Answer
$$\int^{\pi/4}_{-\pi/4}6\tan^4xdx=3\pi-8$$
Work Step by Step
$$A=\int^{\pi/4}_{-\pi/4}6\tan^4xdx=\int^{\pi/4}_{-\pi/4}6\tan^2x\tan^2xdx$$ $$A=\int^{\pi/4}_{-\pi/4}6\tan^2x(\sec^2x-1)dx$$ $$A=6\Big(\int^{\pi/4}_{-\pi/4}\tan^2x\sec^2xdx-\int^{\pi/4}_{-\pi/4}\tan^2xdx\Big)$$ $$A=6\Big(M-N\Big)$$
1) Consider $M$: $$M=\int^{\pi/4}_{-\pi/4}\tan^2x\sec^2xdx$$ $$M=\int^{\pi/4}_{-\pi/4}\tan^2xd(\tan x)$$
We set $u=\tan x$
For $x=\pi/4$, we have $u=1$ and for $x=-\pi/4$, we have $u=-1$
$$M=\int^1_{-1}u^2du=\frac{u^3}{3}\Big]^1_{-1}$$ $$A=\frac{1}{3}-\Big(-\frac{1}{3}\Big)=\frac{2}{3}$$
2) Consider $N$: $$N=\int^{\pi/4}_{-\pi/4}\tan^2xdx=\int^{\pi/4}_{-\pi/4}(\sec^2x-1)dx$$ $$N=\int^{\pi/4}_{-\pi/4}\sec^2xdx-\int^{\pi/4}_{-\pi/4}dx$$ $$N=\tan x\Big]^{\pi/4}_{-\pi/4}-x\Big]^{\pi/4}_{-\pi/4}$$ $$N=[1-(-1)]-\Big[\frac{\pi}{4}-\Big(-\frac{\pi}{4}\Big)\Big]$$ $$N=2-\frac{\pi}{2}$$
Therefore, $$A=6\Big(\frac{2}{3}-2+\frac{\pi}{2}\Big)$$ $$A=4-12+3\pi=3\pi-8$$
