find the antiderivative for each function when c equals 0. a. f(x)= - 6 sin (6x) b. g(x)=5 sin (x) c…

Explanation:

Step1: Recall antiderivative formula for $\sin(ax)$

The antiderivative of $\sin(ax)$ is $-\frac{1}{a}\cos(ax)+C$. When $C = 0$, we focus on the non - constant part.

Step2: Find antiderivative of $f(x)=-6\sin(6x)$

Using the formula, for $a = 6$ in $\sin(6x)$, the antiderivative of $\sin(6x)$ is $-\frac{1}{6}\cos(6x)$. Then for $f(x)=-6\sin(6x)$, the antiderivative $F(x)=-6\times(-\frac{1}{6})\cos(6x)=\cos(6x)$.

Step3: Find antiderivative of $g(x)=5\sin(x)$

For $a = 1$ in $\sin(x)$, the antiderivative of $\sin(x)$ is $-\cos(x)$. Then for $g(x)=5\sin(x)$, the antiderivative $G(x)=5\times(-\cos(x))=- 5\cos(x)$.

Step4: Find antiderivative of $h(x)=\sin(6x)-5\sin(5x)$

The antiderivative of $\sin(6x)$ is $-\frac{1}{6}\cos(6x)$ and the antiderivative of $\sin(5x)$ is $-\frac{1}{5}\cos(5x)$. So the antiderivative of $h(x)$ is $H(x)=-\frac{1}{6}\cos(6x)-5\times(-\frac{1}{5})\cos(5x)=-\frac{1}{6}\cos(6x)+\cos(5x)$.

Answer:

a. $\cos(6x)$ b. $-5\cos(x)$ c. $-\frac{1}{6}\cos(6x)+\cos(5x)$