Question

How to solve for sin(x)/(cos^2(x))?

Answer 1

apparently it's sec(x)tan(x). I understand that 1/cos(x) = sec(x) and 1/cos^2(x) = sec(x) but why with the sin(x)/cos(x) do we get 1 sec and 1 tan...?

Answer 2

isn't it just
\[ \frac{sin(x)}{cos(x)}\cdot \frac{1}{cos(x)}\]

Answer 3

is there a sign in the middle of that?

Answer 4

no, multiplying

Answer 5

I was thinkikng that actually, thanks.

Answer 6

Now that I think about it, I just realized it's just sec(x)tan(x)... I thought it was sec^2(x) tan(x) nvm :p

Answer 7

derpa thanks Phi :P.

Answer 8

yes sin/cos = tan and 1/cos = sec

Answer 9

mhmhm :)

Answer 10

Actually, I was looking at wolfram and the answer key and realized I was konfused...

Answer 11

wolfram said sin(x)/cos^(x) is = sec(x)tan(x).

Answer 12

the actual question was (1)/(1-sin(x))

Answer 13

it got to a point where it was int of 1/cos^2(x) + int sin(x)/cos^2(x)

Answer 14

the professor changed it to - int -sin(x)/cos^2(x)

Answer 15

then did -cos^-1(x)/-1?

Answer 16

which turned into 1/cos(x) which turned into sec(x)....

Answer 17

@phi what did she do?

Answer 18

exactly what was (is) the original question?
is it an integral?

Answer 19

KonradZuse 0
the actual question was int (1)/(1-sin(x))

Answer 20

1/(1-sin(x)) * (1+sin(x))/(1+sin(x))

Answer 21

(1+sin(x))/(1-sin^2(x))

Answer 22

(1+sin(x))/cos^(x)

Answer 23

(1)/cos^2(x) + (sin(x))/(cos^2(x))

Answer 24

first part is sec^2(x) second part would be sec(x)tan(x)... Problem is you need to integrate it...

Answer 25

:(?

Answer 26

I would write sec(x)tan(x) as

\[\int cos^{-2}(x) sin(x) dx \]
now this is in the form
\[ \int u^{-2} du \]

Answer 27

with a minus sign in there

Answer 28

u = cos(x)
du = -sin(x) dx

Answer 29

oic that makes snese now, thanks :)