How Can Trig Substitutions Help Simplify Integrals?

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Homework Statement


Integral of 1/(2+sin(x)) dx


Homework Equations





The Attempt at a Solution


I've been told that you can use trig subs, but I never had to learn that in high school and it hasn't appeared in any of my calculus coursework.

As a side note. I've been wondering if it is possible to solve asin(x) + bcos(x) = c
 
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To solve your side note use this website. it helped me out! good luck

http://www.education2000.com/demo/demo/btnchtml/sinplcos.htm
 
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Wow, that's so cool. Thanks for the link. I'll try to remember how to derive it.
 
n4rush0 said:

Homework Statement


Integral of 1/(2+sin(x)) dx

The standard substitution \tan\frac{x}{2} = t applies for your integral. It will convert it into a integral of an algebraic function for which the method of partial fraction decomposition will get it solved.
 
Thanks, I'll try that. Is that something you just memorized or is there a certain rule that let's you know what to substitute?
 
There are rules. That substitution will apply to an antiderivative of a function a+b\sin x/c+d\cos x and the other 3 ways of interchanging cos with sin and more generally to any algebraic function of sin and cos.
 
Where can I learn all these rules? I usually only see substitutions with x = asint, atant, or asect
 
You're normally taught these rules of substitution in high-school. I wasn't, so I picked them up for myself from books, especially for engineers, because the proofs are missing :)
 
Okay so, given:
integral dx/(2+sinx)

tan(x/2) = t
(1/2)sec^2 (x/2) dx = dt
dx = 2cos^2 (x/2) dt

integral
2cos^2 (x/2) dt / (2+sinx)

Am I supposed to use x = arctan(2t)? If so, is it possible to simplify by drawing a triangle?
 
  • #10
Of course you have to use that. It's the whole purpose of substitution, you need to change every function of x including the dx with the approproate function of t and dt.
 
  • #11
I know how to change sinx to sin 2t/sqrt(1+4t^2)
but I'm not sure how to simplify cos^2 (x/2) since it has the 1/2 in front of the x and I can't use the same trick that I used for sinx.
 
  • #12
But you need sin (2 arctan t) from the initial integral.

\sin (2\arctan t) = 2 (\sin\arctan t) (\cos\arctan t)

\sin\arctan t = \frac{t}{\sqrt{1+t^2}} \, , \, \cos\arctan t = \frac{1}{\sqrt{1+t^2}}

What about the integration element ?
 
  • #13
n4rush0 said:
Okay so, given:
integral dx/(2+sinx)

tan(x/2) = t
(1/2)sec^2 (x/2) dx = dt
dx = 2cos^2 (x/2) dt

integral
2cos^2 (x/2) dt / (2+sinx)

Am I supposed to use x = arctan(2t)? If so, is it possible to simplify by drawing a triangle?

It's not x=arctan(2t), but rather x=2arctan(t). that might help.
 
  • #14
Thank you. I modified the integral to
dt/t^2+t+1)
Are you sure it's partial fractions?
 
  • #15
There, I would actually use completing the square in the denominator, then do another trig sub.
 
  • #16
Thank you. I finally get it now. I'll still have problems with the initial trig substitutions though since I'm not sure how to get tan(x/2) = t.
 
  • #17
Letting t = \tan(x/2) is part of something called a Weierstrass substitution. This is usually a pretty messy substitution, but it's good to have in your toolbox of integration tricks, especially for those pesky integrals where nothing else seems to work.
 
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