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computation of integrals.

 
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Nov13-06, 01:45 PM   #1
 

computation of integrals.


i need to solve/prove the next two integrals:
[tex]\int\frac{dx}{u^2+u+4}[/tex]
and i need to show that:
[tex]\int_{0}^{\pi}\sqrt{1+sinx}dx=4[/tex]
the problem is that i have a clue to substitute u=sinx and then sin(pi)=0=sin0 so the integral should be equal zero, is it not?
ofcourse the integrand becomes: sqrt(1+u)/sqrt(1-u^2)
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Nov13-06, 01:58 PM   #2
 
For the first you can use the method of partial fractions.

You're right about the second. With the given limits, the integral is equal to zero.
Nov13-06, 01:59 PM   #3
 
substitute u = 1 + sin x. Are you sure the integral is equal to 4, not -4?
Nov13-06, 02:05 PM   #4
 

computation of integrals.


But the plot (area) of the function sqrt(1+sin(x)) from 0 to pi seems to be non-zero!
Nov13-06, 02:06 PM   #5
 
It's non-zero.
Nov13-06, 02:06 PM   #6
 
neutrino, how would i use partail fractions here?
i mean i need to decompose u^2+u+4 into a product of terms, but i have complex roots here.
Nov13-06, 02:07 PM   #7
 
complete the square.
Nov13-06, 02:11 PM   #8
 
you mean something like this: u^2+u+4=(u-2)^2+5u
i still dont get an appropiate term to integrate.
Nov13-06, 02:12 PM   #9
 
More like (u +0.5)^2 + 15/4
Nov13-06, 02:15 PM   #10
 
ok, thanks.
btw, what about the second integral does it equal zero or it really does equal 4?
Nov13-06, 02:17 PM   #11
 
u^2+u+4= (u + 0.5)^2 + 15/4.

edit: too slow, the second integral should equal minus -4, I guess they're defining is it as area so you just need the modulus.
Nov13-06, 02:21 PM   #12
 
Not -4. I just put the function through the Integrator and substituted the values, and I got 4. This graph is completely above the x-axis.
Nov13-06, 02:22 PM   #13
 
Btw, you will need to know what the derivative of the inverse tangent is.
Nov13-06, 02:28 PM   #14
 
Oops, I'm missing/added a minus somewhere. Didn't have the common senese to think about the graph :).
Nov15-06, 06:16 AM   #15
 
wait a minute, then integral does converge to 4, care to explain how, where did i get it wrong?
Nov15-06, 06:35 AM   #16
 
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For the integral try the sub

[tex] \tan\frac{x}{2}=t [/tex]

Daniel.
Nov15-06, 06:53 AM   #17
 
but what's wrong with the substitution that im given a hint to use here?
i.e
sinx=u?
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