evaluate $int_{0}^{4pi}sin^{5}\frac{x}{8}dx$\n$int_{0}^{4pi}sin^{5}\frac{x}{8}dx=square$ (type an exact…

Explanation:

Step1: Use substitution

Let $u = \frac{x}{8}$, then $x = 8u$ and $dx=8du$. When $x = 0$, $u = 0$; when $x = 4\pi$, $u=\frac{\pi}{2}$. The integral becomes $\int_{0}^{\frac{\pi}{2}}8\sin^{5}u\ du$.

Step2: Recall the reduction - formula for $\int\sin^{n}x\ dx$

The reduction - formula for $\int\sin^{n}x\ dx=-\frac{1}{n}\sin^{n - 1}x\cos x+\frac{n - 1}{n}\int\sin^{n - 2}x\ dx$. For $n = 5$, we have: $\int\sin^{5}u\ du=-\frac{1}{5}\sin^{4}u\cos u+\frac{4}{5}\int\sin^{3}u\ du$. For $\int\sin^{3}u\ du=-\frac{1}{3}\sin^{2}u\cos u+\frac{2}{3}\int\sin u\ du$. And $\int\sin u\ du=-\cos u + C$. So, $\int\sin^{3}u\ du=-\frac{1}{3}\sin^{2}u\cos u-\frac{2}{3}\cos u + C$. Then $\int\sin^{5}u\ du=-\frac{1}{5}\sin^{4}u\cos u+\frac{4}{5}\left(-\frac{1}{3}\sin^{2}u\cos u-\frac{2}{3}\cos u\right)+C$.

Step3: Evaluate the definite integral

$8\int_{0}^{\frac{\pi}{2}}\sin^{5}u\ du=8\left[-\frac{1}{5}\sin^{4}u\cos u-\frac{4}{15}\sin^{2}u\cos u-\frac{8}{15}\cos u\right]_{0}^{\frac{\pi}{2}}$. When $u=\frac{\pi}{2}$, $\sin u = 1,\cos u = 0$: $-\frac{1}{5}\sin^{4}u\cos u-\frac{4}{15}\sin^{2}u\cos u-\frac{8}{15}\cos u = 0$. When $u = 0$, $\sin u=0,\cos u = 1$: $-\frac{1}{5}\sin^{4}u\cos u-\frac{4}{15}\sin^{2}u\cos u-\frac{8}{15}\cos u=-\frac{8}{15}$. So, $8\times\left(0-\left(-\frac{8}{15}\right)\right)=\frac{64}{15}$.

Answer:

$\frac{64}{15}$