

Originally Posted by kyou
INTEGRATE (x sec x)^2 dx

\displaystyle \begin{align*} \int{\left(x \sec{x}\right)^2\,dx} &= \int{x^2\sec^2{x}\,dx} \end{align*}

Use integration by parts with \displaystyle \begin{align*} u = x^2 \implies du = 2x\,dx \end{align*} and \displaystyle \begin{align*} dv = \sec^2{x}\,dx \implies v = \tan{x} \end{align*} and the integral becomes

\displaystyle \begin{align*} \int{x^2\sec^2{x}\,dx} &= x^2\tan{x} - \int{2x\tan{x}\,dx} \\ &= x^2\tan{x} - 2\int{x\tan{x}\,dx} \end{align*}

Now use integration by parts again with \displaystyle \begin{align*} u = x \implies du = dx \end{align*} and \displaystyle \begin{align*} dv = \tan{x}\,dx \implies v = \ln{|\sec{x}|} \end{align*} and the integral becomes

\displaystyle \begin{align*} x^2\tan{x} - 2\int{x\tan{x}\,dx} &= x^2\tan{x} - 2\left(x\ln{|\sec{x}|} - \int{\ln{|\sec{x}|}\,dx}\right) \\ &= x^2\tan{x} - 2x\ln{|\sec{x}|} - \int{\ln{|\sec{x}|}\,dx} \end{align*}

This does not have a closed form solution.
• May 3rd 2012, 02:08 AM
biffboy
Re: Indefinite integration
That leaves me requiring integral of xtanx (?)
• May 3rd 2012, 02:15 AM
Prove It
Re: Indefinite integration
Quote:

Originally Posted by biffboy
That leaves me requiring integral of xtanx (?)

Did you even read my post? I showed you that the integral does not have a closed form solution...
• May 3rd 2012, 02:52 AM
biffboy
Re: Indefinite integration
My post was before yours :-)
• May 3rd 2012, 02:54 AM
biffboy
Re: Indefinite integration
No it wasn't. Apologies.
• May 3rd 2012, 02:56 AM
Prove It
Re: Indefinite integration
Quote:

Originally Posted by biffboy
My post was before yours :-)

Was it really?

http://i22.photobucket.com/albums/b3...n/ohreally.jpg
• May 3rd 2012, 02:58 AM
kyou
Re: Indefinite integration
Well, even if the indefinite integral does not have a closed form evaluation , it does have an infinite series representation , cause we can express tan x as an infinite series in terms of x and then we can obviously multiply the series by x and integrate term by term , owing to the convergence of the series , thus we get the indefinite integral
of xtanx and hence that of (xsecx)^2
• May 3rd 2012, 03:15 AM
biffboy
Re: Indefinite integration
My reply was to the post from Spain.