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How to integrate the following

  • Thread starter dashhh
  • Start date
  • #1
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Homework Statement


∫[itex]\frac{sin\Theta}{1+sin^2\Theta}[/itex] d[itex]\Theta[/itex] on [0,[itex]\frac{\pi}{2}[/itex]]


Homework Equations


I substituted:
u=cos[itex]\Theta[/itex]
du=-sin[itex]\Theta[/itex]


The Attempt at a Solution


= ∫ [itex]\frac{sin\Theta}{1+1-cos^2\Theta}[/itex]

= ∫ [itex]\frac{sin\Theta}{2-cos\Theta}[/itex]

= ∫ [itex]\frac{sin\Theta}{2-u^2}[/itex]

= [itex]\frac{1}{(u+\sqrt{2})(u-\sqrt{2})}[/itex] <--- I'm pretty sure I shouldn't have done this.

And that's as far as I've gotten.

Any tips greatly appreciated.
 

Answers and Replies

  • #2
vela
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You can use partial fractions or a trig substitution.
 
  • #3
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You can use partial fractions or a trig substitution.
We haven't done trig substitution, I think partial fractions is covered in the next section of my notes though so I'll check it out. Thanks :)
 
  • #4
HallsofIvy
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Homework Statement


∫[itex]\frac{sin\Theta}{1+sin^2\Theta}[/itex] d[itex]\Theta[/itex] on [0,[itex]\frac{\pi}{2}[/itex]]


Homework Equations


I substituted:
u=cos[itex]\Theta[/itex]
du=-sin[itex]\Theta[/itex]


The Attempt at a Solution


= ∫ [itex]\frac{sin\Theta}{1+1-cos^2\Theta}[/itex]

= ∫ [itex]\frac{sin\Theta}{2-cos\Theta}[/itex]

= ∫ [itex]\frac{sin\Theta}{2-u^2}[/itex]
You've dropped the "[itex]d\Theta[/itex]" in each of those- don't do that!

= [itex]\frac{1}{(u+\sqrt{2})(u-\sqrt{2})}[/itex] <--- I'm pretty sure I shouldn't have done this.
You need, of course, "du" and you have the sign wrong in the denominator.

And that's as far as I've gotten.

Any tips greatly appreciated.
 
  • #5
eumyang
Homework Helper
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(The LaTeX was a little messy, so I cleaned them up for you.)
I substituted:
[itex]u = \cos \theta[/itex]
[itex]du = -\sin \theta \;[/itex]
Don't forget the dθ. Also, from here, move the negative sign:
[itex]-du = \sin \theta \; d\theta[/itex]

[itex]= \int\frac{\sin\theta}{1+1-\cos^2\theta}[/itex]
Again, don't forget the dθ!
[itex]= \int \frac{\sin\theta}{2-\cos^2 \theta}[/itex]
[itex]= \int \frac{\sin\theta}{2-u^2}[/itex]
Substitute in -du for sin θ dθ:
[tex]= \int \frac{-du}{2-u^2}[/tex]
Can you take it from there?

EDIT: Beaten to it. :wink:
 

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