What is the integral of sec2x tan2x dx?

Question

Please explain how you got the answer. First one to get it right gets the 10 pts!

Answer 1

try to replace 2x with some other variable like t = 2x
once you do that its pretty clear.. :) else try something like thinking backwards... :)

int (sec2x.tan2x.dx) = int( d/dx(sec2x) *(1/2).dx)
hence the answer is

= (Sec2x)/2 + C

Answer 2

Here's another way to do this problem.

Use the fact that sec2x = 1/cos2x and tan2x=sin2x/cos2x.
Now you have the integral of sin2x/(cos2x)^2.
Let u = cos2x, du = -2sin2x dx, so you have -1/2 *integral of
u^(-2) dx.
Use the power rule, get the answer 1/2 * u^(-1) +c,
substitute u = cos2x and you have your answer.

Answer 3

let 2x=y
int. sec2x tan2x dx
= (1/2) * int. secy tany dy
= (1/2) * sec y +c
= sec 2x /2 +c

Answer 4

Let us write f(y) = sec y tan y.
If now y be a function of x, then integral of f(y) dx = (integral of f(y) dy) / (dy/dx).

Now in our problem, y=2x and dy/dx=d(2x)/dx = 2.
Therefore, integral of sec 2x tan 2x dx = (sec 2x)/2.
Since, integral of sec y tan y dy =sec y.

Answer 5

f=ma if you sat in a gyro that spun instanious you would not spin

Answer 6

1/2*sec(2x)