Integral check

1. Jul 19, 2009

mikemichiel

Hey i have this integral √1+y^2-(cothφ)ydy(with the square root on consisting of 1+y^2. I evaluated it but I just wanna you guys to check and see if you think its good. For the radical part I used a trig substitution y=tanφ
The integral of tanφ is
ln |sec x|
And for the (cothφ)y part I figured it would just be (cothφ)y^2/2 because the function is in terms of dy and I figured I just integrate with respect to y. So this gives me the final answer
ln |sec x|-(cothφ)y^2/2 for my final. Does this look good to you guys? My only concern about this is since my y=tanφ would I have to substitute it for the the y after the cothφ?
Or I might be thinking of this completely wrong. Need advice!!

2. Jul 19, 2009

rock.freak667

$$\int ( \sqrt{1+y^2} -y coth \phi) dy$$

For

$$\int \sqrt{1+y^2} dy$$

if y=tan$\theta$, then what is dy= ?
Then you will need to integrate more than tan$\theta$

3. Jul 19, 2009

mikemichiel

so my dy would be sec^2φ.
So if i plug that in il get
ln |sec x|-ycothφsec^2φ. So would i have to plug in my y=tanφ for my y in front of the cothφ? If not would it just be ln |sec x|-y^2/2cothφsec^2φ?

4. Jul 19, 2009

rock.freak667

Well firstly it is not wise to put the substituting variable as a symbol in the the integrand.

So if you put y=tan$\theta$, use $\theta$ as the new variable and not $\phi$

In your original post, -(cothφ)y2/2 this part is correct. It is the first anti-derivative that is wrong.

so back to it now. Yes dy= sec$^2 \theta \ d\theta$

$$\int \sqrt{1+y^2} dy \equiv \int (\sqrt{1+tan^2\theta} )sec^2\theta d\theta$$

Now do you know an identity for $1+tan^2 \theta$ ?

5. Jul 20, 2009

mikemichiel

yes it would be sec^2φ. Then I would square that and it would become secφ then multiply it be the sec^2φ I had for my dy. So im basically integrating sec^3φ?

6. Jul 20, 2009

rock.freak667

yep.

7. Jul 20, 2009

thanks!!