Question

Prove that:

`int 1/sqrt(x^2 - a^2) dx = log |x + sqrt(x^2 - a^2)| + c`.

Answer 1

Let I = `int (dx)/sqrt(x^2 - a^2)`

Put x = a sec θ   

∴ dx = a sec θ tan θ dθ and sec θ = `x/a`

Also, `sqrt(x^2 - a^2)`

= `sqrt(a^2 sec^2 θ - a^2)`

= `asqrt(sec^2θ - 1)`

= a tan θ

∴ tan θ = `sqrt(x^2 - a^2)/a`

∴ I = `int(asecθtanθ)/(atanθ) dθ = int secθ dθ`

= `log|secθ + tanθ| + c_1`

= `log|x/a + sqrt(x^2 - a^2)/a| + c_1`

= `log|(x + sqrt(x^2 - a^2))/a| + c_1`

= `log|x + sqrt(x^2 - a^2)| - log a + c_1`

= `log|x + sqrt(x^2 - a^2) + c`, where c = – log a + c₁

∴ `int 1/sqrt(x^2 - a^2) dx = log|x + sqrt(x^2 - a^2)| + c`.