Question

integrate sec2x[tan2x]

Answer 1

youre missing a 2

Answer 2

sec(2x)/2

Answer 3

nope

Answer 4

you are for it to be a normal looking derivative, wo use a s from 2/2 and integrate

Answer 5

sec[2x]tan[2x]

Answer 6

..... my keyboard hate me

Answer 7

use parts

Answer 8

i know

Answer 9

Just a better solution

Answer 10

anybody?

Answer 11

sec2x= a/(cos2x), tan2x=(sin2x/cos2x)
sec2x[tan2x] = a/(cos2x)(sin2x/cos2x)=asin2x/(cos2x)^2,let u=cos2x, du=-2sin2xdx
dx=(-1/2)(1/sin2x)du from here substitute this into (asin2x/(cos2x)^2)dx and replace all x with u and dx with du, so (asin2x/(cos2x)^2)(-1/2)(1/sin2x)du = (-1/2)(a)(1/u^2)du...then multiply (-1/2)(a) after integrating u^(-2) which is (-1/u) so (-1/2)(a) (-1/u)= (1/2u)=(a/2cos2x), so the solution y=(a/2cos2x)
to confirm this is correct take the derivative dy/dx of this and you will get
dy/dx=4asin2x/(4(cos2x)(cos2x))=asin2x/(cos2x)^2 which confirms solution is correct

Answer 12

hey piyere please look at my solution and let me know if there is any confusion about it...thanks