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?