Evaluating an Integral: √1+y^2-(cothφ)ydy

[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 want to 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!

 

[rock.freak667]

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

if that is your integral, then your final answer is wrong I believe.

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

 

[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?

 

[rock.freak667]

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?

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 ?

 

[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 I am basically integrating sec^3φ?

 

[rock.freak667]

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 I am basically integrating sec^3φ?

yep.

 

[mikemichiel]

thanks!