To determine [tex]\(\int \sin^2 x \, dx\)[/tex], we'll utilize a common trigonometric identity to simplify the integrand before integrating.
1. Rewrite [tex]\(\sin^2 x\)[/tex] using the trigonometric identity:
There exists a useful trigonometric identity:
[tex]\[
\sin^2 x = \frac{1 - \cos(2x)}{2}
\][/tex]
2. Substitute this identity into the integral:
[tex]\[
\int \sin^2 x \, dx = \int \frac{1 - \cos(2x)}{2} \, dx
\][/tex]
3. Simplify the integral:
Split the integral into two separate terms:
[tex]\[
\int \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \int (1 - \cos(2x)) \, dx
\][/tex]
4. Integrate each term individually:
[tex]\[
= \frac{1}{2} \left( \int 1 \, dx - \int \cos(2x) \, dx \right)
\][/tex]
Start with the simpler integrals:
[tex]\[
\int 1 \, dx = x
\][/tex]
For [tex]\(\int \cos(2x) \, dx\)[/tex], let [tex]\(u = 2x\)[/tex]. Then [tex]\(\frac{du}{dx} = 2\)[/tex] or [tex]\(dx = \frac{du}{2}\)[/tex]:
[tex]\[
\int \cos(2x) \, dx = \int \cos(u) \cdot \frac{du}{2} = \frac{1}{2} \int \cos(u) \, du
\][/tex]
We know that:
[tex]\[
\int \cos(u) \, du = \sin(u)
\][/tex]
Substituting back [tex]\(u = 2x\)[/tex]:
[tex]\[
\int \cos(2x) \, dx = \frac{1}{2} \sin(2x)
\][/tex]
5. Combine the integrated results:
[tex]\[
\frac{1}{2} \left( x - \frac{1}{2} \sin(2x) \right)
\][/tex]
6. Simplify the expression:
[tex]\[
= \frac{x}{2} - \frac{\sin(2x)}{4}
\][/tex]
Therefore, the indefinite integral [tex]\(\int \sin^2 x \, dx\)[/tex] is:
[tex]\[
\int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C
\][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
