Find $\int \sin ^{2}x\,dx$.

Solution Since the exponent of $\sin x$ is an even integer, we use the identity $$\sin ^{2}x=\dfrac{1-\cos (2x) }{2}$$

Then $$\begin{eqnarray*} \int \sin ^{2}x\,dx&=&\dfrac{1}{2}\int \,[ 1-\cos (2x) ] \,dx =\dfrac{1}{2}\int\, dx-\dfrac{1}{2}\int \cos (2x) \,dx \\[4pt] &=& \dfrac{1}{2} x + C_1 - \dfrac{1}{2} \int \cos u \dfrac{du}{2} \qquad {\color{#0066A7}{\hbox{\(u=2x, du=2\, dx\).}}} \\[4pt] &=& \dfrac{1}{2} x + C_1 - \dfrac{1}{4} \sin (2x) + C_2 \end{eqnarray*}$$

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Since $C_1$ and $C_2$ are constants, we write the solution as $$\int \sin^2 x \, dx =\dfrac{1}{2}x -\dfrac{1}{4} \sin (2x) +C$$

where $C=C_1+C_2$.