Question

ma 114 exam 1

1. (5 points) find $intcos^{3}(7x)dx$.

a. $\frac{1}{7}sin(7x)-\frac{1}{21}sin^{3}(7x)+c$
b. $\frac{1}{28}cos^{4}(7x)+c$
c. $\frac{1}{7}x-\frac{1}{21}sin^{3}(7x)+c$
d. $\frac{7}{2}x+\frac{7}{4}sin(2x)+c$
e. $\frac{1}{7}-\frac{1}{7}sin^{2}(7x)+c$

Explanation:

Step1: Rewrite $\cos^{3}(7x)$

We know that $\cos^{3}(u)=\cos(u)(1 - \sin^{2}(u))$. Let $u = 7x$, so $\cos^{3}(7x)=\cos(7x)(1-\sin^{2}(7x))$. Then the integral becomes $\int\cos(7x)(1 - \sin^{2}(7x))dx$.

Step2: Use substitution

Let $t=\sin(7x)$, then $dt = 7\cos(7x)dx$, and $\cos(7x)dx=\frac{1}{7}dt$. The integral $\int\cos(7x)(1 - \sin^{2}(7x))dx$ is rewritten as $\frac{1}{7}\int(1 - t^{2})dt$.

Step3: Integrate term - by - term

We know that $\int(1 - t^{2})dt=\int 1dt-\int t^{2}dt=t-\frac{1}{3}t^{3}+C$.

Step4: Substitute back $t = \sin(7x)$

Substituting $t=\sin(7x)$ back into the result, we get $\frac{1}{7}(\sin(7x)-\frac{1}{3}\sin^{3}(7x))+C=\frac{1}{7}\sin(7x)-\frac{1}{21}\sin^{3}(7x)+C$.

Answer:

A. $\frac{1}{7}\sin(7x)-\frac{1}{21}\sin^{3}(7x)+C$