MHB What is the identity used to rewrite fractions in calculus?

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The discussion centers on rewriting the integral $$\int \frac{\tan^3x}{\cos^3x} \, dx$$ as $$\int \tan^3x \sec^3x \, dx$$ using trigonometric identities. The key identity utilized is that $$\sec x = \frac{1}{\cos x}$$, which allows for the transformation of $$\frac{1}{\cos^3 x}$$ into $$\sec^3 x$$. Participants clarify that while $$\frac{\tan^3 x}{\cos^3 x}$$ can be expressed as $$\tan^3 x \sec^3 x$$, it is essential to understand the steps involved in this rewriting process. The discussion emphasizes the importance of correctly applying trigonometric definitions in calculus. This understanding is crucial for accurately manipulating integrals involving trigonometric functions.
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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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