How to prove the following equation about $\tan^h(x)$

Question

$$\int \tan^h x \, dx = \frac{tan^{h-1}\ x}{h-1} - \int \tan^{h-2} x \,dx$$ I just cannot find a way to somehow combine the two integral's result together....

Answer 1

Hint:

Rewrite $\tan^h x$ as $$\tan^{h-2} x(1+\tan^2 x)-\tan^{h-2} x$$ and split the integral in two.

Answer 2

\begin{eqnarray*} \int(\tan^h (x) + \tan^{h-2} (x) )dx =\int(\tan^{h-2}(x) \sec^2(x) dx \underbrace{=}_{u=\tan(x)} \int u^{h-2} du = \frac{u^{h-1}}{h-1}+C = \frac{\tan^{h-1}(x)}{h-1}+C. \end{eqnarray*} So your equation should be \begin{eqnarray*} \int\tan^h (x) dx =\frac{\tan^{h-1}(x)}{h-1}-\int \tan^{h-2} (x) dx \end{eqnarray*}