Задача 65203 ...

62b415697ce1dc540585723f

23.06.2022 10:28:31

Найти неопределенный интеграл сведением к табличному. ∫ (tg^2x-1)dx, ∫ (dx)/(16+x^2)

626fab9a354ab65fb2f43abb

23.06.2022 10:51:23

★

[m]\int (tg^2 x - 1) dx = \int (\frac{sin^2 x}{cos^2 x} - 1) dx = \int \frac{sin^2 x - cos^2 x}{cos^2 x} dx= \int \frac{sin^2 x + cos^2 x - 2cos^2 x}{cos^2 x} dx= [/m]
[m]= \int \frac{1 - 2cos^2 x}{cos^2 x} dx= \int (\frac{1}{cos^2 x} - 2) dx = tg x - 2x + C[/m]

[m]\int \frac{dx}{16+x^2} =\int \frac{dx}{16(1+x^2/16)} =\frac{1}{16}\int \frac{dx}{1+(x/4)^2} = \frac{1}{16}\int \frac{4*d(x/4)}{1+(x/4)^2} = \frac{4}{16}\int \frac{d(x/4)}{1+(x/4)^2} = \frac{1}{4}arctg\frac{x}{4} + C[/m]