Question

evaluate the integral using integration by parts.
int x^{2}cosleft(\frac{1}{6}x
ight)dx
(c)(6x^{2}sinleft(\frac{1}{6}x
ight)-int 12xsinleft(\frac{1}{6}x
ight)dx)
(d)(12xsinleft(\frac{1}{6}x
ight)+int 6x^{2}sinleft(\frac{1}{6}x
ight)dx)
evaluate the integral.
int x^{2}cosleft(\frac{1}{6}x
ight)dx=square

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 $x$ to get $\mathrm{d}u = 2x\mathrm{d}x$. 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 integration - by - parts

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

Step4: Apply integration - by - parts again on $\int12x\sin(\frac{1}{6}x)\mathrm{d}x$

Let $u = 12x$, $\mathrm{d}v=\sin(\frac{1}{6}x)\mathrm{d}x$. Then $\mathrm{d}u = 12\mathrm{d}x$, and $v=- 72\cos(\frac{1}{6}x)$.
$\int12x\sin(\frac{1}{6}x)\mathrm{d}x=12x\times(-72\cos(\frac{1}{6}x))-\int(-72\cos(\frac{1}{6}x))\times12\mathrm{d}x=-864x\cos(\frac{1}{6}x)+864\int\cos(\frac{1}{6}x)\mathrm{d}x=-864x\cos(\frac{1}{6}x)+5184\sin(\frac{1}{6}x)+C$.
So $\int x^{2}\cos(\frac{1}{6}x)\mathrm{d}x=6x^{2}\sin(\frac{1}{6}x)-(-864x\cos(\frac{1}{6}x)+5184\sin(\frac{1}{6}x))+C=6x^{2}\sin(\frac{1}{6}x)+864x\cos(\frac{1}{6}x)-5184\sin(\frac{1}{6}x)+C$.

Answer:

$6x^{2}\sin(\frac{1}{6}x)+864x\cos(\frac{1}{6}x)-5184\sin(\frac{1}{6}x)+C$