evaluate the integral.\n int sin(18x)cos(11x)dx \n int sin(18x)cos(11x)dx=square

Explanation:

Step1: Use product - to - sum formula

We know that $\sin A\cos B=\frac{1}{2}[\sin(A + B)+\sin(A - B)]$. Here $A = 18x$ and $B=11x$, so $\sin(18x)\cos(11x)=\frac{1}{2}[\sin(18x + 11x)+\sin(18x-11x)]=\frac{1}{2}(\sin(29x)+\sin(7x))$.

Step2: Integrate term - by - term

$\int\sin(18x)\cos(11x)dx=\frac{1}{2}\int(\sin(29x)+\sin(7x))dx=\frac{1}{2}\left(\int\sin(29x)dx+\int\sin(7x)dx\right)$. For $\int\sin(ax)dx=-\frac{1}{a}\cos(ax)+C$ ($a\neq0$). When $a = 29$, $\int\sin(29x)dx=-\frac{1}{29}\cos(29x)+C_1$, and when $a = 7$, $\int\sin(7x)dx=-\frac{1}{7}\cos(7x)+C_2$.

Step3: Combine the results

$\frac{1}{2}\left(-\frac{1}{29}\cos(29x)-\frac{1}{7}\cos(7x)\right)+C=-\frac{1}{58}\cos(29x)-\frac{1}{14}\cos(7x)+C$.

Answer:

$-\frac{1}{58}\cos(29x)-\frac{1}{14}\cos(7x)+C$