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Old 2008-10-01, 15:57   #1
Primeinator
 
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Default Challenging Integral

I was doing a homework problem involving the indefinate integral of sin(2x)/sqr(4cos(x) -1) but unable to find a way to solve it. I have tried several approaches involving a number of trig identities, but a solution still remains elusive. Can anyone give me a hint that I am probably missing? Thanks.
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Old 2008-10-01, 17:19   #2
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Quote:
Originally Posted by Primeinator View Post
I was doing a homework problem involving the indefinate integral of sin(2x)/sqr(4cos(x) -1) but unable to find a way to solve it. I have tried several approaches involving a number of trig identities, but a solution still remains elusive. Can anyone give me a hint that I am probably missing? Thanks.
I think some integration by parts does the trick, start by looking at
d/dx(cos(x)sqr(4cos(x) - 1)) and follow your nose.

Chris
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Old 2008-10-01, 19:33   #3
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Split the steps up line by line. You should only highlight the first parts first to see if you can complete it without seeing the whole solution.

Rewrite the numerator using sin(2x)=2*sin(x)*cos(x)

Now do change of variables:

w=4*cos(x)-1
dw=-4*sin(x) dx

You should get -1/2*int[cos(x)/sqrt(w) dw]

But you still have a term involving x, so you rearrange the substitution to get cos(x)=(w-1)/4.

This should give you -1/2*int[(sqrt(w)/4)-(1/(4*sqrt(w))) dw], which is manageable.
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Old 2008-10-02, 01:05   #4
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Thanks, I managed to solve it actually. Two substitutions, let u=4cos x, du = -4sin x dx. Then, let v=u -1, dv =du. It comes out very nicely.
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Old 2008-10-02, 01:56   #5
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Quote:
Originally Posted by Primeinator View Post
Thanks, I managed to solve it actually. Two substitutions, let u=4cos x, du = -4sin x dx. Then, let v=u -1, dv =du. It comes out very nicely.
Actually, if you let u=4*cos(x)-1, you only need to do one substitution.
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Old 2008-10-02, 06:03   #6
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Quote:
Originally Posted by Kevin View Post
Actually, if you let u=4*cos(x)-1, you only need to do one substitution.
I think you can do it without any substitutions:

Code:

d/dx [ cos(x) sqrt(4cos(x) - 1) ] = -2sin(x)cos(x) / sqrt(4cos(x) - 1) 
                                    - sin(x) sqrt(4cos(x) - 1)

                                            = - sin(2x) / sqrt(4cos(x) - 1)
                                              - sin(x) sqrt(4cos(x) - 1)
so

Code:

int[ sin(2x) / sqrt(4cos(x) - 1) ] = -cos(x) sqrt(4cos(x) - 1)
                                     - int [ sin(x) sqrt(4cos(x) - 1) ] 

                                           = -cos(x) sqrt(4cos(x) - 1)
                                             + (4cos(x) - 1)^(3/2) / 6 + constant
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Old 2008-10-02, 09:28   #7
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Quote:
Originally Posted by Chris Card View Post
I think you can do it without any substitutions:
But integration by parts is a lot more work than substitution, especially when it's not at all obvious what your parts should be, and even more so when the result of integration by parts gives you another integral that a student not intimately familiar with the subject would still need to use substitution to evaluate (- int [ sin(x) sqrt(4cos(x) - 1) ]).
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Old 2008-12-21, 16:47   #8
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Default Integral Problem

The integration problem Sin(2x)/ Sqrt [4Cos(x) -1] has a solution according to Mathematica program. Since this is home work I cannot give you the answer
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Old 2008-12-22, 15:37   #9
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Quote:
Originally Posted by S309907 View Post
The integration problem Sin(2x)/ Sqrt [4Cos(x) -1] has a solution according to Mathematica program. Since this is home work I cannot give you the answer
considering that the problem was posted 2.5 months ago i think it would be safe to post the answer
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Old 2008-12-23, 16:40   #10
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If you absolutely want to, the answer is irrelevant at this point but would still be interesting.

Last fiddled with by Primeinator on 2008-12-23 at 16:42
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