caseclosed Posted September 23, 2006 Posted September 23, 2006 integrate (1+x^2)^(1/2) I used trig substitution and gotten integrate sec(theta)^3 but no clue what to do ....
uncool Posted September 23, 2006 Posted September 23, 2006 Have you learned integration by parts yet? =Uncool=
caseclosed Posted September 24, 2006 Author Posted September 24, 2006 yes, I tried setting u=(x^2+1)^(1/2) and dv=dx but I am not getting any log in the answer which means wrong. using the parts on sec(theta)^3 gets me nowhere, I change to (tan(x)^2+1)sec(x) and then set u=tan(x)^2 and dv=sec(x) dx which lead me to very complicated. I know I did this before but I don't remember how I did it.
JustStuit Posted September 24, 2006 Posted September 24, 2006 Use trig substitution such that [math]x[/math] = [math]\tan{\theta}[/math] Then it will be easy. [edit] When you get to [math]\int{\sec{\theta}^3 dx}[/math], use parts with [math]u = \sec{\theta} [/math] and [math] dv = \sec{\theta}^2 dx [/math] and then use trig identities to get the [math]\int{\sec{\theta}^3 dx}[/math] again. Once you have that, move it to the left side and you're done. (Sub in x) [/edit]
caseclosed Posted September 24, 2006 Author Posted September 24, 2006 ok, that's why... I should just leave it as sec^3 instead of changing using trig indentity. thank you very much.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now