Задача 31811 Решение задач...

vk380208912

2018-12-11 14:03:38

Решение задач

sova

2018-12-11 15:02:20

1.
tg^25x=sin^25x/cos^25x=(1-cos^25x)/cos^25x=(1/cos^25x) - 1

∫ tg^25x dx= ∫ ((1/cos^25x) - 1)dx = (1/5)tg(5x)- x + C

2.

5sin^2x-3cos^2x+4*(sin^2x+cos^2x)=9sin^2x+cos^2x

= ∫ dx/(9sin^2x+cos^2x)=(1/9) ∫ 1/(tg^2x+(1/9)) * dx/(cos^2x)=

[b][tgx=t; dx/cos^2x=dt][/b]

=(1/9) ∫ dt/(t^2+(1/9))=(cм. 13) (1/9) * 1/(1/3) atctg t/(1/3) + C=

=(1/3) arctg(3tgx)+C

3.
= ∫ dx/(2cos^2x-sin^2x)=∫ 1/(2-tg^2x) * dx/(cos^2x)=

[b][tgx=t; dx/cos^2x=dt][/b]

= ∫ dt/(2- t^2)= - ∫ dt/( t^2- 2) = ( см. 10)

=(1/2sqrt(2))ln|(t-sqrt(2))/(t+sqrt(2))| + C=

=(1/2sqrt(2))ln|(tgx-sqrt(2))/(tgx+sqrt(2))| + C