calculus

dx
I=∫ --------- (1)
x2 - a2

Substitute x = a sec 𝜃 in the expression x 2 - a 2

x 2 - a 2 = (a sec 𝜃) 2 - a 2 = a 2 sec 2 𝜃 - a 2 = a 2 sec 2 𝜃 - 1 ------- (2)

Since
1 + tan 2 𝜃 = sec 2 𝜃
tan 2 𝜃 = sec 2 𝜃 - 1 ----------- (3)
Putting equation (3) in equatio (2)

x 2 - a 2 = a 2 tan 2 𝜃
-

x2 - a2 = a 2 tan 2 𝜃 = a tan 𝜃 --------- (4)

Since
x = a sec 𝜃
Differentiating with respect to 𝜃

dx d d
= (a sec 𝜃) = a (sec 𝜃) = a sec 𝜃 tan 𝜃
d𝜃 d𝜃 d𝜃

dx = a sec 𝜃 tan 𝜃 d𝜃 ------------- (5)

Putting equation (4) & equation (5) in equation (1)


a sec 𝜃 tan 𝜃d𝜃
I=∫ = ∫sec 𝜃d𝜃
a tan

I = ln|sec 𝜃 + tan 𝜃| + C1 ---------- (6)

Since
x = a sec 𝜃