Question

Помогите пожалуйста! Найти неопределенный интеграл:
$\int\limits \frac{sin^2(x)}{1+sin^2(x)} dx$

Answer 1

Ответ:

$\int\limits \frac{ \sin {}^{2} (x) }{ \sin {}^{2} (x) + 1} dx = \int\limits \frac{ \sin {}^{2} (x) + 1 - 1}{ \sin {}^{2} (x) + 1 } dx = \\ = \int\limits \: dx - \int\limits \frac{dx}{ \sin {}^{2} (x) + 1} \\ \\ \sin {}^{2} ( x) = \frac{1}{1 + {ctg}^{2}x } \\ \\ = x - \int\limits \frac{dx}{ \frac{1}{1 + {ctg}^{2}x } + 1} = x - \int\limits \frac{1 + {ctg}^{2}x }{1 + 1 + {ctg}^{2} x} dx = \\ = x - \int\limits \frac{ \frac{1}{ \sin {}^{2} (x) } }{2 + {ctg}^{2}x } dx = x + \int\limits \frac{( - \frac{1}{ \sin {}^{2} (x) } )}{2+ {ctg}^{2}x } dx = \\ = x + \int\limits \frac{d(ctgx)}{2 + {ctg}^{2}x } = x + \int\limits \frac{d(ctgx)}{( \sqrt{2}) {}^{2} + ctg {}^{2} x} = \\ = x + \frac{1}{ \sqrt{2} } arctg( \frac{ctgx}{ \sqrt{2} }) + C$