Задача 78847 ...

673235f95a00e67c51d41502

11/29/2024 6:11:23 PM

Вычислите интеграл тригонамеьрических функций

7. ∫ tg³ x dx

626fab9a354ab65fb2f43abb

11/29/2024 8:42:50 PM

★

[m]\int tg^3\ x\ dx = \int \frac{\sin^3 x}{\cos^3 x}\ dx = \int \frac{\sin^2 x}{\cos^3 x} \cdot \sin x\ dx = \int \frac{1 - \cos^2 x}{\cos^3 x} \cdot \sin x\ dx[/m]
Замена t = cos x; dt = -sin x dx
[m]\int \frac{1 - \cos^2 x}{\cos^3 x} \cdot \sin x\ dx = \int \frac{1 - t^2}{t^3} (-dt) = \int \frac{t^2 - 1}{t^3} dt = \int (\frac{t^2}{t^3} - \frac{1}{t^3})\ dt =[/m]
[m]= \int (\frac{1}{t} - t^{-3})\ dt = \ln |t| - \frac{t^{-2}}{-2} + C = \ln |\cos x| + \frac{1}{2\cos^2 x} + C[/m]