evaluate $intcos^{3}\frac{x}{6}dx$.\n$intcos^{3}\frac{x}{6}dx=square$

Explanation:

Step1: Rewrite $\cos^{3}u$

Use the identity $\cos^{3}u=\cos u(1 - \sin^{2}u)$. Let $u = \frac{x}{6}$, then $du=\frac{1}{6}dx$ and $dx = 6du$. So the integral $\int\cos^{3}\frac{x}{6}dx=6\int\cos^{3}u du=6\int\cos u(1 - \sin^{2}u)du$.

Step2: Use substitution

Let $t=\sin u$, then $dt=\cos udu$. The integral $6\int\cos u(1 - \sin^{2}u)du$ becomes $6\int(1 - t^{2})dt$.

Step3: Integrate term - by - term

$6\int(1 - t^{2})dt=6\left(\int 1dt-\int t^{2}dt\right)$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $6\left(t-\frac{t^{3}}{3}\right)+C$.

Step4: Substitute back

Substitute $t = \sin u$ and $u=\frac{x}{6}$ back. We get $6\left(\sin\frac{x}{6}-\frac{\sin^{3}\frac{x}{6}}{3}\right)+C = 6\sin\frac{x}{6}-2\sin^{3}\frac{x}{6}+C$.

Answer:

$6\sin\frac{x}{6}-2\sin^{3}\frac{x}{6}+C$