Evaluating these integrals

  • #1
Hi,
I'm having a bit of trouble evaluating this integral: (This is not a homework problem btw)
Integral between Pi and 0 (x sin x)/(1 + cos^2 x) d x
I don't even know where to begin with the substitutions or anything...... I was thinking the denominator could be sin^2 x..... only when I realized this is trig and not hyperbolic. :(

Also, something in my book:
Integral (2t^2 + 3t*t^3) d(t^2)
This was after the substitution of dy to d(t^2) when dealing with a line integral; I don't understand... how do you integrate this?

Thanks for any ideas.
 

Answers and Replies

  • #2
Tide
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I don't think you're going to find a closed form for your integral. If it's any help, the numerical value of your integral is approximately -2.4674011...

For the second integral, replace [itex]d (t^2)[/itex] with [itex]2 t dt[/itex].
 
  • #3
Hi,
Thanks for the reply,
My book shows something weird with the integral...
Here's the work.... (Assuming everything I have below has the limits pi and 0)
Let x = pi y, then:
integral (xsin x)/(1 + cos^2 x) dx = ---> integ (pi-y)(sin y)/(1 + cos^2 y) dy = pi * integ (sin y)/(1 + cos^2 y) dy - integ (y sin y)/(1 + cos^2 y) dy = -pi * integ d(cos y)/(1 + cos^2 y) - I = -pi*arctan(cos y) - I = Pi^2 /2 - I
And as it says:
i.e. I = Pi^2 / 2 - I or I = Pi^2 / 4
I got lost at the point where I showed the arrow (And I'm confused about why they chose that particular substitution)...
Also, what's up with this - I thing? Can somebody explain it?
Thanks loads for any replies.
 
Last edited:
  • #4
Tide
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Oh, I like that! Basically, it's a simple transformation of the integral with the result that the original integral = some other integral - the original integral and, luckily "some other integral" can be evaluated!
 
  • #5
dextercioby
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U can part integrate

[tex] \int_{\pi}^{0} x \frac{-d(\cos x)}{1+\cos^{2}x} =...[/tex]

Daniel.
 

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