Evaluate the integral: $\int\tan^5 (4x)\,\mathrm dx$. We have to use a trig identity for this. I think it's $\tan^2 x = \sec^2 x - 1$. 
But I'm having difficulty setting up the integral. 

The secret is to use $1+\tan^2 w=\sec^2 w$. Indeed:
$$\int \tan^5 4x dx=\int \tan^3 4x\cdot \tan^2 4x dx=\int \tan^3 4x(\sec^2 4x-1)dx= $$
$$=\frac{1}{4}\int (\tan 4x)^3(\sec^2 4x \cdot 4 dx)-\int \tan^3 4x dx= $$
$$=\frac{1}{4}\frac{(\tan 4x)^4}{4} -\int \tan 4x\cdot \tan^2 4xdx=$$
$$=\frac{1}{16}\tan^4 4x-\int \tan 4x (\sec^2 4x-1)dx= $$
$$= \frac{1}{16}\tan^4 4x- \frac{1}{4}\int (\tan 4x)^1(\sec^2 4x\cdot 4 dx)+\int \tan 4x dx=$$
$$=  \frac{1}{16}\tan^4 4x- \frac{1}{4}\frac{(\tan 4x)^2}{2}+\frac{1}{4}\int \frac{(\sin 4x\cdot 4 dx)}{\cos 4x}=$$
$$=\frac{1}{16}\tan^4 4x- \frac{1}{4}\frac{(\tan 4x)^2}{2}+\frac{1}{4} \ln |\sin 4x|+c.$$
