Integral question

  • #1
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i added a file with of the integral
and how i "solved" it

on the final stage of my integral i gut stuck on a
complicated fuction

is there a way to change it so it will look simpler

plz help
 

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Answers and Replies

  • #2
StatusX
Homework Helper
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So you're asking if you can simplify [itex]\sin(2\tan^{-1}(x))[/itex]? Use the double angle formula to get it in terms of [itex]\sin(\tan^{-1}(x))[/itex] and [itex]\cos(\tan^{-1}(x))[/itex]. Then if you put [itex]u=\sin(\tan^{-1}(x))[/itex], you have [itex]\cos(\tan^{-1}(x))=\sqrt{1-u^2}[/itex] and:

[tex]\frac{u}{\sqrt{1-u^2}}=\frac{\sin(\tan^{-1}(x)}{\cos(\tan^{-1}(x))}=\tan(\tan^{-1}(x))=x,[/tex]

which you can solve for in terms of u.
 
  • #3
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you said u=sin(arctan(x))
how come i have cos(arctan(x))=(1-u^2)^0.5

i didnt understand what to do step by step

???
 
  • #4
StatusX
Homework Helper
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Because cos^2+sin^2=1.
 
  • #5
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0
but the "u" expression represents x

i heard of some triangle method of solving stuff like
[itex]
\sin(2\tan^{-1}(x))
[/itex]

??
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
956
but the "u" expression represents x
No, u= sin(arctan(x))

i heard of some triangle method of solving stuff like
[itex]
\sin(2\tan^{-1}(x))
[/itex]

??
For something simple, like sin(tan-1(x)), you can imagine a triangle with "opposite side" x and "near side" 1 so that "opposite side over near side" = tan(angle)= x and angle= tan-1(x). Then sin(tan-1(x))= sin(angle) which is "near side over hypotenuse". Use the Pythagorean theorem to find the length of the hypotenus: [itex]\sqrt{x^2+ 1}[/itex] and you get sin(tan-1(x))= [itex]x/\sqrt{x^2+ 1}[/itex].

However, the "2" multiplying tan-1(x) makes that much harder.
 
  • #7
1,395
0
how to build the new integral
i need to build a du for that which i the derivative of
u= sin(arctan(x))

what is the full new integral?
 

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