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- #1

- Thread starter shamieh
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- #1

- Jan 30, 2012

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The definition of $\sec x$.What is the identity they are using to do this?

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- #3

What??? What do you mean?The definition of $\sec x$.

\(\displaystyle \sec x \, dx = ln|\sec x + \tan x| + c\)

- Jan 30, 2012

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- #5

can you show me what's going on?

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even if you re wrote it you would still have tan^3x/sec^3x

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Because you don;t have 1/cosx you have tan^3x/cos^3x

Surely not...even if you re wrote it you would still have tan^3x/sec^3x

$\displaystyle \begin{align*} \frac{\tan^3{(x)}}{\cos^3{(x)}} &= \tan^3{(x)} \left[ \frac{1}{\cos^3{(x)}} \right] \\ &= \tan^3{(x)} \left[ \frac{1}{\cos{(x)}} \right] ^3 \\ &= \tan^3{(x)} \sec^3{(x)} \end{align*}$

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You can't. But you CAN write $\displaystyle \begin{align*} \frac{1}{\cos^3{(x)}} \end{align*}$ as $\displaystyle \begin{align*} \left[ \frac{1}{\cos{(x)}} \right] ^3 \end{align*}$.Oh I see now.. I didn't know you could rewrite cos^3x to 1/cosx

- Jan 30, 2012

- 2,541

By definition,

\[

\frac{1}{\cos x}=\sec x.\]

Taking the cube of both sides,

\[

\frac{1}{\cos^3 x}=\left(\frac{1}{\cos x}\right)^3=(\sec x)^3=\sec^3x.

\]

Multiplying both sides by $\tan^3 x$ we get

\[

\frac{\tan^3 x}{\cos^3 x}=\tan^3 (x)\sec^3x.

\]