# Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 8

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Evaluate $\displaystyle \int x^2 \cos mx dx$ by using Integration by parts.

If we let $u = x^2$ and $dv = \cos mx dx$, then

$du = 2x dx$ and $\displaystyle v = \int \cos mx dx = \frac{1}{m} \sin mx$

So,

\begin{aligned} \int x^2 \cos mx dx = uv - \int vdu &= \frac{x^2}{m} \sin (mx) - \int \left( \frac{1}{m} \sin mx \right) (2x dx)\\ \\ &= \frac{x^2}{m} \sin (mx) - \frac{2}{m} \int x \sin (mx) dx \end{aligned}

To evaluate $\displaystyle \int x \sin (mx) dx$, we must use integration by parts once more, so...

If we let $u_1 = x$ and $dv_1 = \sin (mx) dx$, then

$du_1 = dx$ and $\displaystyle v_1 = \int \sin (mx) dx = \frac{1}{m} \left(-\cos (mx) \right)$

Thus,

\begin{aligned} \int x \sin (mx) dx &= u_1 v_1 - \int v_1 du_1 = \frac{-x}{m} \cos (mx) - \int \frac{-\cos (mx) dx}{m}\\ \\ &= \frac{-x \cos (mx)}{m} + \frac{\sin(mx)}{m^2} + c \end{aligned}

Therefore,

\begin{aligned} \int x^2 \cos mx dx &= \frac{x^2}{m} \sin (mx) - \frac{2}{m} \left[ \frac{-x \cos (mx)}{m} + \frac{\sin (mx)}{m^2} + c\right]\\ \\ &= \frac{x^2 \sin (mx)}{m} + \frac{2x \cos (mx)}{m^2} - \frac{2 \sin (mx)}{m^3} + c \end{aligned}