1. **State the problem:** We want to evaluate the definite integral $$\int_0^{\pi/2} \sin^4 \theta \cos^2 \theta \, d\theta.$$\n\n2. **Use trigonometric identities:** To integrate powers of sine and cosine, it helps to use power-reduction or product-to-sum formulas. Here, rewrite $\sin^4 \theta$ as $(\sin^2 \theta)^2$. Recall that $$\sin^2 \theta = \frac{1 - \cos 2\theta}{2},$$ so $$\sin^4 \theta = \left(\frac{1 - \cos 2\theta}{2}\right)^2 = \frac{1 - 2\cos 2\theta + \cos^2 2\theta}{4}.$$\n\n3. **Rewrite $\cos^2 \theta$ similarly:** $$\cos^2 \theta = \frac{1 + \cos 2\theta}{2}.$$\n\n4. **Express the integrand as:** $$\sin^4 \theta \cos^2 \theta = \frac{1 - 2\cos 2\theta + \cos^2 2\theta}{4} \times \frac{1 + \cos 2\theta}{2} = \frac{(1 - 2\cos 2\theta + \cos^2 2\theta)(1 + \cos 2\theta)}{8}.$$\n\n5. **Expand numerator:**\n$$ (1)(1 + \cos 2\theta) - 2\cos 2\theta (1 + \cos 2\theta) + \cos^2 2\theta (1 + \cos 2\theta) $$\n$$ = 1 + \cos 2\theta - 2\cos 2\theta - 2\cos^2 2\theta + \cos^2 2\theta + \cos^3 2\theta $$\n$$ = 1 - \cos 2\theta - \cos^2 2\theta + \cos^3 2\theta.$$\n\n6. **Final integrand:**\n$$ \frac{1 - \cos 2\theta - \cos^2 2\theta + \cos^3 2\theta}{8}. $$\n\n7. **Rewrite integral:**\n$$ \int_0^{\pi/2} \sin^4 \theta \cos^2 \theta d\theta = \frac{1}{8} \int_0^{\pi/2} \left(1 - \cos 2\theta - \cos^2 2\theta + \cos^3 2\theta\right)d\theta. $$\n\n8. **Integrate each term separately:**\n- $$\int_0^{\pi/2} 1 \, d\theta = \frac{\pi}{2}.$$\n- $$\int_0^{\pi/2} \cos 2\theta \, d\theta = \left[\frac{\sin 2\theta}{2}\right]_0^{\pi/2} = 0.$$\n\n9. **For $\cos^2 2\theta$, use power-reduction:**\n$$ \cos^2 x = \frac{1 + \cos 2x}{2}, $$ so \n$$ \int_0^{\pi/2} \cos^2 2\theta \, d\theta = \int_0^{\pi/2} \frac{1 + \cos 4\theta}{2} d\theta = \frac{1}{2} \int_0^{\pi/2} 1 \, d\theta + \frac{1}{2} \int_0^{\pi/2} \cos 4\theta \, d\theta. $$\n\nCalculate these: \n$$ \frac{1}{2} \times \frac{\pi}{2} + \frac{1}{2} \left[\frac{\sin 4\theta}{4}\right]_0^{\pi/2} = \frac{\pi}{4} + 0 = \frac{\pi}{4}. $$\n\n10. **For $\cos^3 2\theta$, rewrite using $\cos^3 x = \cos x (\cos^2 x)$ and power reduction:**\nUsing $ \cos^2 x = \frac{1 + \cos 2x}{2}$,\n$$ \cos^3 2\theta = \cos 2\theta \times \cos^2 2\theta = \cos 2\theta \times \frac{1 + \cos 4\theta}{2} = \frac{\cos 2\theta + \cos 2\theta \cos 4\theta}{2}. $$\n\nUse product-to-sum for $\cos A \cos B$: \n$$ \cos A \cos B = \frac{\cos (A-B) + \cos (A+B)}{2}, $$\nso\n$$ \cos 2\theta \cos 4\theta = \frac{\cos(-2\theta) + \cos 6\theta}{2} = \frac{\cos 2\theta + \cos 6\theta}{2}. $$\n\nTherefore,\n$$ \cos^3 2\theta = \frac{\cos 2\theta}{2} + \frac{\cos 2\theta + \cos 6\theta}{4} = \frac{2\cos 2\theta + \cos 6\theta}{4}. $$\n\n11. **Integrate $\cos^3 2\theta$:**\n$$ \int_0^{\pi/2} \cos^3 2\theta \, d\theta = \int_0^{\pi/2} \frac{2\cos 2\theta + \cos 6\theta}{4} \, d\theta = \frac{1}{4} \left( 2 \int_0^{\pi/2} \cos 2\theta d\theta + \int_0^{\pi/2} \cos 6\theta d\theta \right). $$\n\nCalculate integrals:\n$$ \int_0^{\pi/2} \cos 2\theta d\theta = 0, $$\n$$ \int_0^{\pi/2} \cos 6\theta d\theta = \left[ \frac{\sin 6\theta}{6} \right]_0^{\pi/2} = 0. $$\n\nThus,\n$$ \int_0^{\pi/2} \cos^3 2\theta \, d\theta = 0. $$\n\n12. **Combine all to compute original integral:**\n$$ \int_0^{\pi/2} \sin^4 \theta \cos^2 \theta d\theta = \frac{1}{8} \left( \frac{\pi}{2} - 0 - \frac{\pi}{4} + 0 \right) = \frac{1}{8} \times \frac{\pi}{4} = \frac{\pi}{32}. $$\n\n**Final answer:**\n$$\boxed{\frac{\pi}{32}}.$$