Solving Undefined Integrals with the Help of Arctan and Integration by Parts"

[ruiwp13]

Homework Statement

Solve the following integral:

∫arctg(1/√x)dx

Homework Equations

∫u'/1+u^2 = arctg(u)+c
∫u.v' = u.v - ∫ u'.v

The Attempt at a Solution

So , I tried to define u as arctg(1/√x) but I'm having trouble finding the du. The derivate of arctg(1/√x) is -1/2*x^(3/2), so it stays -1/2*x^(3/2)/1-(1/√x)^2 (converting to the derivate u'/1+u^2) ? I was thinking about getting the derivate of arctg(1/√x) and doing this by parts. dv = dx , u = arctg(1/√x)

 

[LCKurtz]

The derivative of $\arctan(x^{-\frac 1 2})$ is$$
\frac 1 {1+\frac 1 x}\left(-\frac 1 2 x^{-\frac 3 2}\right )=
-\frac 1 2 \frac{x^{-\frac 1 2}}{x+1}$$That might help your integration by parts.

 

[ruiwp13]

The derivative of $\arctan(x^{-\frac 1 2})$ is$$
\frac 1 {1+\frac 1 x}\left(-\frac 1 2 x^{-\frac 3 2}\right )=
-\frac 1 2 \frac{x^{-\frac 1 2}}{x+1}$$That might help your integration by parts.

Thank you, now I got arctg(1/√x)*x - ∫((√x)/2x+2)*x . Do I need to make the integral by parts again?

 

[Ray Vickson]

Homework Statement

Solve the following integral:

∫arctg(1/√x)dx

Homework Equations

∫u'/1+u^2 = arctg(u)+c
∫u.v' = u.v - ∫ u'.v

The Attempt at a Solution

So , I tried to define u as arctg(1/√x) but I'm having trouble finding the du. The derivate of arctg(1/√x) is -1/2*x^(3/2), so it stays -1/2*x^(3/2)/1-(1/√x)^2 (converting to the derivate u'/1+u^2) ? I was thinking about getting the derivate of arctg(1/√x) and doing this by parts. dv = dx , u = arctg(1/√x)

You should not get u'/1+u^2, which reads as
$$\frac{u'}{1} + u^2.$$ You should have gotten u'/(1+u^2), which reads as
$$\frac{u'}{1+u^2}.$$ Parentheses are important!

RGV

 

[ruiwp13]

yes, I got this x.arctg(1/√x)+∫x/sqrt(x)*(2x+2) but now I'm having trouble on solving the integral :p any help?

 

[haruspex]

yes, I got this x.arctg(1/√x)+∫x/sqrt(x)*(2x+2) but now I'm having trouble on solving the integral :p any help?

Substitute for √x and use partial fractions.

 

[ruiwp13]

I don't remeber how to use the partial fractions , it's that thing with A/(x-1) * B/(x+1) and A*(x+1)*B(x-1) thing? Can't remember quite well

 

[haruspex]

I don't remember how to use the partial fractions , it's that thing with A/(x-1) * B/(x+1) and A*(x+1)*B(x-1) thing?

It's that thing where you have one polynomial divided by another, P(x)/Q(x). You rewrite it as a sum of simpler fractions. http://en.wikipedia.org/wiki/Partial_fractions_in_integration

 

[ruiwp13]

I transformed x/sqrt(x)*(2x+2) into sqrt(x)/2x+2 . Then I passed 2x+2 up and got Int (sqrt(x)*(2x+2)^-1) and multiplied and got Int (1/2sqrt(x)) + Int (sqrt(x)/2) but the second one is incorrect I think... Can't figure out why, tho.

Thanks in advance

 

[haruspex]

I transformed x/sqrt(x)*(2x+2) into sqrt(x)/2x+2 .

As I indicated, you first need to get rid of the surd by substituting a different variable for √x.