find the antiderivative for each function when c equals 0. do as many as you can mentally. check your…

Explanation:

Step1: Recall antiderivative of $\cos(ax)$

The antiderivative of $\cos(ax)$ is $\frac{1}{a}\sin(ax)+C$. When $C = 0$, for $y = a\cos(ax)$, its antiderivative $F(x)=\sin(ax)$.

Step2: Find antiderivative of part a

For $y = 9\cos(9x)$, using the above - rule with $a = 9$, the antiderivative $F(x)=\sin(9x)$.

Step3: Find antiderivative of part b

For $y=\frac{9\pi}{2}\cos\frac{9\pi x}{2}$, here $a=\frac{9\pi}{2}$. The antiderivative $F(x)=\sin\frac{9\pi x}{2}$.

Step4: Find antiderivative of part c

The antiderivative of $\cos\frac{17\pi x}{2}$ is $\frac{2}{17\pi}\sin\frac{17\pi x}{2}$ and the antiderivative of $13\pi\cos x$ is $13\pi\sin x$. So the antiderivative of $y=\cos\frac{17\pi x}{2}+13\pi\cos x$ is $F(x)=\frac{2}{17\pi}\sin\frac{17\pi x}{2}+13\pi\sin x$.

Step5: Check by differentiation

• For $F(x)=\sin(9x)$, $F^\prime(x)=9\cos(9x)$ by the chain - rule.
• For $F(x)=\sin\frac{9\pi x}{2}$, $F^\prime(x)=\frac{9\pi}{2}\cos\frac{9\pi x}{2}$ by the chain - rule.
• For $F(x)=\frac{2}{17\pi}\sin\frac{17\pi x}{2}+13\pi\sin x$, $F^\prime(x)=\cos\frac{17\pi x}{2}+13\pi\cos x$ using the chain - rule for the first term and the derivative of $\sin x$ for the second term.

Answer:

a. $\sin(9x)$ b. $\sin\frac{9\pi x}{2}$ c. $\frac{2}{17\pi}\sin\frac{17\pi x}{2}+13\pi\sin x$