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Use integration by parts to evaluate the given integral

Evaluate the following integral using integration by parts:

    \[ \int x^3 \cos x \, dx. \]


To apply the integration by parts formula we let:

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

Then we use integration by parts

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

Now, from the previous exercise we know

    \[ \int x^2 \sin x \, dx = 2x \sin x + 2 \cos x - x^2 \cos x + C. \]

Thus, we have

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

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