Find the Indefinite Integral dx/(cos(5x)-1)?

Question

I've tried everything I could think of on this thing and it seems to lead nowhere. Right off the bat I don't see any trig identities to use, you could use sin^5x + cos^5x = 1 if the bottom was squared, but it isn't.

The only other thing I can think of is adding 1 to top and bottom and trying to work from there..but I'm not sure if that will work.

I split them and was able to solve the first one, but I am stuck on the second one.

I used the half-angle identity sin^2 x = (1/2)(1-cos2x)

and came up with (1/2) * integral(1/(1-cos(10x) dx) but I don't think that's where I should be going seeing how the cos term is higher than when I started.

A double angle identity was mentioned, but I'm not sure which one to use.

Answer 1

First, let u=5x
du=5*dx
dx=du/5

Resub:

(1/5)int(1/(cosu - 1) du)

Multiply by the conjugate:

(1/5)int((cosu+1)/(cos^2(u)-1) du)

Simplify:

(1/5)int((cosu + 1)/(-sin^2(u)) du)

Break the fraction apart:

(1/5)int(cosu/-sin^2(u) + 1/-sin^2(u))du)

(1/5)int(-cotu*cscu - csc^2(u))du

Integrate

(1/5)(cscu + cotu)

Back substitute for u
(1/5)(csc(5x) + cot(5x)) + C

Using the Integrator (found at http://integrals.wolfram.com/index.jsp) you get:

(1/5)cot(5x/2) + C

If you convert this to trig functions of angle 5x, I'm certain they are the same.

Viola

Answer 2

I don't want to solve this entirely for you, but here's a tip. You could multiply the top and bottom by (cos(5x) + 1). That would get you the integral of (cos(5x) + 1)/(cos^2(5x) - 1) which can be simplified to (cos(5x) + 1)/(sin^2(5x))

You can then solve by splitting up the fraction and using u substitution for the cos/sin and one of the double-angle trig identities for the 1/sin^2.

Hope that helps =)

Answer 3

A useful trick:

cos(2t) = 1 - 2sin^2(t), so

1/(cos(2t) - 1) = (-1/2) csc^2(t).

Now can you integrate csc^2(t)? It should have been taught.

[ I know, I replaced 5x/2 with t, but you'll manage. ;) ]

Answer 4

what about partial integration, did you try that ?

Integral (fg') dx = fg - Integral f'g dx


sin(5x) / [ 5 - 5 cos(5x) ] modulus the contant

Answer 5

http://wims.unice.fr/wims/wims.cgi?session=QSD743C25D.4&+lang=en&+cmd=resume&+module=tool%2Fanalysis%2Ffunction.en

The online calculator there saved my life in so many math/physics courses, its not funny.