evaluate the integral.\n intsin(22x)cos(13x)dx \n intsin(22x)cos(13x)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 $A = 22x$ and $B=13x$, so $\sin(22x)\cos(13x)=\frac{1}{2}[\sin(22x + 13x)+\sin(22x-13x)]=\frac{1}{2}[\sin(35x)+\sin(9x)]$.

Step2: Integrate term - by - term

$\int\sin(22x)\cos(13x)dx=\frac{1}{2}\int[\sin(35x)+\sin(9x)]dx=\frac{1}{2}(\int\sin(35x)dx+\int\sin(9x)dx)$. For $\int\sin(ax)dx=-\frac{1}{a}\cos(ax)+C$ ($a\neq0$), when $a = 35$, $\int\sin(35x)dx=-\frac{1}{35}\cos(35x)+C_1$, and when $a = 9$, $\int\sin(9x)dx=-\frac{1}{9}\cos(9x)+C_2$.

Step3: Simplify the result

$\frac{1}{2}(\int\sin(35x)dx+\int\sin(9x)dx)=\frac{1}{2}(-\frac{1}{35}\cos(35x)-\frac{1}{9}\cos(9x))+C=-\frac{1}{70}\cos(35x)-\frac{1}{18}\cos(9x)+C$.

Answer:

$-\frac{1}{70}\cos(35x)-\frac{1}{18}\cos(9x)+C$