Question

evaluate the integral using integration by parts.
\\(\int x^{2}\cos(\frac{1}{6}x)dx\\)
which of the following shows the result of using integration by parts?
a. \\(12x\sin(\frac{1}{6}x)-\int 6x^{2}\sin(\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\\)

Explanation:

Step1: Recall 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 $\mathrm{d}v=\cos(\frac{1}{6}x)\mathrm{d}x$.

Step2: Find $\mathrm{d}u$ and $v$

Differentiate $u = x^{2}$ with respect to $x$: $\mathrm{d}u = 2x\mathrm{d}x$. Integrate $\mathrm{d}v=\cos(\frac{1}{6}x)\mathrm{d}x$ with respect to $x$. Let $t=\frac{1}{6}x$, then $\mathrm{d}x = 6\mathrm{d}t$, and $\int\cos(\frac{1}{6}x)\mathrm{d}x=6\int\cos t\mathrm{d}t = 6\sin(\frac{1}{6}x)+C$, so $v = 6\sin(\frac{1}{6}x)$.

Step3: Apply the integration - by - parts 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. $6x^{2}\sin(\frac{1}{6}x)-\int12x\sin(\frac{1}{6}x)\mathrm{d}x$