Calculus problem... antiderivative...please help...?

Question

For 0≤x<(π/2), an antiderivative of 2tanx is:

Choices:

a. ln(sec2x)
b. 2(secx)^2
c. ln((secx)^2)
d. 2ln(cosx)
e. ln(2secx)

What I did was, i used u substitution and made u=tanx...
du=(secx)^2 dx
I then put 1/(secx)^2 on the outside of the integral, substituted 2tanx with 2u, found the anti derivative of 2u which is (2u^2)/2...
then I put tanx back in, and got (tanx)^2...which I divided by (secx)^2 since 1/(secx)^2 was on the outside...
((tanx)^2)/((secx)^2)=(sinx)^2 but that is not one of the choices... so I probably did something wrong... not sure how to do this problem correctly... help...
thanks!!!

Answer 1

Let u = tan(x)
du = sec^2(x)dx
= (1 + u^2)dx
dx = du / (1 + u^2).

Then:
int(2tan(x))dx
= int(2u)du / (1 + u^2)
= ln( 1 + u^2)
= ln( 1 + tan^2(x) )
= ln (sec^2(x)).

Answer (c).

Answer 2

When you moved 1/(secx)² outside the integral, you blew it. It's part of the integration and you can only move constants across the integral sign.

FWIW, I get -2*ln|cos(x)| evaluated between 0 and π/2

HTH

Doug

Answer 3

Let's do it a different way.
Write tan x = sin x/cos x.
Then 2 ∫ sin x dx/ cos x = -2 ∫ du/u,
where u = cos x, du = -sin x dx.
So we get -2 ln u = -2 ln cos u = 2 ln sec u= ln(sec² x) + C.

Answer 4

You must take into account the fact that sin(1/x) and cos(1/x) are both functions of a function and therefore aren't just simple differentiations.

Answer 5

The closest answer is d.

However, the answer should be:

-2ln[cosx]