# Tricky trigonometric integral

reaper616
Hi. I'm having troubles calculating this indefinite integral. I have tried a few things, but none seem to work.

Actually, universal substitution does make it possible to integrate, but there has to be some shorter, more elegant way. Anyone?

Count Iblis
Replacing the 1 in the numerator by sin^2(x) + cos^2(x) does seem to work. You then do get the original integral back, but it is multiplied by 1/2 so you can move it back to the left hand side to solve for it.

reaper616
Replacing the 1 in the numerator by sin^2(x) + cos^2(x) does seem to work. You then do get the original integral back, but it is multiplied by 1/2 so you can move it back to the left hand side to solve for it.

I'm sorry, but I'm not sure that I understand what are you trying to say. Could you go a little bit more in-depth?

Count Iblis
Abbreviation: Sin(x) = S, Cos(x) = C

[S^2 + C^2]/[S (2 C^2 - 1)] =

S/[2 C^2 - 1] (easy to integrate as the derivative of C is -S and S is in the numerator)

+

C^2/[S(2C^2 - 1)]

We can rewrite the numerator of the last term as:

C^2 = 1/2 2 C^2 = 1/2 (2 C^2 - 1 + 1)

This means that you can write the last term as:

1/(2S) + 1/2 * Original term you wanted to integrate.

Then you're done if you can integrate 1/S and that you can do using more or less the same trick:

1/Sin(x) = 1/(2Sin(1/2 x) Cos(1/2 x)) and then replace the numerator by Cos^2(1/2 x) + Sin^2(1/2 x) and you're done.

reaper616
It works. Very nice. Thank you very much.

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