Question

evaluate the integral
int 2cos^{3}(6x)dx
int 2cos^{3}(6x)dx=square

Explanation:

Step1: Use trigonometric identity

We know that $\cos^{3}\theta=\cos\theta(1 - \sin^{2}\theta)$. So, $2\cos^{3}(6x)=2\cos(6x)(1 - \sin^{2}(6x))$. Then the integral becomes $\int2\cos(6x)(1 - \sin^{2}(6x))dx$.

Step2: Use substitution

Let $u = \sin(6x)$, then $du=6\cos(6x)dx$ and $\cos(6x)dx=\frac{1}{6}du$. The integral $\int2\cos(6x)(1 - \sin^{2}(6x))dx$ becomes $\int2(1 - u^{2})\frac{1}{6}du=\frac{1}{3}\int(1 - u^{2})du$.

Step3: Integrate term - by - term

We know that $\int(1 - u^{2})du=\int1du-\int u^{2}du$. Since $\int1du = u$ 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

Substitute $u = \sin(6x)$ back into the result. We get $\frac{1}{3}(\sin(6x)-\frac{\sin^{3}(6x)}{3})+C=\frac{1}{3}\sin(6x)-\frac{1}{9}\sin^{3}(6x)+C$.

Answer:

$\frac{1}{3}\sin(6x)-\frac{1}{9}\sin^{3}(6x)+C$