Solve Integral: How to \int\sqrt{x^{2}+2}/x

[930R93]

Problem:
\int\sqrt{x^{2}+2}/x

Attempt:
Let x= \sqrt{2} sin\vartheta
dx= \sqrt{2} cos\vartheta d\vartheta

from this I got

\int\sqrt{2}sin\theta\sqrt{2}cos\theta/\sqrt{2}cos\theta

I think inverse substitution was not the right way to solve this problem...
any help would be greatly appreciated! Thanks!

 

[quantumdude]

Problem:
\int\sqrt{x^{2}+2}/x

Attempt:
Let x= \sqrt{2} sin\vartheta
dx= \sqrt{2} cos\vartheta d\vartheta

That's fine.

from this I got

\int\sqrt{2}sin\theta\sqrt{2}cos\theta/\sqrt{2}cos\theta

That's not fine. If x=\sqrt{2}\sin(\theta) then shouldn't there be a sine function in the denominator? And why is there a sine function in the numerator?

 

[930R93]

That's not fine. If x=\sqrt{2}\sin(\theta) then shouldn't there be a sine function in the denominator? And why is there a sine function in the numerator?

Yikes! I saw (2-x^2)^(1/2) in the numerator and used a trig identity to simplify the top...
so after you pointed this out i corrected the mistake and am left with nothing (from my point of view) i can simplify... i have root(2)*root(sin^2(O) +1)*root(2)*cos(O)dO over (root(2)*sinO)

umm...

 

[quantumdude]

Yikes! I saw (2-x^2)^(1/2)

That's not what you had in the original problem. What's going on here?

 

[aostraff]

If the posted question is the one you want to solve, think about the the trig identity relating tan and sec.

 

[930R93]

If the posted question is the one you want to solve, think about the the trig identity relating tan and sec.

Hey thanks! I am not sure why I didn't see this. I got it! thanks again!

 

[930R93]

That's not what you had in the original problem. What's going on here?

Sorry, it was a mistake on my end; I didn't give a very clear question. I used sine and cosine instead of tan and sec. whoops! it got it though but thanks for trying to help me, ill have to get better at asking if i want any help lol.

-930R93