How to get (secx)(tanx) from (1/cosx)(sinx/cosx)?

Quick question:

h(x) = sinx/cos^(2) x

= (1/cosx)(sinx/cosx)

Then you get (secx)(tanx)..

I do not get how they get secx x tanx?? Anyone?? Thanks
 

chroot

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1/cos(x) is also called sec(x).

sin(x)/cos(x) is also called tan(x).

- Warren
 
so the answer would be

(secx)(tanx) + c

Correct??
 

chroot

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All you've done so far is convert the function you gave me into a slightly simpler form.

sin(x) / cos^2(x) = sec(x) tan(x).

Since you didn't actually post the problem as it was given to you, I don't know if h(x) is a function of which you need to find the antiderivative, or whether you've already done that step. You probably need to actually perform the antiderivative now.

- Warren
 
sorry how would i go about doing this.. Iam so confused.
 

chroot

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Find the function which has a derivative of sec(x) tan(x). You should have a list of such facts in your book.

- Warren
 
alright.. I get it now.. Sorry.. I dont know why this problem was causing me problems.. Thanks for clearing it up.
 

dextercioby

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Do you know the method of substitution to find antiderivatives ? If so, just plug

[tex] \cos x = t [/tex]

and c what u get.

Daniel.
 

HallsofIvy

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It would have helped if you had told us from the beginning that you were trying to find an anti-derivative! All you said was that you couldn't see how they had gone from Quick question:

h(x) = sinx/cos^(2) x

to h(x)= (secx)(tanx).
 

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