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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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