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Hard integral

  • Thread starter evol_w10lv
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  • #1
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Homework Statement


How to integrate:
ww682rqwias6riyy5m97.png



Homework Equations





The Attempt at a Solution


I used formula: sin^2(t) = ( 1-cos^2(t))
and now it's:
9lkj89s3d5cesi48a3ax.png

Then:
u=cos(t)
du=-sin(t)
8y6dyp0r0o162zbcibba.png


What to do next?
 
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Answers and Replies

  • #2
arildno
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We may now rewrite the integrand as:
[tex]-6\sqrt{s^{2}+1}, s=\frac{\sqrt{5}u}{2}[/tex]
Now, utilize the trigonometric identity:
[tex]\tan^{2}(y)+1=\frac{1}{\cos^{2}(y)}[/tex]
in a creative way.
 
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  • #3
CAF123
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Where did the 3 come from?
Try using the identity 1 + tan2x = sec2x after some initial manipulation of the integrand.
 
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  • #4
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Actually the task sounds like this:
v4jlxgd41iy2lui365kt.png


Maybe I have to use polar coordinates? Any sugestions? Before I tried with diferent way, but I guess that integration without polar coordinates is too hard.
 
  • #5
arildno
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L is a segment of an ellipse. You might go polar, but you can eliminate t as well, and express y as a function of x(Hint: sin^+cos^2=1)
 
  • #6
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rz4k8k32eudm9k2wr91e.png

Seems to me that variant when we use y=y(x) is more complicated than variant with polar coordinates.

We may now rewrite the integrand as:
[tex]-6\sqrt{s^{2}+1}, s=\frac{\sqrt{5}u}{2}[/tex]
Now, utilize the trigonometric identity:
[tex]\tan^{2}(y)+1=\frac{1}{\cos^{2}(y)}[/tex]
in a creative way.
Not clear, how did you get there: [tex]s=\frac{\sqrt{5}u}{2}[/tex]
We didn't learn about triple substitution, but I want to understand, how to get the final result. Can you explain some how?
 

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