Integrals involving inverse trig functions

[kidmo87]

Hello everyone. I am having trouble with finding out a certain part of an equation. I've looked online, in my text book, and through my class notes, and my teacher really didn't show use an equation with this part. So I'm looking for some assistance if that's possible, thanks. I mainly need help with the "u" part, thanks.

 

[DrPhil]

Hello everyone. I am having trouble with finding out a certain part of an equation. I've looked online, in my text book, and through my class notes, and my teacher really didn't show use an equation with this part. So I'm looking for some assistance if that's possible, thanks. I mainly need help with the "u" part, thanks.

If you make the substitution

u = x - 2, du = -dx

then the integral becomes

\(\displaystyle \displaystyle \int\dfrac{-du}{3 + u^2}\)

then you can say \(\displaystyle a = \sqrt{3}\) and you are ready to plug in to the formula. Be sure to make the reverse transformation x = (u+2) for your final result.

 

[lookagain]

If you make the substitution

u = x - 2, du = > > -dx <<

then the integral becomes

\(\displaystyle \displaystyle \int\dfrac{-du}{3 + u^2}\)

then you can say \(\displaystyle a = \sqrt{3}\) and you are ready to plug in to the formula. Be sure to make the reverse transformation x = (u+2) for your final result.

This should be just "dx."

And then in the integral it must also be changed.