Question

Решите плизз
$\frac{2}{l} \int\limits^\frac{l}{5} _0 {sin^{2} (\frac{x\pi }{l} } )\, dx$

Answer 1

$\dfrac{2}{l} \displaystyle \int\limits^{l/5}_{0} {\sin^{2}\left(\dfrac{x\pi}{l} \right)} \, dx$

Сделаем замену: $\dfrac{x\pi}{l} = t$, откуда $x = \dfrac{tl}{\pi}$ и $dx = \dfrac{l}{\pi}dt$

Если $x = 0$, то $t = 0$

Если $x = \dfrac{l}{5}$, то $t = \dfrac{\pi}{5}$

$\dfrac{2}{l} \displaystyle \int\limits^{\pi/5}_{0} {\sin^{2}t} \cdot \dfrac{l}{\pi} \, dt =\dfrac{2}{l} \cdot \dfrac{l}{\pi} \displaystyle \int\limits^{\pi/5}_{0} {\sin^{2}t} \, dt = \dfrac{2}{\pi} \displaystyle \int\limits^{\pi/5}_{0} {\dfrac{1 - \cos 2t}{2} } \, dt =$

$\displaystyle = \dfrac{1}{\pi} \int\limits^{\pi /5}_{0} {(1 - \cos 2t)} \, dt = \dfrac{1}{\pi} \cdot \left(t - \dfrac{\sin 2t}{2} \right) \bigg | ^{\pi/5}_{0} =$

$= \dfrac{1}{\pi} \left(\dfrac{\pi}{5} - \dfrac{1}{2} \sin \dfrac{2\pi}{5} - 0 + \dfrac{\sin 0}{2} \right) = \dfrac{1}{5} - \dfrac{1}{2\pi} \sin \dfrac{2\pi}{5}$

Ответ: $\dfrac{1}{5} - \dfrac{1}{2\pi} \sin \dfrac{2\pi}{5}$