evaluate the following integral using integration by parts. $int xsin 7xmathrm{d}x$ b. $-\frac{1}{7}xcos…

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 $u = x$ and $\mathrm{d}v=\sin(7x)\mathrm{d}x$.

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

Differentiate $u = x$ with respect to $x$ to get $\mathrm{d}u=\mathrm{d}x$. Integrate $\mathrm{d}v=\sin(7x)\mathrm{d}x$ with respect to $x$. Let $t = 7x$, then $\mathrm{d}t=7\mathrm{d}x$ and $\int\sin(7x)\mathrm{d}x=\frac{1}{7}\int\sin(t)\mathrm{d}t=-\frac{1}{7}\cos(7x)+C$, so $v =-\frac{1}{7}\cos(7x)$.

Step3: Apply integration - by - parts formula

$\int x\sin(7x)\mathrm{d}x=u v-\int v\mathrm{d}u=x\left(-\frac{1}{7}\cos(7x)\right)-\int\left(-\frac{1}{7}\cos(7x)\right)\mathrm{d}x=-\frac{1}{7}x\cos(7x)+\frac{1}{7}\int\cos(7x)\mathrm{d}x$.

The correct first - step result of integration by parts for $\int x\sin(7x)\mathrm{d}x$ is $-\frac{1}{7}x\cos(7x)-\int\left(-\frac{1}{7}\cos(7x)\right)\mathrm{d}x$, so the answer for the first part is B.

Now, integrate $\int\cos(7x)\mathrm{d}x$. Let $t = 7x$, $\mathrm{d}t = 7\mathrm{d}x$, then $\int\cos(7x)\mathrm{d}x=\frac{1}{7}\sin(7x)+C$.

So $\int x\sin(7x)\mathrm{d}x=-\frac{1}{7}x\cos(7x)+\frac{1}{49}\sin(7x)+C$.

Answer:

1. For the multiple - choice part: B. $-\frac{1}{7}x\cos(7x)-\int\left(-\frac{1}{7}\cos(7x)\right)\mathrm{d}x$
2. For the integral evaluation part: $-\frac{1}{7}x\cos(7x)+\frac{1}{49}\sin(7x)+C$