Inegration with Substitution

[Unregistered]

how do i ∫1 / (1 + x ) to ∫cos θ .δ θ when x = tanθ ?

[KatieB]

∫1 / (1 + x ) to ∫cos θ .δ θ when x = tanθ


x = tanθ
diff wrt θ

dx/dθ = secθ
= 1+tanθ
so
dx = (1+tanθ) dθ


∫1 / (1 + x )dx = ∫[1 / (1+tanθ)] (1+tanθ) dθ

cancel (1+tanθ)

to give
∫1 / (1 + x )dx = ∫[1 / (1+tanθ) ]dθ
=∫[1 / (secθ) ]dθ
=∫cos θ dθ