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Find the integral of (2+3x) sin (5x)

by RoRi

Evaluate the following integral:

    \[ \int(2+3x) \sin (5x) \, dx. \]

First, we multiply out to get

    \[ \int (2+3x) \sin (5x) \, dx = 2 \int \sin (5x) \, dx + 3 \int x \sin (5x) \, dx. \]

For the integral on the left we make the substitution u = 5x, du = 5 \, dx. Therefore,

    \begin{align*} 2 \int \sin (5x) \, dx &= \frac{2}{5} \int \sin u \, du \\ &= -\frac{2}{5} \cos u + C \\ &= -\frac{2}{5} \cos (5x) + C. \end{align*}

For the integral on the right we integrate by parts letting

    \begin{align*} u &= x & du &= dx \\ dv &= \sin (5x)\, dx & v &= -\frac{1}{5} \cos (5x). \end{align*}

Therefore,

    \begin{align*} 3 \int x \sin (5x) \, dx &= 3 \int u \, dv \\ &= 3 \left( uv - \int v \, du \right) \\ &= 3 \left( -\frac{x}{5} \cos (5x) + \frac{1}{5} \int \cos (5x) \, dx \right) \\ &= 3 \left( -\frac{x}{5} \cos (5x) + \frac{1}{25} \sin (5x) \right) + C\\ &= \frac{3}{25} \sin (5x) - \frac{3x}{5} \cos (5x) + C. \end{align*}

Putting these together we have

    \[ \int (2+3x) \sin (5x) \, dx = -\frac{2}{5} \cos (5x) + \frac{3}{25} \sin (5x) - \frac{3}{5} x \cos (5x) + C. \]

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