Tips on Integrating Trig Functions
Date: 02/01/2001 at 00:47:35 From: Karan Subject: Integration Dear Math, I've been trying to integrate this funtion for quite some time now: dx/(3+4sin x)^2 I've tried to use u = 1/(3+4sinx), but the answer doesn't come out right. When I saw the solution, it was solved in a way that required you to practically know the answer beforehand. They used y = cosx/(3+4sinx), differentiated it, separated the desired integral and integrated the rest for the answer. Is there any other way? Why isn't my method working? Thanks, Karan
Date: 02/01/2001 at 11:28:12 From: Doctor Rob Subject: Re: Integration Thanks for writing to Ask Dr. Math, Karan. Such tricks are clever, but hard to discover. There is another way that works whenever you have a rational function of trigonometric functions, as you do here. The rational function is 1/(3+4*u)^2 and the trigonometric function is u = sin(x). The systematic method is to use the substitution: z = tan(x/2). Then sin(x) = 2*z/(1+z^2) cos(x) = (1-z^2)/(1+z^2) dx = 2*dz/(1+z^2) Substitute this in, and you'll end up needing to integrate a rational function of z: INTEGRAL dx/(3+4*sin[x])^2 = INTEGRAL 1/(3+4*2*z/[1+z^2])^2 * 2*dz/(1+z^2) = INTEGRAL (1+z^2)^2/(3*z^2+8*z+3)^2 * 2*dz/(1+z^2) = INTEGRAL 2*(1+z^2)/(3*z^2+8*z+3)^2 * dz Now you can split this using the fact that 3*z^2 + 8*z + 3 = 3*(z + [4+sqrt(7)]/3)*(z + [4-sqrt(7)]/3) and partial fractions to break it into parts which you can easily integrate. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Date: 02/24/2001 at 14:09:11 From: Karan Subject: Integration Please integrate: 1. ((tanx)^2+2)^0.5 2. (1-x^2)^n from (0,1) where n is an integer. Do I have to get a reduction formula? Is there on other way? Thanks, Karan
Date: 02/26/2001 at 09:23:42 From: Doctor Rob Subject: Re: Integration Thanks for writing back, Karan. 1. Use the very non-obvious substitution tan(x) = sqrt(2)*sinh(y) to reduce this to the integral of 1 + sech(2*y) dy. This may be a bit easier to deal with. 2. You can avoid the reduction formulas by using the Binomial Theorem to expand (1-x^2)^n into a sum, and then integrating term by term. I don't think you can get the integrated sum into a simple closed form for general n. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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