Question

How can solve this integral
(2x^7-x^5+2x^3-x+1)/(cos^2x)dx..

Answer 1

\[\Large \int\limits\frac{2x^7-x^5+2x^3-x+1}{\cos^2x}~~dx~~~~~~~~~~~~~?\]

Answer 2

For starters, consider the two easiest results:
\[\int\sec^2x\,\mathrm{d}x=\text{should be obvious}\]
Next,\(\displaystyle\int x\sec^2x\,\mathrm{d}x\) can be obtained via integration by parts. Set
\[\begin{matrix}u=x&&&\mathrm{d}v=\sec^2x\,\mathrm{d}x\\[1ex]
\mathrm{d}u=\mathrm{d}x&&&v=\text{(see previous result)}\end{matrix}\]and continue from there.

The remaining terms will make use of a function you likely aren't familiar with:
http://mathworld.wolfram.com/Polylogarithm.html

Answer 3

Just to get an idea of how it comes into play, I'll partially work out the next simplest term:
\[\int x^3\sec^2x\,\mathrm{d}x\]Integrate by parts, using
\[\begin{matrix}u=x^3&&&\mathrm{d}v=\sec^2x\,\mathrm{d}x\\[1ex]
\mathrm{d}u=3x^2\,\mathrm{d}x&&&v=\tan x\end{matrix}\](alright, so I gave it away... :P)

Now
\[\int x^3\sec^2x\,\mathrm{d}x=x^3\tan x-3\int x^2\tan x\,\mathrm{d}x\]Substitute \(x=\arctan y\), so that \(\tan x=y\) and \(\mathrm{d}x=\dfrac{y}{1+y^2}\,\mathrm{d}y\), giving you
\[\int x^3\sec^2x\,\mathrm{d}x=x^3\tan x-3\underbrace{\int(\arctan y)^2\frac{y}{1+y^2}\,\mathrm{d}y}_{\mathcal{I}}\]Rewrite the inverse tangent in its logarithmic form:
\[\arctan y=\frac{i}{2}\Big(\ln(1-iy)-\ln(1+iy)\Big)\]and squaring this, you are left with
\[\mathcal{I}=-\frac{1}{4}\int\frac{y\Big(\ln^2(1+iy)-2\ln(1+iy)\ln(1-iy)+\ln^2(1-iy)\Big)}{1+y^2}\,\mathrm{d}y\]Let's take a closer look at the first term of this integrand:
\[\mathcal{J}_1=\int\frac{y\ln^2(1+iy)}{1+y^2}\,\mathrm{d}y\]Substitute \(z=\ln(1+iy)\) so that \(y=-i(e^z-1)\) and \(\mathrm{d}y=-ie^z\,\mathrm{d}z\). Now,
\[\mathcal{J}_1=\int \frac{z^2\Big(-i(e^z-1)\Big)}{2e^z-e^{2z}}(-ie^z)\,\mathrm{d}z=\int\frac{z^2(e^z-e^{2z})}{2e^z-e^{2z}}\,\mathrm{d}z\]Add and subtract \(z^2e^z\) in the numerator, so that you can split up the integral like so,
\[\mathcal{J}_1=\int\frac{z^2(2e^z-e^{2z})}{2e^z-e^{2z}}\,\mathrm{d}z-\int\frac{z^2e^z}{2e^z-e^{2z}}\,\mathrm{d}z=\int z^2\,\mathrm{d}z-\underbrace{\int\frac{z^2}{2-e^z}\,\mathrm{d}z}_{\mathcal{K}}\]Next, substitute \(w=2-e^z\) so that \(z=\ln(2-w)\) and \(\mathrm{d}z=-\dfrac{\mathrm{d}w}{2-w}\), giving
\[\mathcal{K}=-\int\frac{\ln^2(2-w)}{w(2-w)}\,\mathrm{d}w=-\frac{1}{2}\int\left(\frac{\ln^2(2-w)}{2-w}+\frac{\ln^2(2-w)}{w}\right)\,\mathrm{d}w\]The first term is easy to handle, as a substitution of \(t=\ln(2-w)\) yields \(\displaystyle\frac{1}{2}\int t^2\,\mathrm{d}t\).

