1. Problem statement: Compute the integral $\int \sin^2 x\, dx$. 2. Formula and rule: Use the power-reduction identity $\sin^2 x = \frac{1-\cos(2x)}{2}$ and the linearity of the integral. 3. Apply the identity: $$\int \sin^2 x\, dx = \int \frac{1-\cos(2x)}{2}\, dx = \frac{1}{2}\int 1\, dx - \frac{1}{2}\int \cos(2x)\, dx$$. 4. Integrate each term: The first term gives $\frac{1}{2}x$. 5. For the second term, use $\int \cos(2x)\, dx = \frac{1}{2}\sin(2x)$, so $-\frac{1}{2}\int \cos(2x)\, dx = -\frac{1}{2}\cdot\frac{1}{2}\sin(2x) = -\frac{1}{4}\sin(2x)$. 6. Combine results and add the constant of integration: $\int \sin^2 x\, dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C$. 7. Alternative equivalent forms: Using $\sin(2x)=2\sin x\cos x$ we get $\frac{x}{2} - \frac{\sin(2x)}{4} = \frac{1}{2}x - \frac{1}{2}\sin x\cos x + C = \frac{1}{2}(x - \sin x\cos x) + C$. Final answer: $\boxed{\frac{x}{2} - \frac{\sin(2x)}{4} + C}$