1. We are asked to evaluate the integral $$\int \cos^7\left(\frac{2x}{5}\right) \cdot \sin^6\left(\frac{2x}{5}\right) \, dx.$$\n\n2. The integral involves powers of sine and cosine functions with the same argument. A common strategy is to use substitution and trigonometric identities.\n\n3. Let $$t = \sin\left(\frac{2x}{5}\right).$$ Then, $$dt = \frac{2}{5} \cos\left(\frac{2x}{5}\right) dx,$$ so $$dx = \frac{5}{2} \frac{dt}{\cos\left(\frac{2x}{5}\right)}.$$\n\n4. Rewrite the integral in terms of $t$ and $dt$:\n$$\int \cos^7\left(\frac{2x}{5}\right) \sin^6\left(\frac{2x}{5}\right) dx = \int \cos^7\left(\frac{2x}{5}\right) t^6 \cdot \frac{5}{2} \frac{dt}{\cos\left(\frac{2x}{5}\right)} = \frac{5}{2} \int \cos^6\left(\frac{2x}{5}\right) t^6 dt.$$\n\n5. Using the Pythagorean identity, $$\cos^2\theta = 1 - \sin^2\theta,$$ so\n$$\cos^6\left(\frac{2x}{5}\right) = \left(\cos^2\left(\frac{2x}{5}\right)\right)^3 = (1 - t^2)^3.$$\n\n6. Substitute back into the integral:\n$$\frac{5}{2} \int (1 - t^2)^3 t^6 dt.$$\n\n7. Expand $(1 - t^2)^3$ using the binomial theorem:\n$$(1 - t^2)^3 = 1 - 3t^2 + 3t^4 - t^6.$$\n\n8. Multiply by $t^6$:\n$$t^6 - 3t^8 + 3t^{10} - t^{12}.$$\n\n9. The integral becomes:\n$$\frac{5}{2} \int \left(t^6 - 3t^8 + 3t^{10} - t^{12}\right) dt = \frac{5}{2} \left( \int t^6 dt - 3 \int t^8 dt + 3 \int t^{10} dt - \int t^{12} dt \right).$$\n\n10. Integrate each term:\n$$\int t^n dt = \frac{t^{n+1}}{n+1} + C.$$\n\nSo,\n$$\int t^6 dt = \frac{t^7}{7}, \quad \int t^8 dt = \frac{t^9}{9}, \quad \int t^{10} dt = \frac{t^{11}}{11}, \quad \int t^{12} dt = \frac{t^{13}}{13}.$$\n\n11. Substitute back:\n$$\frac{5}{2} \left( \frac{t^7}{7} - 3 \frac{t^9}{9} + 3 \frac{t^{11}}{11} - \frac{t^{13}}{13} \right) + C = \frac{5}{2} \left( \frac{t^7}{7} - \frac{t^9}{3} + \frac{3 t^{11}}{11} - \frac{t^{13}}{13} \right) + C.$$\n\n12. Finally, substitute back $t = \sin\left(\frac{2x}{5}\right)$:\n$$\boxed{\frac{5}{2} \left( \frac{\sin^7\left(\frac{2x}{5}\right)}{7} - \frac{\sin^9\left(\frac{2x}{5}\right)}{3} + \frac{3 \sin^{11}\left(\frac{2x}{5}\right)}{11} - \frac{\sin^{13}\left(\frac{2x}{5}\right)}{13} \right) + C}.$$