evaluate the $intcos(3x)cos(4x)dx$.

Explanation:

Step1: Use product - to - sum formula

We know that $\cos A\cos B=\frac{1}{2}[\cos(A + B)+\cos(A - B)]$. Here $A = 3x$ and $B=4x$, so $\cos(3x)\cos(4x)=\frac{1}{2}[\cos(3x + 4x)+\cos(3x-4x)]=\frac{1}{2}[\cos(7x)+\cos(-x)]$. Since $\cos(-x)=\cos(x)$, then $\cos(3x)\cos(4x)=\frac{1}{2}[\cos(7x)+\cos(x)]$.

Step2: Integrate term - by - term

$\int\cos(3x)\cos(4x)dx=\int\frac{1}{2}[\cos(7x)+\cos(x)]dx=\frac{1}{2}\int\cos(7x)dx+\frac{1}{2}\int\cos(x)dx$. For $\int\cos(7x)dx$, let $u = 7x$, then $du=7dx$ and $\int\cos(7x)dx=\frac{1}{7}\int\cos(u)du=\frac{1}{7}\sin(u)+C_1=\frac{1}{7}\sin(7x)+C_1$. And $\int\cos(x)dx=\sin(x)+C_2$. So $\frac{1}{2}\int\cos(7x)dx+\frac{1}{2}\int\cos(x)dx=\frac{1}{2}\times\frac{1}{7}\sin(7x)+\frac{1}{2}\sin(x)+C=\frac{1}{14}\sin(7x)+\frac{1}{2}\sin(x)+C$.

Answer:

$\frac{1}{14}\sin(7x)+\frac{1}{2}\sin(x)+C$