Question

$\int\cos(7x)^{3}dx =$

Explanation:

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

We know that $\cos^{3}\theta=\cos\theta(1 - \sin^{2}\theta)$. 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 $u = \sin(7x)$, then $du=7\cos(7x)dx$, and $\cos(7x)dx=\frac{1}{7}du$. The integral $\int\cos(7x)(1 - \sin^{2}(7x))dx$ is rewritten as $\frac{1}{7}\int(1 - u^{2})du$.

Step3: Integrate term - by - term

We know that $\int(1 - u^{2})du=\int 1du-\int u^{2}du$. Since $\int 1du = u$ and $\int u^{2}du=\frac{u^{3}}{3}+C$, then $\int(1 - u^{2})du=u-\frac{u^{3}}{3}+C$.

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

Substituting $u = \sin(7x)$ into $\frac{1}{7}\int(1 - u^{2})du$, we get $\frac{1}{7}(\sin(7x)-\frac{\sin^{3}(7x)}{3})+C$.

Answer:

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