Need help with math
Not asking for a step-by-step process of how to do this, but does anyone know how to approach this integral?
∫ 1/[sq.rt(x)*(x +6)] dx
I have no idea how to integrate it. After putting it in wolframalpha, all I could think of would be that integrating it involves trig substitution, but I gave it a shot and it seemed like hell trying to simplify what came out. The only other thing that comes to mind is partial fractions, but I can only do that with linear factors and irreducible quadratics.
September 26, 2012
4 Comments • Newest first
Someone's doing inverse trig integrals~
I'm on that section in cal 2, as well. Good look with it, man. Trig's never been my strong point.
[quote=Oyster]If you rewrite the (x+6) -> 6(1+ x/6)
You'll begin to see that the answer is some constant*arctan[sqrt(x/6)]
Easiest method is [b]trig substitution[/b]:
sqrt[x]=sqrt[6]*tan[t]
......x=6*tan[t]^2
.....dx=12*tan[t]*sec[t]^2 dt
(1+[x/6]) = sec[t]^2
After trig substitution your integral should look like: [1/6]*Integral{ {12*tan[t]*sec[t]^2 dt} / {sqrt[6]*tan[t]*sec[t]^2} }
Which all simplifies to sqrt[2/3]*Integral[1*dt] and t=arctan[sqrt{x/6}][/quote]
:/ I distributed in the beginning and it made the substitution a lot harder than it needed to be.
Thanks a ton to both of you guys
substitution seems to be the way to go
try to see if you ain't messing up a number/letter/sign on the way
If you rewrite the (x+6) -> 6(1+ x/6)
You'll begin to see that the answer is some constant*arctan[sqrt(x/6)]
Easiest method is [b]trig substitution[/b]:
sqrt[x]=sqrt[6]*tan[t]
......x=6*tan[t]^2
.....dx=12*tan[t]*sec[t]^2 dt
(1+[x/6]) = sec[t]^2
After trig substitution your integral should look like: [1/6]*Integral{ {12*tan[t]*sec[t]^2 dt} / {sqrt[6]*tan[t]*sec[t]^2} }
Which all simplifies to sqrt[2/3]*Integral[1*dt] and t=arctan[sqrt{x/6}]