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# Help with this integral please. watch

1. Can someone help me integrate (cot x) / (rt(1-cos 2x)) please.

I think it's possible to rewrite it as (1/ rt2) cosec x cot x or something very similar, but have no idea how to get there, or perform the subsequent integration.

2. I'm guessing rt is a constant so just take that outside of the integral

double angle formula - cos 2x = 1-2sin^2(x)
substitution, let u=sinx => du/cosx = dx

Work that through and you should get a result
3. so getting from the first part to the second part, it is just algebraic manipulation, when there is a cos2x, what comes to mind?

and to integrate it, it is a standard integral (if you remember what the derivative of the cosec, sec, and cot is)
4. (Original post by Ewan)
I'm guessing rt is a constant so just take that outside of the integral

double angle formula - cos 2x = 1-2sin^2(x)
substitution, let u=sinx => du/cosx = dx

Work that through and you should get a result
rt is short for "square root".

Thanks for the other points, I'll have another go!
5. (Original post by baffled_mathman)
rt is short for "square root".

Thanks for the other points, I'll have another go!
Ah fair enough, don't need to edit my post in that case xD (you'll see why)
6. (Original post by 1729)
so getting from the first part to the second part, it is just algebraic manipulation, when there is a cos2x, what comes to mind?

and to integrate it, it is a standard integral (if you remember what the derivative of the cosec, sec, and cot is)

I could rewrite is as cot x / (rt (1-2cos^2x-1) which gives cot x / -2 cos x.

I'm not sure where to take it from the though. I can recall cot, sec and cosec derivatives, but they aren't helping yet!

7. writing cos2x as in what you did is one way, but is there another way of writing cos2x?
8. oh and btw, 1 - cos2x is not 1-2cos^2x-1

(Original post by baffled_mathman)
I could rewrite is as cot x / (rt (1-2cos^2x-1) which gives cot x / -2 cos x.

I'm not sure where to take it from the though. I can recall cot, sec and cosec derivatives, but they aren't helping yet!

9. (Original post by baffled_mathman)
I could rewrite is as cot x / (rt (1-2cos^2x-1) which gives cot x / -2 cos x.

I'm not sure where to take it from the though. I can recall cot, sec and cosec derivatives, but they aren't helping yet!

No you can't, check your minus signs.
10. (Original post by 1729)
oh and btw, 1 - cos2x is not 1-2cos^2x-1
(Original post by Ewan)
No you can't, check your minus signs.
Thanks, no idea what I was thinking there!

(Original post by 1729)
writing cos2x as in what you did is one way, but is there another way of writing cos2x?
Thanks! I didn't think to use the cos 2x = 1-2sin^2 form. It then becomes cot x / (rt 2 sin x). Then take out 1/rt2 leaving cosec x cot x. The integral of which is -cosec x. So the solution is (-1/rt2 ) cosec x + c.

I appreciate it people.

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