int_0^pi sin^5left( frac{x}{2} right) , dx eq | vedclass

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(A) Let $I = \int_0^\pi \sin^5\left( \frac{x}{2} \right) \, dx$. Substitute $t = \frac{x}{2}$,then $dx = 2 \, dt$. When $x = 0$,$t = 0$. When $x = \pi

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$\frac{16}{15}$

Vedclass · Answer

(A) Let $I = \int_0^\pi \sin^5\left( \frac{x}{2} \right) \, dx$. Substitute $t = \frac{x}{2}$,then $dx = 2 \, dt$. When $x = 0$,$t = 0$. When $x = \pi$,$t = \frac{\pi}{2}$. So,$I = \int_0^{\pi/2} \sin^5(t) \cdot 2 \, dt = 2 \int_0^{\pi/2} \sin^5(t) \, dt$. Using Wallis' Formula,$\int_0^{\pi/2} \sin^n(x) \, dx = \frac{(n-1)(n-3)\dots(1)}{n(n-2)\dots(2)} \cdot \frac{\pi}{2}$ (if $n$ is even) or $\frac{(n-1)(n-3)\dots(2)}{n(n-2)\dots(1)}$ (if $n$ is odd). For $n = 5$,$\int_0^{\pi/2} \sin^5(t) \, dt = \frac{4 \cdot 2}{5 \cdot 3 \cdot 1} = \frac{8}{15}$. Therefore,$I = 2 \cdot \frac{8}{15} = \frac{16}{15}$.

$\int_0^\pi \sin^5\left( \frac{x}{2} \right) \, dx$ equals

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