Есть тригонометрический интеграл - подскажите способ решения:
int dx/(sin x + cos x)
int dx/(sin x + cos x) = int dx/(2 * sin (x/2) * cos (x/2) + cos^2 (x/2) - sin^2 (x/2)) =
= -int dx/(cos^2 (x/2) * (tg^2 (x/2) - 2 * tg (x/2) - 1)) =
= -2 * int d(tg (x/2))/(tg^2 (x/2) -2 * tg (x/2) - 1) = |tg (x/2) = t | =
= -2 * int dt/(t^2 - 2 * t - 1) = - 2 * int dt/((t - 1)^2 - 2) =
= -2 * int d(t - 1)/((t - 1)^2 - 2) = | u = t - 1 | =
= -2 * int du/(u^2 - 2) = 2 * int du/(2 - u^2) =
= 2 * 1/(2 * 2^(1/2) * ln |(2^(1/2) + u)/(2^(1/2) - u)| + C = | u = t - 1 | =
= 1/2^(1/2) * ln |(2^(1/2) + t - 1)/(2^(1/2) - t + 1)| + C = | t = tg (x/2) | =
= 1/2^(1/2) * ln |(2^(1/2) + tg (x/2) - 1)/(2^(1/2) - tg (x/2) + 1)| + C
Способ понял, но сам бы никогда не догадался... Спасибо огромное!!!)
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