hw part 1: 7.2 trigonometric integrals\nscore: 2.1/8 answered: 3/8\nprogress saved done\nquestion 4\n0/1 pt…

Explanation:

Step1: Use product - to - sum formula

We know that $\sin A\cos B=\frac{1}{2}[\sin(A + B)+\sin(A - B)]$. Here $A = 6v$ and $B=9v$, so $\sin(6v)\cos(9v)=\frac{1}{2}[\sin(6v + 9v)+\sin(6v-9v)]=\frac{1}{2}[\sin(15v)+\sin(- 3v)]$. Since $\sin(-x)=-\sin(x)$, then $\sin(6v)\cos(9v)=\frac{1}{2}[\sin(15v)-\sin(3v)]$.

Step2: Integrate term - by - term

$\int\sin(6v)\cos(9v)dv=\frac{1}{2}\int[\sin(15v)-\sin(3v)]dv=\frac{1}{2}\left(\int\sin(15v)dv-\int\sin(3v)dv\right)$. For $\int\sin(15v)dv$, let $u = 15v$, then $du=15dv$ and $\int\sin(15v)dv=\frac{1}{15}\int\sin(u)du=-\frac{1}{15}\cos(u)=-\frac{1}{15}\cos(15v)$. For $\int\sin(3v)dv$, let $t = 3v$, then $dt = 3dv$ and $\int\sin(3v)dv=\frac{1}{3}\int\sin(t)dt=-\frac{1}{3}\cos(t)=-\frac{1}{3}\cos(3v)$.

Step3: Combine the results

$\frac{1}{2}\left(-\frac{1}{15}\cos(15v)-\left(-\frac{1}{3}\cos(3v)\right)\right)=\frac{1}{6}\cos(3v)-\frac{1}{30}\cos(15v)+C$.

Answer:

$\frac{1}{6}\cos(3v)-\frac{1}{30}\cos(15v)+C$