How to integrate $1-\tan^4(x)$ without using secant? How to integrate such function:
$\int_{-\pi/3}^{\pi/3}1-\tan^4(x)$
I already found a solution using the trigonometric function secant. It looks like this:
$$u=\tan(x),\quad \frac{du}{dx}=\sec^2(x)$$
$$
\begin{align}
\int 1-\tan^4(x) & = \int 1 \;dx - \int\tan^4(x) \; dx \\

& = x-\int\tan^4(x) \; dx \\

& = x-\int\tan^2(x) \tan^2(x) \; dx \\

& = x-\int\tan^2(x) (\sec^2(x) - 1) \; dx \\

& = x-\int\tan^2(x) \sec^2(x) \; dx - \int\tan^2(x) \; dx \\

& = x-\int u^2 \; du - (\tan(x)-x) \\

& = x-\frac{1}{3}u^3 - (\tan(x)-x) \\

& = 2x-\frac{1}{3}\tan^3(x)-\tan(x)
\end{align}
$$
The only problem now is, that I have to find a solution without using secant. Can someone say me how to solve this without secant?

Use
$$
  \frac{\mathrm{d} }{\mathrm{d} x} \tan(x) = 1 + \tan^2(x)
$$
Then
$$ \begin{eqnarray}
  \int \left(1-\tan^4(x)\right)\mathrm{d} x &=& \int \left(1-\tan^2(x)\right) \left(1+\tan^2(x)\right) \mathrm{d} x = \int \left(1-\tan^2(x)\right) \mathrm{d} \tan(x)\\  
&=& \tan(x) - \frac{1}{3} \tan^3(x) + \color{gray}{\text{const}}
\end{eqnarray}
$$