The second is where the polylogarithm starts to take shape. Integrating by parts yet again, with
\[\begin{matrix}u=\ln^2(2-w)&&&\mathrm{d}v=\dfrac{\mathrm{d}w}{w}\\[1ex]\mathrm{d}u=-\dfrac{2\ln(2-w)}{2-w}\,\mathrm{d}w&&&v=\ln w\end{matrix}\]you get
\[\int\frac{\ln^2(2-w)}{w}\,\mathrm{d}w=\ln^2(2-w)\ln w+\underbrace{2\int\frac{\ln(2-w)\ln w}{2-w}\,\mathrm{d}w}_{\mathcal{L}}\]Using the reflection formula,
\[\mathrm{Li}_2(x)-\mathrm{Li}_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x)\]we can rewrite the numerator as
\[\begin{align*}\ln(2-w)\ln w&=\ln\left(2\left(1-\frac{w}{2}\right)\right)\ln\left(2\left(\frac{w}{2}\right)\right)\\[1ex]
&= \left(\ln2+\ln\left(1-\frac{w}{2}\right)\right)\left(\ln2+\ln\frac{w}{2}\right)\\[1ex]
&=\ln^22+\ln2\left(\ln\left(1-\frac{w}{2}\right)+\ln\frac{w}{2}\right)+\ln\left(1-\frac{w}{2}\right)\ln\frac{w}{2}\\[1ex]
&=\ln^22+\frac{\pi^2}{6} +\ln2\left(\ln\left(1-\frac{w}{2}\right)+\ln\frac{w}{2}\right)\\[1ex]
&\quad\quad -\mathrm{Li}_2\left(\frac{w}{2}\right)-\mathrm{Li}_2\left(1-\frac{w}{2}\right)\end{align*}\]which gives us
\[\begin{align*}\mathcal{L}&=2\left(\frac{\pi^2}{6}+\ln^22\right)\int\frac{\mathrm{d}w}{2-w}+\ln2\int\frac{\ln\left(1-\frac{w}{2}\right)}{1-\frac{w}{2}}\,\mathrm{d}w \\[1ex]
&\quad\quad +\ln2\int\frac{\ln\frac{w}{2}}{1-\frac{w}{2}}\,\mathrm{d}w-\int\frac{\mathrm{Li}_2\left(\frac{w}{2}\right)}{1-\frac{w}{2}}\,\mathrm{d}w-\int\frac{\mathrm{Li}_2\left(1-\frac{w}{2}\right)}{1-\frac{w}{2}}\,\mathrm{d}w \end{align*}\]The first integral is easy, as are the second and fifth, as we have the following neat property of the polylogarithm,
\[\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{Li}_{n+1}(x)=\frac{\mathrm{Li}_n(x)}{x}\]which yields
\[\begin{align*}\mathcal{L}&=-2\left(\frac{\pi^2}{6}+\ln^22\right)\ln(2-w)+\ln2\,\mathrm{Li}_2\left(1-\frac{w}{2}\right)-\mathrm{Li}_3\left(1-\frac{w}{2}\right) \\[1ex]
&\quad\quad +\ln2\int\frac{\ln\frac{w}{2}}{1-\frac{w}{2}}\,\mathrm{d}w-\int\frac{\mathrm{Li}_2\left(\frac{w}{2}\right)}{1-\frac{w}{2}}\,\mathrm{d}w \end{align*}\]That's not to say the remaining two are difficult, as one can be handled with a substitution \(t=1-\dfrac{w}{2}\), which gives
\[\int \frac{\ln\frac{w}{2}}{1-\frac{w}{2}}\,\mathrm{d}w=-2\int\frac{\ln(1-t)}{t}\,\mathrm{d}t=2\int\frac{\mathrm{Li}_1(t)}{t}\,\mathrm{d}t=2\mathrm{Li}_2(t)\]Unfortunately, the final one is eluding me at the moment:\[\int \frac{\mathrm{Li}_2\left(\frac{w}{2}\right)}{1-\frac{w}{2}}\,\mathrm{d}w=\cdots\]

Answer 4

As you can probably guess, this gets very tedious, very quickly. I can't imagine that the other parts of \(\mathcal{I}\) are computed in a drastically different way, but even so I'm stopping right here. Hopefully it gives you an idea of how one might go about finding a closed form for the antiderivative.