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Use integration by parts to evaluate the given integral

by RoRi

Evaluate the following integral using integration by parts:

    \[ \int x^3 \sin x \, dx. \]

To apply the integration by parts formula we let:

    \begin{align*} u &= x^3 & \implies && du &= 3x^2 \, dx \\ dv &= \sin x \, dx & \implies && v &= -\cos x. \end{align*}

Then we use integration by parts

    \begin{align*} \int x^3 \sin x \, dx &= \int u \, dv \\ &= uv - \int v \, du \\ &= -x^3 \cos x + 3 \int x^2 \cos x \, dx \\ \end{align*}

Now, we use integration by parts again to evaluate the integral of x^2 \cos x. Let

    \begin{align*} u &= x^2 & \implies && du &= 2x \, dx \\ dv &= \cos x \, dx & \implies && v &= \sin x. \end{align*}

Then we have

    \begin{align*} \int x^2 \cos x \, dx &= \int u \, dv \\ &= uv - \int v \, du \\ &= x^2 \sin x - 2 \int x \sin x \, dx \end{align*}

Then from the first exercise of this section we know

    \[ \int x \sin x \, dx = \sin x - x \cos x + C. \]

So we then have

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

Plugging this back into our expression for \int x^3 \sin x \, dx we obtain

    \begin{align*} \int x^3 \sin x \, dx &= -x^3 \cos x + 3 \int x^2 \cos x \, dx \\ &= -x^3 \cos x + 3 (x^2 \sinx + 2x \cos x - 2 \sin x) + C \\ &= -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C \end{align*}

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