Answer to Question #3749 in Calculus for noor

Question

What is the integral of
& sec[sup]5[/sup] x dx ?

Answer 1

I = ∫ sec^5(x) dx = ∫ (sin^2(x) + cos^2(x))*sec^5(x) dx = ∫ sin^2(x)*sec^5(x) dx + ∫ sec^3(x) dx

Note: d(sin(x)*sec^4(x)) = cos(x)*sec^4(x) dx + sin(x)*4sec^3(x)*sec(x)*tan(x) dx = sec^3(x) dx + 4*sin^2(x)*sec^5(x) dx
so sin^2(x)*sec^5(x) dx = [d(sin(x)*sec^4(x)) - sec^3(x) dx]/4

so I = sin(x)*sec^4(x)/4 - (1/4) ∫ sec^3(x) dx + ∫ sec^3(x) dx
= (1/4)sin(x)*sec^4(x) + (3/4) ∫ sec^3(x) dx
= (1/4)sin(x)*sec^4(x) + (3/4) ∫ [sin^2(x) + cos^2(x)]*sec^3(x) dx
= (1/4)sin(x)*sec^4(x) + (3/4) ∫ sin^2(x) *sec^3(x) dx + (3/4) ∫ sec(x) dx

Note: d(sin(x)*sec^2(x)) = cos(x)*sec^2(x) dx + sin(x)*2*sec(x)*sec(x)*tan(x) dx = sec(x) dx + 2sin^2(x)*sec^3(x) dx
so sin^2(x)*sec^3(x) dx = (1/2) d(sin(x)*sec^2(x)) - (1/2) sec(x) dx

I = (1/4)sin(x)*sec^4(x) + (3/4)*(1/2) sin(x)*sec^2(x) - (3/4)*(1/2)∫ sec(x) dx + (3/4) ∫ sec(x) dx
= (1/4)sin(x)*sec^4(x) + (3/8)sin(x)*sec^2(x) + (3/8) ∫ sec(x) dx

∫ sec(x) dx = ∫ cos(x) dx / cos^2(x) = ∫ dsin(x) / [1 - sin^2(x)]
let sin(x) = y
∫ sec(x) dx = ∫ dy/(1 - y^2) = (1/2) ∫ [dy/(1-y) + dy/(1+y)] = (1/2) [-ln(1-y) + ln(1+y)] = (1/2) ln[(1+y)/(1-y)] = (1/2) ln[(1+sin(x))/(1-sin(x))]

So,

I = (1/4)sin(x)*sec^4(x) + (3/8)sin(x)*sec^2(x) + (3/16) ln[(1+sin(x))/(1-sin(x))].