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So for one my exercise question is this:

$$\int x \sec(x)\,dx$$

I tried every way that I can think of. I believe this could only be done in integration by parts. My class only learned three techniques which is PFD, $u$ -sub, and integration by parts. I tried $u = x$, $u = \sec(x)$ and then, $u = x\sec(x)$, which didn't help.

Any help would be great.

  • 2
    The answer is not neat. It is not in terms of elementary functions. In fact, the solution to the integral contains [polylogarithm functions](http://mathworld.wolfram.com/Polylogarithm.html) $\operatorname*{Li}_2(x)$. – 2017-02-12
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    What do you get when you try integration by parts? – 2017-02-12
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    "The answer is not neat. It is not in terms of elementary functions" Hmm, I don't think my prof would give us something like that. Guess ill skip it. – 2017-02-12
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    @MichaelMcGovern Looks harder than the original function in all the integration by parts that I came up with. – 2017-02-12
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    @MichaelMcGovern Integration by parts won't be a great idea since you will end up with multiple of expressions. – 2017-02-12
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    If I were you, I would express \(x\sec(x)\) in terms of sums. Then, using @projectilemotion 's clue, try to express in terms of \(\text{Li}_n(x)\). – 2017-02-12
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    From the methods that I have learned a u-substitution would work either because of sec(x). I could try that but i honestly dont think im suppose to do this question and it was just put on accident – 2017-02-12
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    if you use int by parts the real problem is $\iint \sec xdx$ – 2017-02-12
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    Whenever you have a product of functions, integration by parts is a good place to start. However, if after trying several different ways of dividing up the integrand, you aren't able to find a suitable $u$ and $v$, integration by parts most probably will not work, as seems to be the case in this problem. – 2017-02-12
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    Wolframalpha gives this answer x (Log[1 - I E^(I x)] - Log[1 + I E^(I x)]) + I (PolyLog[2, (-I) E^(I x)] - PolyLog[2, I E^(I x)]) if someone understands this cuz idk what PolyLog is. – 2017-02-12
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    I asked one of my friends he said he asked the teacher and it was a typo. The real question is $\int xsec^2(x)dx$ which I know how to do. – 2017-02-12
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    @mettled mike https://en.wikipedia.org/wiki/Polylogarithm – 2017-02-12

1 Answers 1

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Well, use the identity:

$$\sec\left(x\right):=\frac{1}{\cos\left(x\right)}=\frac{2}{e^{xi}+e^{-xi}}\tag1$$

So, we get that:

$$\mathcal{I}\left(x\right):=\int x\sec\left(x\right)\space\text{d}x=2\int\frac{x}{e^{xi}+e^{-xi}}\space\text{d}x=2\int\frac{xe^{xi}}{1+e^{2xi}}\space\text{d}x\tag2$$

Now, substittue $\text{u}=xi$:

$$2\int\frac{xe^{xi}}{1+e^{2xi}}\space\text{d}x=-2\int\frac{\text{u}e^\text{u}}{1+e^{2\text{u}}}\space\text{d}\text{u}\tag3$$

Now, substitute $\text{v}=e^\text{u}$:

$$-2\int\frac{\text{u}e^\text{u}}{1+e^{2\text{u}}}\space\text{d}\text{u}=-2\int\frac{\ln\left(\text{v}\right)}{1+\text{v}^2}\space\text{d}\text{v}\tag4$$

Now, we know that:

$$1+\text{v}^2=\left(\text{v}-i\right)\left(\text{v}+i\right)\tag5$$

So, using partial fraction decomposition:

$$-2\int\frac{\ln\left(\text{v}\right)}{1+\text{v}^2}\space\text{d}\text{v}=-2\int\frac{\ln\left(\text{v}\right)}{\left(\text{v}-i\right)\left(\text{v}+i\right)}\space\text{d}\text{v}=i\left\{\int\frac{\ln\left(\text{v}\right)}{\text{v}-i}\space\text{d}\text{v}-\int\frac{\ln\left(\text{v}\right)}{\text{v}+i}\space\text{d}\text{v}\right\}\tag6$$