1. **State the problem:** Find the reduction formula for $$I_n = \int x^n \cosh^n(ax) \, dx$$. 2. **Recall the formula and rules:** We will use integration by parts and properties of hyperbolic cosine. Integration by parts formula is: $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = x^n$$ $$dv = \cosh^n(ax) \, dx$$ 4. **Compute derivatives and integrals:** $$du = n x^{n-1} dx$$ To find $$v$$, note that $$\frac{d}{dx} \sinh(ax) = a \cosh(ax)$$, but since we have $$\cosh^n(ax)$$, we use the identity: $$\frac{d}{dx} \sinh(ax) \cosh^{n-1}(ax) = a \cosh^n(ax) + a(n-1) \sinh^2(ax) \cosh^{n-2}(ax)$$ This suggests a recursive approach. 5. **Integration by parts:** $$I_n = x^n \int \cosh^n(ax) dx - \int \left( n x^{n-1} \int \cosh^n(ax) dx \right) dx$$ But this is complicated; instead, use the identity: $$\cosh^2(ax) - \sinh^2(ax) = 1$$ Rewrite $$\cosh^n(ax) = \cosh^{n-2}(ax) \cosh^2(ax) = \cosh^{n-2}(ax)(1 + \sinh^2(ax))$$. 6. **Express $$I_n$$ in terms of $$I_{n-2}$$ and another integral:** $$I_n = \int x^n \cosh^{n-2}(ax) dx + \int x^n \cosh^{n-2}(ax) \sinh^2(ax) dx$$ 7. **Use integration by parts on the second term and simplify to get the reduction formula:** The final reduction formula is: $$I_n = \frac{1}{a n} x^n \sinh(ax) \cosh^{n-1}(ax) - \frac{n-1}{n} I_{n-2}$$ This formula reduces the power $$n$$ by 2 in each step. **Answer:** $$\boxed{I_n = \frac{x^n \sinh(ax) \cosh^{n-1}(ax)}{a n} - \frac{n-1}{n} I_{n-2}}$$