Question

evaluate the integral using integration by parts.
\\(\int x^{2}\cos(\frac{1}{6}x)dx\\)

b. \\(6x^{2}\sin(\frac{1}{6}x)+\int 12x\sin(\frac{1}{6}x)dx\\)

c. \\(6x^{2}\sin(\frac{1}{6}x)-\int 12x\sin(\frac{1}{6}x)dx\\)

d. \\(12x\sin(\frac{1}{6}x)+\int 6x^{2}\sin(\frac{1}{6}x)dx\\)

evaluate the integral.
\\(\int x^{2}\cos(\frac{1}{6}x)dx = \square\\)

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = x^{2}$ and $\mathrm{d}v=\cos(\frac{1}{6}x)\mathrm{d}x$.
First, find $\mathrm{d}u$ and $v$. Differentiating $u = x^{2}$ gives $\mathrm{d}u = 2x\mathrm{d}x$. Integrating $\mathrm{d}v=\cos(\frac{1}{6}x)\mathrm{d}x$, we have $v = 6\sin(\frac{1}{6}x)$ (since $\int\cos(ax)\mathrm{d}x=\frac{1}{a}\sin(ax)+C$, here $a=\frac{1}{6}$).

Step2: Substitute into the formula

$\int x^{2}\cos(\frac{1}{6}x)\mathrm{d}x=x^{2}\times6\sin(\frac{1}{6}x)-\int6\sin(\frac{1}{6}x)\times2x\mathrm{d}x=6x^{2}\sin(\frac{1}{6}x)-\int12x\sin(\frac{1}{6}x)\mathrm{d}x$

Answer:

C. $6x^{2}\sin(\frac{1}{6}x)-\int12x\sin(\frac{1}{6}x)\mathrm{d}x$