1. The problem is to evaluate the integral $$\int_0^{\frac{\pi}{4}} \tan^6 x \, dx$$. 2. We use the reduction formula for integrals of powers of tangent: $$\int \tan^n x \, dx = \frac{1}{n-1} \tan^{n-1} x - \int \tan^{n-2} x \, dx$$ for $n > 1$. 3. Applying this formula for $n=6$, we get: $$\int \tan^6 x \, dx = \frac{1}{5} \tan^5 x - \int \tan^4 x \, dx$$ 4. Next, apply the formula again for $\int \tan^4 x \, dx$: $$\int \tan^4 x \, dx = \frac{1}{3} \tan^3 x - \int \tan^2 x \, dx$$ 5. We know that: $$\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \tan x - x + C$$ 6. Substitute back: $$\int \tan^4 x \, dx = \frac{1}{3} \tan^3 x - (\tan x - x) + C = \frac{1}{3} \tan^3 x - \tan x + x + C$$ 7. Substitute into the original integral: $$\int \tan^6 x \, dx = \frac{1}{5} \tan^5 x - \left( \frac{1}{3} \tan^3 x - \tan x + x \right) + C = \frac{1}{5} \tan^5 x - \frac{1}{3} \tan^3 x + \tan x - x + C$$ 8. Evaluate definite integral from $0$ to $\frac{\pi}{4}$: At $x=\frac{\pi}{4}$, $\tan \frac{\pi}{4} = 1$: $$\frac{1}{5} (1)^5 - \frac{1}{3} (1)^3 + 1 - \frac{\pi}{4} = \frac{1}{5} - \frac{1}{3} + 1 - \frac{\pi}{4}$$ At $x=0$, $\tan 0 = 0$: $$0 - 0 + 0 - 0 = 0$$ 9. Simplify the result: $$\frac{1}{5} - \frac{1}{3} + 1 - \frac{\pi}{4} = \left( \frac{3}{15} - \frac{5}{15} + \frac{15}{15} \right) - \frac{\pi}{4} = \frac{13}{15} - \frac{\pi}{4}$$ Final answer: $$\int_0^{\frac{\pi}{4}} \tan^6 x \, dx = \frac{13}{15} - \frac{\pi}{4}$$