Question

Find the antideravative for the given function:

f(x) = sinx/(cosx)^2

Answer 1

\[u=\cos x\]what then is \(du\) ?

Answer 2

(cosx)^2?

Answer 3

Oh wait, du would be, -sinx

Answer 4

\[du=-\sin xdx\]the dx is actually very important, but yes

Answer 5

Okay

Answer 6

the integral you have is\[\int{\sin xdx\over\cos^2x}\]now substitute\[u=\cos x\]\[du=-\sin xdx\implies\sin xdx=-du\]into the integral
what do you get?

Answer 7

I'm sorry, I'm actually confused.

Answer 8

It's just that we have been introduced to this method yet, so I'm not sure what you're doing.

Answer 9

have not been *

Answer 10

we are making use of the fact that the numerator is almost the derivative of the function in the denominator
do yo agree that\[du=-\sin xdx\]?

Answer 11

yes

Answer 12

so then can you see that\[\sin xdx=-du\]
?

Answer 13

yes

Answer 14

so we can substitute in\[\int{\sin xdx\over\cos^2x}\]\[\cos x=u\]\[\sin xdx=-du\]and we get\[\int\frac{-du}{u^2}=-\int u^{-2}du\]which you should be able to integrate now

Answer 15

so the answer is -1/cosx + c?

Answer 16

i thinnk the -

Answer 17

...goes away

Answer 18

Okay, got it.Thank you! :)

Answer 19

welcome :)