Calculus II: ∫ sec³(x) dx

0. ∫ sec³(x) dx

I = ∫ sec³(x) dx

1. ∫ sec(x) dx

K = ∫ sec(x) dx

sec(x) = 1/cos(x) = A/B

A = (1 + sin (x))/cos²(x)
B = (1 + sin (x))/cos(x)

dB/dx = [cos²(x) + (1 + sin (x))sin(x)]/cos²(x) =
[(1 + sin (x)]/cos²(x) = A

Then

dB/dx = A = B/cos(x) dx = B sec(x) dx

dB/B = dx/cos(x)

∫ dx/cos(x) = ∫ dB/B = ln|B| + cst.

Therefore

K = ∫ sec(x) dx = ∫ dx/cos(x) = ln|(1 + sin (x))/cos(x)| + cst =

ln|(sec(x) + tan(x))| + cst

K = ∫ sec(x) dx = ln|(sec(x) + tan(x))| + cst

2. ∫ tan(x). sin(x) sec²(x) dx

J = ∫ tan(x). sin(x) sec²(x) dx =
∫ sin²(x) sec³(x) dx =
∫ [1 - cos²(x)] sec³(x) dx =
∫sec³(x) dx - ∫cos²(x) sec³(x) dx
= I - ∫ dx /cos(x) = I - ∫ sec(x) dx = I - K

J = I - K

J = I - K = ∫ tan(x). sin(x) sec²(x) dx =
∫ sec³(x) dx - ∫ sec(x) dx

3. ∫ sec³(x) dx

I = ∫ sec³(x) dx = ∫ sec(x) sec²(x) dx

Let

u = sec(x), so du = sin(x)/cos²(x) dx = sin(x) sec²(x) dx
dv = sec²(x) dx , so v = tan(x)

Therefore

I = ∫ sec(x) sec²(x) dx = ∫ u dv = u v - ∫ v du =
sec(x). tan(x) - ∫ tan(x). sin(x) sec²(x) dx =
sec(x). tan(x) - J

so

I = sec(x). tan(x) - I + K

Threfore

2 I = sec(x). tan(x) + K
I = (1/2) sec(x). tan(x) + (1/2)ln|(sec(x) + tan(x))| + cst

∫ sec³(x) dx = (1/2) sec(x). tan(x) + (1/2)ln|(sec(x) + tan(x))| + cst