help with this.?

Question

tan^4 x dx
: tan ^ 2 x ( sec ^2 x -1 )
tan^2 x sec^2 x dx - tan^2 x
= 1 / 3 tan^3 x - tan x + x + c

why is the integration of tan^2 x sec^2 x is 1/3 tan^3 x?

and how do u integrate tan^5 x then?

Answer 1

Well, your first example looks fine - I don't know if you need any help with that or not.

For your second question, the derivative of tan(x) = sex^2(x)dx. So ∫tan^2(x)sex^2(x)dx = ∫u^2du. The integral of that is (1/3)u^3 + C...and since u = tan(x), the answer is (1/3)tan^3(x).

For your third question, I'll work it out for you.

∫tan^5(x)dx

∫[tan^3(x)][tan^2(x)]dx

∫[tan^3(x)][sec^2(x) - 1]dx

∫tan^3(x)sec^2(x)dx - ∫tan^3(x)dx

∫tan^3(x)sec^2(x)dx - ∫[tan(x)][tan^2(x)]dx

∫tan^3(x)sec^2(x)dx - ∫[tan(x)][sec^2(x) - 1]dx

∫tan^3(x)sec^2(x)dx - ∫tan(x)sec^2(x)dx + ∫tan(x)dx

(1/4)tan^4(x) - (1/2)tan^2(x) - ln|cos(x)| + C

You could have gotten sex^2(x) for the second term instead, but this is just 1 way of solving.

Answer 2

Integral ( tan² x sec² x ) dx

let u = tan x
du = sec² x dx => dx = du / sec² x

=
Integral u² sec² x (du / sec² x)
=
Integral u² du
=
u³/3 + c

Substitute tan x back in:

=
1/3 tan³ x + C


To find tan^5 x, say, it would probably be easiest to use the reduction method.