To solve the integral

$\int \sin^2 x \, dx,$

we can use the identity for $\sin^2 x$:

$\sin^2 x = \frac{1 - \cos(2x)}{2}.$

Substituting this into the integral gives:

$\int \sin^2 x \, dx = \int \frac{1 - \cos(2x)}{2} \, dx.$

Now, we can split the integral:

$= \frac{1}{2} \int (1 - \cos(2x)) \, dx = \frac{1}{2} \left( \int 1 \, dx - \int \cos(2x) \, dx \right).$

Calculating each integral:

1. $\int 1 \, dx = x$.
2. For $\int \cos(2x) \, dx$, we use the substitution $u = 2x$, $du = 2dx$, so $dx = \frac{du}{2}$:

$\int \cos(2x) \, dx = \frac{1}{2} \int \cos(u) \, du = \frac{1}{2} \sin(u) + C = \frac{1}{2} \sin(2x) + C.$

Combining these results, we have:

$\int \sin^2 x \, dx = \frac{1}{2} \left( x - \frac{1}{2} \sin(2x) \right) + C.$

Thus, the final answer is:

$\int \sin^2 x \, dx = \frac{x}{2} - \frac{1}{4} \sin(2x) + C.$

If you have any questions or need further details, feel free to ask! Here are some related questions you might consider:

1. What is the integral of $\cos^2 x$?
2. How can we evaluate definite integrals involving $\sin^2 x$?
3. What are the applications of integrals involving trigonometric functions?
4. How does the substitution method work in integrals?
5. Can you explain the derivation of the double angle formula for sine?

Tip: When dealing with trigonometric integrals, using identities can simplify the process significantly.