How to integrate inverse functions?

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Discussion Overview

The discussion revolves around the integration of inverse trigonometric functions, specifically focusing on methods such as integration by parts. Participants explore various approaches and clarify their understanding of the topic.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks for a brief explanation on how to integrate inverse trigonometric functions like inverse tangent.
  • Another participant initially misinterprets the question as asking for differentiation rather than integration, suggesting a method involving rewriting variables.
  • Several participants propose using integration by parts, with some suggesting different assignments for u and dv.
  • One participant introduces the acronym LIPET to help determine the choice of u in integration by parts, noting that it may not always apply.
  • Another participant mentions the derivative of the inverse tangent function and suggests calculating it before attempting the integral.
  • A later reply indicates that a different approach worked for them, confirming the flexibility in methods used for integration.
  • Another participant asks about integrating the inverse cosine function, indicating a broader interest in inverse trigonometric integrals.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to take when integrating inverse trigonometric functions, with no consensus reached on a single method. Some methods are suggested, but participants have not agreed on a definitive approach.

Contextual Notes

There are unresolved aspects regarding the choice of u and dv in integration by parts, and participants express uncertainty about the existence of a definite integral for inverse trigonometric functions.

Who May Find This Useful

Readers interested in mathematical integration techniques, particularly those involving inverse trigonometric functions, may find this discussion relevant.

KAS90
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hi there.. I want to know how to integrate inverse trigonometric functions?like inverse tanx for example?
thanx a lot..I just want a brief explanation?
 
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Welcome to PF!

KAS90 said:
hi there.. I want to know how to integrate inverse trigonometric functions?like inverse tanx for example?
thanx a lot..I just want a brief explanation?

Hi KAS90! Welcome to PF! :smile:

You want dy/dx for y = tan-1x.

So rewrite it x = tany, dx = sec2y dy

so dy/dx = cos2y

and then convert that back to a function of x.

Same method for any inverse fucnction! :smile:
 
I thought he asked abt integration! :huh?

Hint: integrate by parts
 


tiny-tim said:
Hi KAS90! Welcome to PF! :smile:

You want dy/dx for y = tan-1x.

So rewrite it x = tany, dx = sec2y dy

so dy/dx = cos2y

and then convert that back to a function of x.

Same method for any inverse fucnction! :smile:

I think the OP wanted to integrate, not differentiate. In that case may I suggest integration by parts.
 
… oops!

d_leet said:
I think the OP wanted to integrate, not differentiate. In that case may I suggest integration by parts.

oops! :redface:

yes … integrate by parts (starting with xtan-1x) :smile:
 
hey tiny-tim!
I thank u first for ur fast response and willingness to help..:)
but yes, I want to know about the integration of inverse trig functions..
for example, the integration of xtan-1x will be solved using integration by parts.. the question is:
u=x
du=dx
dv=tan-1x dx
v=?
or shall i solve it the other way round..
u=tan-1x dx
dv=xdx?
I mean, is there really a definite integral for inverse trig functions?
Thanx again 2 u, da_vinci,d_leet...
 
Yes you should try it the other way around. You know that finding the antiderivative of an inverse trig function (which is the subject of this post) is going to be harder than finding that of a polynomial. Thus, it makes sense that you want u = tan-1x so you'll be taking the derivative of it when you integrate by parts.

I use the acronym LIPET where L stands for natural logarithm, I stands for inverse trig function, P stands for polynomial function, E stands for exponential function, and T stands for trignometric function. Basically the letter that comes first in the acronym takes precedence for determining what u should equal. Of course this won't always work but if you look at why it's in that order, it might be of use.
 
KAS90 said:
the integration of xtan-1x will be solved using integration by parts.. the question is:
u=x
du=dx
dv=tan-1x dx
v=?
or shall i solve it the other way round..
u=tan-1x dx
dv=xdx?
I mean, is there really a definite integral for inverse trig functions?
Thanx again 2 u, da_vinci,d_leet...


This is quite confusing pile of differentials. Use the formula

[tex] D_x \textrm{tan}^{-1}x = \frac{1}{1+x^2}[/tex]

to first calculate

[tex] D_x\big(x\;\textrm{tan}^{-1}x\big) = \cdots[/tex]

and then

[tex] \int\limits_0^x \textrm{tan}^{-1}u\; du = \cdots[/tex]

should start look like more easy.
 
hi snipez
yeah, that system is useful most of the time... but it's not the case all the time..
thanx for ur response..

hi jostpuur
I haven't tried yet the way u suggested..I solved it by taking u as tan-1 x..so ,
u=tan-1x
dv=xdx

and it worked after manipulating numbers!
Thanx a lot ..I'll try ur way too :)
 
  • #10
hello somebody tell me how to intregate cos inverse x Dx
 
  • #11
images?q=tbn:ANd9GcQrFvW1r2bTk9GA1D4AIu5dhpS0Onf7NO6yCJTORSxkNSBWZRIk.jpg
 
  • #12
welcome to pf!

hello divyanshu9! welcome to pf! :smile:

make the obvious substitution :wink:
 

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