evaluate the integral.\n int cos(12x)cos(9x)dx \n int cos(12x)cos(9x)dx=square

Explanation:

Step1: Use product - to - sum formula

We know that $\cos A\cos B=\frac{1}{2}[\cos(A + B)+\cos(A - B)]$. Here $A = 12x$ and $B=9x$, so $\cos(12x)\cos(9x)=\frac{1}{2}[\cos(12x + 9x)+\cos(12x-9x)]=\frac{1}{2}(\cos(21x)+\cos(3x))$.

Step2: Integrate term - by - term

$\int\cos(12x)\cos(9x)dx=\frac{1}{2}\int(\cos(21x)+\cos(3x))dx=\frac{1}{2}(\int\cos(21x)dx+\int\cos(3x)dx)$. For $\int\cos(ax)dx=\frac{1}{a}\sin(ax)+C$ ($a\neq0$). So $\int\cos(21x)dx=\frac{1}{21}\sin(21x)+C_1$ and $\int\cos(3x)dx=\frac{1}{3}\sin(3x)+C_2$.

Step3: Combine the results

$\frac{1}{2}(\int\cos(21x)dx+\int\cos(3x)dx)=\frac{1}{2}(\frac{1}{21}\sin(21x)+\frac{1}{3}\sin(3x))+C=\frac{1}{42}\sin(21x)+\frac{1}{6}\sin(3x)+C$

Answer:

$\frac{1}{42}\sin(21x)+\frac{1}{6}\sin(3x)+C$