Q. \[ \int \sin^2(x)\,dx \]

We want to compute the indefinite integral

\[
\int \sin^2(x)\,dx.
\]

Because \(\sin^2(x)\) is a power of a trig function, a standard strategy is to use a trig identity to rewrite it in a simpler form.

Step 1: Use the power-reduction identity

Use the identity

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

This rewrites the integrand as a combination of a constant term and a cosine term.

Step 2: Substitute the identity into the integral

Substitute \(\sin^2(x)=\frac{1-\cos(2x)}{2}\) into the integral:

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

Factor out the constant \(\frac{1}{2}\):

\[
\int \sin^2(x)\,dx=\frac{1}{2}\int \left(1-\cos(2x)\right)\,dx.
\]

Step 3: Split the integral

Split into two simpler integrals:

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

Step 4: Integrate each part

First integral:

\[
\int 1\,dx = x.
\]

Second integral: \(\int \cos(2x)\,dx\). Use the substitution method conceptually, or recall the rule:

If you know that

\[
\int \cos(bx)\,dx=\frac{1}{b}\sin(bx)+C,
\]

then with \(b=2\):

\[
\int \cos(2x)\,dx=\frac{1}{2}\sin(2x).
\]

Step 5: Combine results

Substitute these back into the expression:

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

Distribute the \(\frac{1}{2}\):

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

Final Answer

The integral of \(\sin^2(x)\) is

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