1. **State the problem:** Evaluate the integral $$\int \frac{\sin^2 n}{\sin^6 n + \cos^4 n} \, dn$$. 2. **Analyze the integrand:** The denominator is $$\sin^6 n + \cos^4 n$$ and the numerator is $$\sin^2 n$$. 3. **Rewrite the denominator:** Note that $$\sin^6 n = (\sin^2 n)^3$$ and $$\cos^4 n = (\cos^2 n)^2$$. 4. **Try substitution:** Let $$x = \sin^2 n$$, then $$dx = 2 \sin n \cos n \, dn$$ or $$dn = \frac{dx}{2 \sin n \cos n}$$. But we need to express everything in terms of $$x$$. Since $$\cos^2 n = 1 - \sin^2 n = 1 - x$$, then $$\cos^4 n = (1 - x)^2$$. 5. **Rewrite the integral in terms of $$x$$:** The integral becomes $$\int \frac{x}{x^3 + (1 - x)^2} \cdot \frac{dn}{dn}$$ But we must express $$dn$$ in terms of $$dx$$: $$dn = \frac{dx}{2 \sin n \cos n} = \frac{dx}{2 \sqrt{x} \sqrt{1 - x}}$$. 6. **Substitute and simplify:** $$\int \frac{x}{x^3 + (1 - x)^2} \cdot \frac{dx}{2 \sqrt{x} \sqrt{1 - x}} = \int \frac{x}{x^3 + (1 - x)^2} \cdot \frac{dx}{2 \sqrt{x} \sqrt{1 - x}}$$ Simplify numerator: $$\frac{x}{2 \sqrt{x}} = \frac{\sqrt{x}}{2}$$ So the integral is $$\int \frac{\sqrt{x}}{2 \sqrt{1 - x} (x^3 + (1 - x)^2)} \, dx$$ 7. **Rewrite denominator inside integral:** $$x^3 + (1 - x)^2 = x^3 + 1 - 2x + x^2 = x^3 + x^2 - 2x + 1$$ 8. **Final integral in $$x$$:** $$\int \frac{\sqrt{x}}{2 \sqrt{1 - x} (x^3 + x^2 - 2x + 1)} \, dx$$ This integral is complicated and does not simplify easily with elementary functions. 9. **Conclusion:** The integral $$\int \frac{\sin^2 n}{\sin^6 n + \cos^4 n} \, dn$$ can be transformed into an integral in terms of $$x = \sin^2 n$$ as above, but it does not have a simple closed form in elementary functions. Hence, the integral is expressed as $$\int \frac{\sqrt{x}}{2 \sqrt{1 - x} (x^3 + x^2 - 2x + 1)} \, dx$$ where $$x = \sin^2 n$$. This is the simplified form for further numerical or special function evaluation.