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Very Difficult Trig Sub?

  • Thread starter n4rush0
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  • #1
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Homework Statement


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


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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
 

Answers and Replies

  • #2
To solve your side note use this website. it helped me out! good luck

http://www.education2000.com/demo/demo/btnchtml/sinplcos.htm [Broken]
 
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  • #3
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Wow, that's so cool. Thanks for the link. I'll try to remember how to derive it.
 
  • #4
dextercioby
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Homework Statement


Integral of 1/(2+sin(x)) dx
The standard substitution [itex] \tan\frac{x}{2} = t [/itex] 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.
 
  • #5
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Thanks, I'll try that. Is that something you just memorized or is there a certain rule that lets you know what to substitute?
 
  • #6
dextercioby
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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.
 
  • #7
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Where can I learn all these rules? I usually only see substitutions with x = asint, atant, or asect
 
  • #8
dextercioby
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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 :)
 
  • #9
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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
dextercioby
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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
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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
dextercioby
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But you need sin (2 arctan t) from the initial integral.

[tex] \sin (2\arctan t) = 2 (\sin\arctan t) (\cos\arctan t) [/tex]

[tex] \sin\arctan t = \frac{t}{\sqrt{1+t^2}} \, , \, \cos\arctan t = \frac{1}{\sqrt{1+t^2}} [/tex]

What about the integration element ?
 
  • #13
Char. Limit
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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
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Thank you. I modified the integral to
dt/t^2+t+1)
Are you sure it's partial fractions?
 
  • #15
Char. Limit
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There, I would actually use completing the square in the denominator, then do another trig sub.
 
  • #16
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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
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Letting [itex]t = \tan(x/2)[/itex] 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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