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July 2022
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помогите пожалуйста решить интегралы
1)Сверху 2пи
Снизу - пи
Х/3 dx
2)Сверху 3Пи
Снизу 0
Dx/cos^2 x/9
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ПрофиЗнания
A) -cos x/3 | (-п;2п) = -cos 2п/3 - (-cos -п/3) = -(-1/2) + cos п/3 = 1/2 +1/2=1
b) tg x/9 | (0;3п) = tg 3п/9 - tg0 = tg п/3 - tg0 = √3-0=√3
2 votes
Thanks 1
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Answers & Comments
b) tg x/9 | (0;3п) = tg 3п/9 - tg0 = tg п/3 - tg0 = √3-0=√3