Question: Integrate (sec^3 x) dx

I = int sec^3(x) dx

Let u = secx

Then du = secx.tanx dx = sinx/cos^2x = √(1 - 1/u^2).u^2 dx = u√(u^2 - 1) dx

i.e. dx = du/[u.√(u^2 - 1)]

I = int u^3/[u.√(u^2 - 1)] du = int u^2/[√(u^2 - 1)] du

I = int u^2/[√(u^2 - 1)] du

Let u = cosh(t)

then du = sinh(t) dt = √(u^2 - 1) dt

So, I = int u^2/[√(u^2 - 1)] du = int cosh^2(t) dt

I = int cosh^2(t) dt = (1/2) int 1 + cosh(2t) dt = (1/2)(t + (1/2)sinh(2t))

I = (1/2).t + (1/2).sinh(t).cosh(t)

I = (1/2).t + (1/2).√(cosh^2(t) - 1).cosh(t)

I = (1/2).t + (1/2).√(u^2 - 1).u

I = (1/2)arccosh(u) + (1/2).√(sec^2(x) - 1).sec(x)

I = (1/2)arccosh(sec(x)) + (1/2).√(tan^2(x)).sec(x)

**I = (1/2){arccosh(sec(x)) + tan(x).sec(x)}**