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Substitution

Why would I use a trigonometric substitution when n is even and not when n is odd?
Original post by moz4rt
Why would I use a trigonometric substitution when n is even and not when n is odd?


If nn is even, you will get even powers of trigonometric functions, which are nice to work with because you can convert between them using the identity sin2x+cos2x=1\sin^{2}{x} + \cos^{2}{x} = 1 (e.g. sin6x=(1cos2x)3=...\sin^{6}{x} = (1-\cos^{2}{x})^3 = ..., so you can get an expression entirely in terms of cosx\cos{x}). Odd powers of trigonometric functions are less easy to work with since you can't reduce things to only one trigonometric function in this way.
Reply 2
Original post by Prasiortle
If nn is even, you will get even powers of trigonometric functions, which are nice to work with because you can convert between them using the identity sin2x+cos2x=1\sin^{2}{x} + \cos^{2}{x} = 1 (e.g. sin6x=(1cos2x)3=...\sin^{6}{x} = (1-\cos^{2}{x})^3 = ..., so you can get an expression entirely in terms of cosx\cos{x}). Odd powers of trigonometric functions are less easy to work with since you can't reduce things to only one trigonometric function in this way.


Oh thanks, I did it using constants and got to that but didn't think it was of much use as you would still get large powers and use de moivre;s theorem of something like that.
Original post by moz4rt
Oh thanks, I did it using constants and got to that but didn't think it was of much use as you would still get large powers and use de moivre;s theorem of something like that.


Yes, with large powers of trig functions, you would use De Moivre's Theorem and the Binomial Theorem to express them in terms of trig functions of compound angles, which can then be integrated.

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