Use substitution x=2tan(y) to integrate (37dx)/(x^2*sqrt(x^2+4)) in y

  • Thread starter beanryu
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  • #1
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I feel so :cry: doing this problem.
PLEASE HELP!!!! AND TEACH ME How TO DO IT.
My question is this.

use the substitution x=2tan(y)

(37dx)/(x^2*sqrt(x^2+4))

give the answer in terms of y.

I did the substition, but it looked more complicated.
It doesn't look like a u*du thing, but it does look like some arcsin thing... i couldn't figure it out. The x^2 is bugging me.
PLease help!!!
 

Answers and Replies

  • #2
benorin
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Consider that

[tex]\sqrt{4+x^2}=\sqrt{4+4\tan^2(y)}=2\sqrt{1+\tan^2(y)}[/tex]

and then use the approiate Pythagorean identity.
 
  • #3
benorin
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namely [tex]1+\tan^2(y)=\sec^2(y)[/tex] so that

[tex]2\sqrt{1+\tan^2(y)}=2\sqrt{\sec^2(y)}=2\sec (y)[/tex]
 
  • #4
benorin
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Also, for [tex]x=2\tan (y)[/tex] we have [tex]dx=2\sec^2 (y)dy[/tex]
 
  • #5
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i got

original expression= (37/4)*((sec(y))/((sec(y))^2-1)*dy

but I am stuck.

I tried to do it so that its integral is ln(u)*du, but then sec's derivative would come up...
 
Last edited:
  • #6
HallsofIvy
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I don't see how you could have gotten that!
Your integral is
[tex]\int\frac{37dx}{x^2\sqrt{x^2+4}}[/tex]
Letting x= 2 tan y, gives,as you were told before, dx= 2 sec2y dy, and [itex]\sqrt{x^2+4}= \sqrt{4tan^2y+4}= 2sec y[/itex]. Of course x2= 4tan2 y. I don't see how you could have gotten that "-1" in "sec2y- 1".
 
  • #7
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1+(tany)^2=(secy)^2
(tany)^2=(secy)^2-1

4(tany)^2=4((secy)^2-1)

this is how i got it... what's wrong?
 
  • #8
HallsofIvy
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Do you mean you changed that x2 inthe denominator into
4tan2 y and then into 4(sec2y-1)?? Why?

As was point out before, if x= 2 tan y, then dx= 2 sec2 y
and x2+ 4= 4tan2y+ 4= 4 sec2y so that [itex]\sqrt{x^2y+ 4}= 2 sec y[/itex]. Your integral becomes
[tex]37\int \frac{2 sec^2 y dy}{(4tan^2 y)(2 sec y)}[/tex]
which reduces to
[tex]\frac{37}{4}\int\frac{sec y dy}{tan^2 y}[/tex]
Because of the odd power of sec y, I would be inclined to convert to sine and cosine now.
 
  • #9
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thanx for the guide!!! I did it.
 

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