Question

Решите плизз
$frac{2}{l} intlimits^frac{l}{5} _0 {sin^{2} (frac{xpi }{l} } ), dx$

Answer 1

$dfrac{2}{l} displaystyle intlimits^{l/5}_{0} {sin^{2}left(dfrac{xpi}{l} right)} , dx$

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

Если $x = 0$, то $t = 0$

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

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

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

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

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