Indefinite Integral?

sausu

New member
Joined
Jan 25, 2011
Messages
1
integral 1/(x(sqrt(x^2 - 4)))

It's some secant inverse...Here's what I got so far:

Int(1/(x(Sqrt(4(x^2 /4)-1)
Int(1/(2x(Sqrt((x^2 /4)-1)
1/2 int(1/(x(Sqrt((x /2)^2)-1)
U-sub
u=x/2
du=1/2 dx
2du= dx
(This is where I hit a wall..I have no clue what I'm doing)

Please help me and show step by step.
 
There are various ways to tackle it. Do you have to use a certain method?.

Trig sub works well with this one.

1xx24dx\displaystyle \int\frac{1}{x\sqrt{x^{2}-4}}dx

Let x=2sec(t),   dx=2sec(t)tan(t)dt\displaystyle x=2sec(t), \;\ dx=2sec(t)tan(t)dt

Make the subs:

2sec(t)tan(t)2sec(t)(2sec(t))24dt\displaystyle \int\frac{2sec(t)tan(t)}{2sec(t)\sqrt{(2sec(t))^{2}-4}}dt

tan(t)4(sec2(t)1)dt\displaystyle \int\frac{tan(t)}{\sqrt{4(sec^{2}(t)-1)}}dt

Remember the identity tan2(t)=sec2(t)1\displaystyle tan^{2}(t)=sec^{2}(t)-1

12dt=t2\displaystyle \frac{1}{2}\int dt=\frac{t}{2}

Now, resub. Noting that t=sec1(x2)\displaystyle t=sec^{-1}(\frac{x}{2})

12sec1(x2)=12cos1(2x)\displaystyle \boxed{\frac{1}{2}sec^{-1}(\frac{x}{2})=\frac{1}{2}cos^{-1}(\frac{2}{x})}

You may see this written in some form that looks completely different, but is actually equivalent.

Such as, 12tan1(x242)\displaystyle \frac{1}{2}tan^{-1}\left(\frac{\sqrt{x^{2}-4}}{2}\right)
 
Top