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)}