# Rewriting fractions

#### shamieh

##### Active member
How can I rewrite $$\displaystyle \int \frac{tan^3x}{cos^3x} \, dx$$ to $$\displaystyle \int tan^3x sec^3x \, dx$$

What is the identity they are using to do this?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
What is the identity they are using to do this?
The definition of $\sec x$.

#### shamieh

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

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

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
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.

#### shamieh

##### Active member
Yes, but $$\displaystyle secx = \frac{1}{cosx}$$ not $$\displaystyle tan^3sec^3x$$

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

#### Prove It

##### Well-known member
MHB Math Helper
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*}

#### shamieh

##### Active member
Oh I see now.. I didn't know you could rewrite cos^3x to 1/cosx

#### Prove It

##### Well-known member
MHB Math Helper
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*}.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
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
$\frac{\tan^3 x}{\cos^3 x}=\tan^3 (x)\sec^3x.$