1-cosx=t dt=-sinx это ошибка распишите полностью пример плиз. Само решение

1-cosx=t
dt=-sinx это ошибка
распишите полностью пример плиз.
Само решение правильное,но подчеркнутое палками нет
- Предмет:
  Алгебра
- Автор:
  dominiquedorsey
- 6 лет назад

Ответы 1

$\int\limits^{2 \pi} _ \frac{3 \pi }{2} {( \sqrt{1-cosx}\times sin x)} \, dx = \int\limits^{2 \pi} _ \frac{3 \pi }{2} {\sqrt{1-cosx}} \, d(-cosx )=\\\\=\int\limits^{2 \pi} _ \frac{3 \pi }{2} {\sqrt{1-cosx}} \, d(1-cosx )= \frac{2}{3} \sqrt{(1-cosx)^3} |^{2 \pi} _ \frac{3 \pi }{2}=\\\\=0- \frac{2}{3} *1=- \frac{2}{3}$ в принципе можно делать замену t = 1-cosx, тогда d(1-cosx) = dtпределы 1 и 0, но думаю можно и без замены, и так очевидно
- Автор:
  josiahgutierrez
- 6 лет назад
- 0
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1-cosx=t dt=-sinx это ошибка распишите полностью пример плиз. Само решение правильное,но подчеркнутое палками нет

Ответы 1

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1-cosx=t
dt=-sinx это ошибка
распишите полностью пример плиз.
Само решение правильное,но подчеркнутое палками нет