Indefinite integration

**kyou** · May 2nd 2012, 11:55 PM

INTEGRATE (x sec x)^2 dx

**FernandoRevilla** · May 3rd 2012, 12:45 AM

Express $\int (x\sec x)^2\;dx=\int x^2\sec^2 x\;dx$ and use integration by parts with $u=x^2,\;dv=\sec^2 x\;dx.$

**Prove It** · May 3rd 2012, 12:51 AM

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.

**biffboy** · May 3rd 2012, 01:08 AM

That leaves me requiring integral of xtanx (?)

**Prove It** · May 3rd 2012, 01:15 AM

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...

**biffboy** · May 3rd 2012, 01:52 AM

My post was before yours :-)

**biffboy** · May 3rd 2012, 01:54 AM

No it wasn't. Apologies.

**Prove It** · May 3rd 2012, 01:56 AM

Originally Posted by biffboy

My post was before yours :-)

Was it really?

**kyou** · May 3rd 2012, 01:58 AM

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

**biffboy** · May 3rd 2012, 02:15 AM

My reply was to the post from Spain.

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