Question

решить неопределенный интеграл:

$\int\limits{\frac{Sinx}{3Cos^2x-1} } \, dx$

Answer 1

$\int\limits \frac{ \sin(x) }{3 \cos {}^{2} (x) - 1} dx \\ \\ \cos(x) = t \\ - \sin(x) dx = dt \\ \sin(x) dx = - dt \\ \\ - \int\limits \frac{dt}{3 {t}^{2} - 1} = - \int\limits \frac{dt}{ {( \sqrt{3} t)}^{2} - 1 {}^{2} } = \\ = - \frac{1}{ \sqrt{3} } \int\limits \frac{d( \sqrt{3}t )}{( \sqrt{3} t) {}^{2} - {1}^{2} } = \\ = - \frac{1}{ \sqrt{3} } \times \frac{1}{2} ln | \frac{ \sqrt{3} t - 1}{ \sqrt{3}t + 1 }| + c = \\ = - \frac{1}{2 \sqrt{3} } ln | \frac{ \sqrt{3} \cos(x) - 1 }{ \sqrt{3} \cos(x) + 1 } | + c$