How to Derive the Integral of Sqrt[x^2-a] Using Trigonometric Substitution?

[coverband]

I believe the answer is 0.5(x(x^2-a)^0.5)-0.5aln(x+(x^2-a)^0.5)

does anyone know how to get this/derive this i.e. not take from tables !

Thanks

 

[flatmaster]

I believe what you can do is write out the expression and take the log of both sides. WHen taking the derivative, you must remember y is a function of x.

y = Sqrt[x^2 - a]
y = [x^2 - a]^(1/2)

Take natural log

Ln[y] = Ln{ [x^2 - a]^(1/2) }
Ln[y] = (1/2) Ln{[x^2 - a]}

Take derivative wrt x. Don't forget you must apply chain rule to right hand side. Where y' comes out anyway.

(1/y) (y') = 2x/ (2 (x^2 - a))

writing more neatly, cancel a two

(y'/y) = x / (x^2 - a)

Finally, multiply a y back over and resub your original y.

y' = yx/ (x^2 - a) y = [x^2 - a]^(1/2)


Thus
y' = x [x^2 - a]^(1/2) / (x^2 - a)

y' = x(x^2 - a)^(-1/2)

 

[flatmaster]

I'm an idiot, I took a deravitive. Not even the easiest way either.

 

[Defennder]

Use a trigo substitution. Specifically, look at the ones for sec and tan.