Help with integrals?

Question

what is the integral of (sec^2)3x?

i'd appreciate any help. thanks,

Answer 1

d/dx (tanx) = (secx)^2

proved by the quotient rule:

let y = tan x
y = sinx/cosx
quotient rule: dy/dx = (vdu/dx-udv/dx)/(v^2)
dy/dx = [(cosx)^2+(sinx)^2]/[(cosx)^2]
dy/dx = 1/[(cosx)^2] = (secx)^2

so d/dx (tanx) = (secx)^2 ... memorise it mate xD... everyone seems to use substitution, though this works just as well

so integral (sec3x)^2 = 1/3 (tan 3x)

Answer 2

Let u = 3x x = u/3 dx 1/3 du.
So we get 1/3 ∫ sec² u du = 1/3 tan u = 1/3 tan 3x + C.

Answer 3

integral of sec^2 (3x)
∫sec^2(3x) dx

u = 3x
du = 3 dx
1/3 du =dx

1/3 ∫sec^2(u) du

integral of sec^2 is tan

1/3 tan(u)

replace u with 3x:


1/3 tan(3x) +C

Answer 4

use substitution

S sec^2 (3x) dx
u=3x, du=3dx, dx=1/3du
1/3 S sec^2 (u) du
1/3 tan u
1/3 tan 3x