What is the identity used to rewrite fractions in calculus?

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SUMMARY

The discussion centers on the identity used to rewrite the integral $$\int \frac{\tan^3x}{\cos^3x} \, dx$$ as $$\int \tan^3x \sec^3x \, dx$$. Participants clarify that the transformation relies on the definition of the secant function, specifically that $$\sec x = \frac{1}{\cos x}$$. By cubing this identity, $$\frac{1}{\cos^3 x}$$ can be expressed as $$\sec^3 x$$, allowing the original fraction to be rewritten correctly. This understanding is crucial for manipulating integrals involving trigonometric functions.

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shamieh
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How can I rewrite $$\int \frac{tan^3x}{cos^3x} \, dx$$ to $$\int tan^3x sec^3x \, dx$$

What is the identity they are using to do this?
 
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shamieh said:
What is the identity they are using to do this?
The definition of $\sec x$.
 
Evgeny.Makarov said:
The definition of $\sec x$.

What? What do you mean?

$$\sec x \, dx = ln|\sec x + \tan x| + c$$
 
I am saying they used the definition of $\sec x$ to rewrite the first expression in post #1 to the second one. This is not related to integration; they rewrote purely the function being integrated.
 
Yes, but $$secx = \frac{1}{cosx}$$ not $$tan^3sec^3x$$

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

Surely not...

$\displaystyle \begin{align*} \frac{\tan^3{(x)}}{\cos^3{(x)}} &= \tan^3{(x)} \left[ \frac{1}{\cos^3{(x)}} \right] \\ &= \tan^3{(x)} \left[ \frac{1}{\cos{(x)}} \right] ^3 \\ &= \tan^3{(x)} \sec^3{(x)} \end{align*}$
 
Oh I see now.. I didn't know you could rewrite cos^3x to 1/cosx
 
shamieh said:
Oh I see now.. I didn't know you could rewrite cos^3x to 1/cosx

You can't. But you CAN write $\displaystyle \begin{align*} \frac{1}{\cos^3{(x)}} \end{align*}$ as $\displaystyle \begin{align*} \left[ \frac{1}{\cos{(x)}} \right] ^3 \end{align*}$.
 
Somehow the following post did not show earlier. I must have accidentally closed it before posting.

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

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