1. Let us solve the integral $\int \tan^5 x \, dx$. 2. Rewrite $\tan^5 x$ as $\tan^4 x \cdot \tan x = (\tan^2 x)^2 \cdot \tan x$. 3. Use the identity $\tan^2 x = \sec^2 x - 1$, so $\tan^4 x = (\sec^2 x - 1)^2$. 4. Substitute into the integral: $\int (\sec^2 x - 1)^2 \tan x \, dx$. 5. Notice that $\frac{d}{dx}(\sec x) = \sec x \tan x$, so try the substitution $u = \sec x$. 6. Then $du = \sec x \tan x \, dx \implies \tan x \, dx = \frac{du}{u}$. 7. Express the integral in terms of $u$: $$\int (u^2 - 1)^2 \cdot \tan x \, dx = \int (u^2 - 1)^2 \cdot \frac{du}{u} = \int \frac{(u^2 - 1)^2}{u} \, du$$ 8. Expand the numerator: $$ (u^2 - 1)^2 = u^4 - 2u^2 + 1 $$ 9. Divide each term by $u$: $$ \frac{u^4}{u} - \frac{2u^2}{u} + \frac{1}{u} = u^3 - 2u + \frac{1}{u} $$ 10. The integral becomes: $$ \int \left(u^3 - 2u + \frac{1}{u} \right) du = \int u^3 du - 2 \int u du + \int \frac{1}{u} du $$ 11. Integrate each term: $$ \frac{u^4}{4} - 2 \cdot \frac{u^2}{2} + \ln |u| + C = \frac{u^4}{4} - u^2 + \ln |u| + C $$ 12. Substitute back $u = \sec x$: $$ \frac{\sec^4 x}{4} - \sec^2 x + \ln |\sec x| + C $$ Final answer: $$ \int \tan^5 x \, dx = \frac{\sec^4 x}{4} - \sec^2 x + \ln |\sec x| + C $$