mersenneforum.org Challenging Integral
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 2008-10-01, 15:57 #1 Primeinator     "Kyle" Feb 2005 Somewhere near M50..sshh! 2×3×149 Posts 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.
2008-10-01, 17:19   #2
Chris Card

Aug 2004

8216 Posts

Quote:
 Originally Posted by Primeinator 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

Chris

 2008-10-01, 19:33 #3 Kevin     Aug 2002 Ann Arbor, MI 6618 Posts 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.
 2008-10-02, 01:05 #4 Primeinator     "Kyle" Feb 2005 Somewhere near M50..sshh! 2×3×149 Posts 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.
2008-10-02, 01:56   #5
Kevin

Aug 2002
Ann Arbor, MI

1B116 Posts

Quote:
 Originally Posted by Primeinator 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.

2008-10-02, 06:03   #6
Chris Card

Aug 2004

100000102 Posts

Quote:
 Originally Posted by Kevin 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

Chris

2008-10-02, 09:28   #7
Kevin

Aug 2002
Ann Arbor, MI

1B116 Posts

Quote:
 Originally Posted by Chris Card 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) ]).

 2008-12-21, 16:47 #8 S309907   C2816 Posts 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
2008-12-22, 15:37   #9
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

10110011110012 Posts

Quote:
 Originally Posted by S309907 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

 2008-12-23, 16:40 #10 Primeinator     "Kyle" Feb 2005 Somewhere near M50..sshh! 2×3×149 Posts 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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