Help, what is ∫tanxsec²xdx?

Question

Is it (sec²x)/2 of (tan²x)/2? I tried two ways of solving this one:

The first one I let u=tanx,

du=sec²xdx

we get ∫udu, where the answer is (tan²x)/2 + C.

The second one I change the form ∫tanxsec²xdx = ∫secx(secxtanxdx).

Then I let u=secx

du=secxtanxdx

We also get ∫udu, but with u=secx, we get an answer of (sec²x)/2+C.

Help.

Answer 1

Both are correct. Odd! Perhaps the mystery lies in the value of C. Try evaluating both results with chosen limits and see what happens.

Answer 2

I = ∫ tan x sec ² x dx

Let u = tan x

du = sec ² x dx

I = ∫ u du

I = u ² / 2 + C

I = (1/2) tan ² x + C

Answer 3

both of your answers are corect and you should khnow we hand many answers for each integral and the diffrences between these answersare in the c value . in your answers secx^2 +c1=1+tanx^2 +c1 =tanx^2+1+c1=tanx^2+c2 ( c1 and c2 are diffrente values

Answer 4

As far as I know, letting u = tan x and du = sec^2 x dx is correct. I got tan^2(x) / 2 and differentiation appears to verify this answer.