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Use integration by parts to evaluate the given integral

Evaluate the following integral using integration by parts:

    \[ \int \sin x \cos x \, dx. \]


To apply the integration by parts formula we let:

    \begin{align*}  u &= \sin x & \implies && du &= \cos x \, dx \\  dv &= \cos x \, dx& \implies && v &= \sin x. \end{align*}

Then we use integration by parts

    \begin{align*}  &&\int \sin x \cos x \, dx &= \int u \, dv  \\ &&&= uv - \int v \, du \\ &&&= \sin^2 x - \int \sin x \cos x \, dx + C \\ \implies&& 2 \int \sin x \cos x \, dx &= \sin^2 x + C \\ \implies&& \int \sin x \cos x \, dx &= \frac{1}{2} \sin^2 x + C. \end{align*}

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