- #1

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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

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- Thread starter helpm3pl3ase
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In summary: WarrenIn summary, the conversation discusses how to find the anti-derivative of a given function, specifically h(x) = sinx/cos^(2) x. The process involves converting the function into a simpler form, using substitution to find the anti-derivative, and recognizing that sin(x)/cos(x) is also known as tan(x) and 1/cos(x) is known as sec(x). The final answer is (secx)(tanx) + c.

- #1

- 79

- 0

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

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- #2

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

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

- Warren

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

- Warren

- #3

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so the answer would be

(secx)(tanx) + c

Correct??

(secx)(tanx) + c

Correct??

- #4

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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

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

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- Warren

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

and c what u get.

Daniel.

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

to h(x)= (secx)(tanx).

The formula for (secx)(tanx) is (1/cosx)(sinx/cosx).

To simplify (1/cosx)(sinx/cosx), you can use the identity sinx/cosx = tanx. This will give you (1/cosx)(tanx), which is equivalent to (secx)(tanx).

(secx)(tanx) is equivalent to (1/cosx)(sinx/cosx) because of the trigonometric identity secx = 1/cosx. This means that (secx)(tanx) can be rewritten as (1/cosx)(tanx), which can then be further simplified to (1/cosx)(sinx/cosx).

Yes, you can also use the identity secx = 1/cosx and tanx = sinx/cosx to get (secx)(tanx) from (1/cosx)(sinx/cosx). This will give you (1/cosx)(1/cosx)(sinx/cosx), which can be simplified to (1/cosx)(sinx/cosx).

Yes, you can think of (secx)(tanx) as the product of two fractions: (1/cosx) and (sinx/cosx). When multiplied, the denominator of (1/cosx) and the numerator of (sinx/cosx) cancel out, leaving you with (1/cosx)(sinx/cosx) or (secx)(tanx).

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