What Formula Solves the Integral of dy / (1+y^2)?

[JJ420]

integrate with a formula??

Homework Statement

dy / (1+y^2)

Homework Equations

The Attempt at a Solution

i tried using substitution technique but that did not yield the correct answer...
u = 1 + y^2
du = 2ydy

is there a formula i should be looking for maybe??
any help or suggestions would be great

 

[HallsofIvy]

You can't use that substitution because there is no "y" in the integral.
What is the derivative of z= arctan(y)?

 

[ranger]

You could use trigonometric substitution.

http://en.wikipedia.org/wiki/Trigonometric_substitution

 

[JJ420]

sorry I'm not sure...u mean i have the derivative of arctan y??
so the correct solution would be arctan y^2 + C??

 

[JJ420]

?......

 

[ranger]

JJ420, your solution is incorrect. Did you refer to the link I gave you? It explicitly uses an intergal similar to yours as an example. You need to use trigonometric substitution to solve this. Sometimes trig substitution is better understood by drawing a right triangle.

 

[Dr Game]

im confused could the integrual be $$\frac{1}{2y^2}ln|\frac{1-y^2}{1+y^2}| + C$$ ?

 

[ranger]

im confused could the integrual be $$\frac{1}{2y^2}ln|\frac{1-y^2}{1+y^2}| + C$$ ?

How did you get that The integral has a trigonometric function.

 

[Dr Game]

then its got to be

$$\frac{1}{y^2}Tan^-^1+C$$

 

[ranger]

then its got to be

$$\frac{1}{y^2}Tan^-^1+C$$

That is also incorrect. I wish I could post the correct answer, but that would go against PF guidelines of offering help in the form of a direct answer. Here is a hint - refer to the wiki link I posted a few replies up.

 

[dextercioby]

Homework Statement

dy / (1+y^2)

Homework Equations

The Attempt at a Solution

i tried using substitution technique but that did not yield the correct answer...
u = 1 + y^2
du = 2ydy

is there a formula i should be looking for maybe??
any help or suggestions would be great

Use partial fractions:

$$\frac{1}{1+y^{2}}=\frac{1}{2}\left(\frac{1}{1+iy}+\frac{1}{1-iy}\right)$$

and then some simple substitutions and finally express the answer as the natural logarithm of a fraction times $\frac{1}{2i}$.

Daniel.

 

[mjsd]

trig substitution is the most instructive way of doing it although the complex number way suggested by dextercioby is probably a good trick to learn.
Way of thinking:
you have something you can't do, look nothing like what u used to seeing, so you need something new, some subtle substitution...
why trig substition?
simply because you have something in the denominator $$1+y^2$$ which looks like one side of a trig identity! Observe that $$1-y^2$$ also looks like one side of a trig identity too. Now you go to the book and look at all your trig identities between $$\sin, \cos, \tan, \cot, \sec, \csc$$, which one do you think it could be useful? Remember we have $$y^2$$, so...?

now once you have picked the right one (and change of variable, that's why it is called trig substitution), you do the integrals (typically in terms of just a combination of $$\sin, \cos, \tan, \cot, \sec, \csc$$. But if you pick the right identity to start with, the identity itself will help you simplifies this expression and it should be quite easily done given that you know how to do these integrals with trig functions. Final step is to convert everything back to the variable $$y$$ and that's where you get your inverse tan...

 

[HallsofIvy]

sorry I'm not sure...u mean i have the derivative of arctan y??
so the correct solution would be arctan y^2 + C??

I asked before "what is the derivative of arctan y?" I was asking you to determine that the derivative of arctan y is [itex]frac{1}{1+ y^2} so that the anti-derivative of $\frac{1}{1+y^2}$ is arctan y. NOT <br /> arctan y²+ C, just arctan y+ C.<br /> <br /> While that can be derived using trig substitutions, it occurs often enough, for example, in integrations using "partial fractions" that it is worth memorizing as a specific integral formula.[/itex]

 

[JJ420]

thanx to everyone for the help...i'm definitely on the right course now...i don't know about DrGame tho