Evaluate the integral: I = integral sin² x dx

Evaluate the integral: $$ I = \int \sin^2 x\,dx. $$

Explanation of Method

The power-reduction identity $\sin^2x=\dfrac{1-\cos2x}{2}$ replaces the squared trig function with a linear combination of a constant and $\cos2x$, both of which integrate directly.

Rewrite $\sin^2x$ using the power-reduction identity.
Split the integral into two simple terms.
Integrate each term.

Step 1 · Apply the power-reduction identity

Recall that:

$$ \sin^2x = \frac{1-\cos2x}{2}. $$

Substituting this into the integral:

$$ I = \int \frac{1-\cos2x}{2}\,dx. $$

Step 2 · Split the integral

$$ I = \frac12\int 1\,dx - \frac12\int \cos2x\,dx. $$

Step 3 · Integrate each term

$$ \int 1\,dx = x, \qquad \int \cos2x\,dx = \frac{\sin2x}{2}. $$

Combining:

$$ I = \frac{x}{2} - \frac{\sin2x}{4} + C. $$

Summary Table

Term	Antiderivative
$1/2$	$x/2$
$-\cos2x/2$	$-\sin2x/4$

Additional Notes

The integrand is defined for all real $x$.
This power-reduction technique is the standard first move for any even power of sine or cosine alone; odd powers instead use the Pythagorean-identity splitting seen in Problems 131–132.
The result can be checked by differentiating $\tfrac{x}{2}-\tfrac{\sin2x}{4}$ and confirming it returns $\sin^2x$.

Result

The solution to the integral is:

$$ I = \frac{x}{2} - \frac{\sin2x}{4} + C. $$