# Integration of x*sec(x)

1. Nov 6, 2004

### heman

i am struggling with this problem for 3 years and still not able to think anything how to integrate it.plz. anyone tell me how to integrate it.

2. Nov 6, 2004

### Tide

There's a good reason why you're having trouble with the integral - there is no simple expression for it! :-)

3. Nov 6, 2004

### josephcollins

Hi ppl,
One suggestion I would make, if the integral is indeed so difficult, is to obtain firstly the maclaurin series for secx, using the binomial expansion of (cosx)^-1 where cosx is also written down as a power series. Then multiply through by x(x*secx) and integrate the x terms one by one to obtain an approximation to this integral. Of course this is only an approximation, yet if all you desire is to work out a definite integral it might prove useful, the indefinite integral is not so straightforward I imagine.
Regards,
Joe

4. Nov 7, 2004

### Zurtex

Well if you put: x * Sec[x] in to: http://integrals.wolfram.com/ it returns:

$$x \left( \log \left[ \frac{1 - ie^{ix}}{1 + ie^{ix}} \right] \right) + i \left( \text{polylog} \left[2, -ie^{ix} \right] - \text{polylog} \left[2, ie^{ix} \right] \right)$$

5. Nov 7, 2004

### heman

i could not understand the solution ,how can i solve the question

6. Nov 7, 2004

### Zurtex

What's the question?

7. Nov 7, 2004

### heman

to integrate x*sec(x)

8. Nov 7, 2004

### phreak

Integration by parts, my friend. Here's the formula:

$$\int (u)(dv) = (uv) - \int (v)(du)$$

[Pardon the parentheses. I'm new to Tex.]

I'm assuming you know how to do the rest. Integrate, differentiate, complete.

Last edited: Nov 7, 2004
9. Nov 8, 2004

### heman

dear phreak i have tried integration by parts so many times,but nothing is solved.actually,i think some advanced theorem is involved in it.

10. Nov 8, 2004

### Zurtex

Quite simply this integration seems to be beyond your ability as well as beyond mine. I had a quick look around to try and explain this better. But the best I can do is say if you differentiate this with respect to x:

$$-x \text{arctanh} \left( ie^{-x} \right) + i \left( \text{polylog} \left[2, -ie^{ix} \right] - \text{polylog} \left[2, ie^{ix} \right] \right)$$

You get $x \sec x$.

If it is of any help:

$$\text{polylog} (n,z) = \sum_{k=1}^{\infty} \frac{z^k}{k^n}$$

So:

$$\text{polylog} (2,z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2}$$

Furthermore:

$$\frac{d}{dx} \left( \text{polylog} (n,x) \right) = \frac{1}{x} \text{polylog}(n-1,x)$$

And:

$$\text{polylog} (1,x) = -\ln (1-x)$$

Best I can do sorry.

11. Nov 9, 2004

### heman

please any guru of integration tell me the pathway to this solution.i will be highly thankful.

12. Nov 9, 2004

### HallsofIvy

Staff Emeritus
How about paying attention to what people HAVE been telling you :
.

Like the great majority of integrable functions, the integral of x sec(x) cannot be written in terms of elementary functions.

13. Nov 9, 2004

### heman

actually dear why i am so much worried about these is that my school teacher knows the solution of this problem and he challenged all the guys of school for 2000 bugs and he is sure to pay if anyone brings the soln and if i will come to know, my little finance prob.s will be solved,our school teacher gave it to us when we were in highschool.that means that teacher is a fraud becoz he asks such tough questions from studs. of 12th class.i will tell this to him.

even than
thanx for urscoepration and suggesting the pathways.