Question

интегралы...........​

Answer 1

Ответ:

Пошаговое объяснение:

1.

$\displaystyle \int{ctg^3(6x)*csc^2(6x)} \, dx =\left[\begin{array}{ccc}u=6x\\du=dx\\\end{array}\right] =\frac{1}{6} \int {ctg^3u*csc^2u} \, du =$

$\displaystyle = \left[\begin{array}{ccc}s=ctgu\\ds=-csc^2du\\\end{array}\right] =-\frac{1}{6}\int{s^3} \, ds=-\frac{1}{24} ctg^4(6x) +C$

2/

$\displaystyle \int\limits^{\pi /2}_{-\pi /2} {2sin^2(x /4)} \, dx =\left[\begin{array}{ccc}u=x/4 \quad du = (1/4)dx\\u_1=-(\pi /8)\hfill\\u_2=\pi/8\hfill\end{array}\right] =-8\int\limits^{\pi /8}_{-\pi /8} {sin^2u} \, du =$

$\displaystyle 8\int\limits^{\pi /8}_{-\pi /8} {(\frac{1}{2}-\frac{1}{2}cos2u)} } \, du=4u\bigg |_{-\pi n/8}^{\pi /8}-4\left[\begin{array}{ccc}s=2u\quad ds=2du\\s_1={-\pi /4}\hfill\\s_2={\pi /4}\hfill\end{array}\right] =$

$\displaystyle = \pi -2sins \bigg |_{-\pi /4}^{\pi /4}=\pi -2\sqrt{2}$