How to integrate inverse functions?

[KAS90]

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?

 

[tiny-tim]

Welcome to PF!

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!

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!

 

[da_vinci]

I thought he asked abt integration! :huh?

Hint: integrate by parts

 

[d_leet]

Hi KAS90! Welcome to PF!

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!

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

 

[tiny-tim]

… oops!

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

oops!

yes … integrate by parts (starting with xtan^-1x)

 

[KAS90]

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...

 

[snipez90]

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.

 

[jostpuur]

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

<br /> D_x \textrm{tan}^{-1}x = \frac{1}{1+x^2}<br />

to first calculate

<br /> D_x\big(x\;\textrm{tan}^{-1}x\big) = \cdots<br />

and then

<br /> \int\limits_0^x \textrm{tan}^{-1}u\; du = \cdots<br />

should start look like more easy.

 

[KAS90]

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 :)

 

[divyanshu9]

hello somebody tell me how to intregate cos inverse x Dx

 

[divyanshu9]

 

[tiny-tim]

welcome to pf!

hello divyanshu9! welcome to pf!

make the obvious substitution ​