(2)=∫4/sin? 2xdx=2∫csc? 2xd2x=-2cot2x C
(3)=1/4∫(2x? 3)?(1/2)d(2x? 3)=(1/6)(2x? 3)?( 3/2) C
(4)=1/4∫pecado? 2xdx = 1/8∫(1-cos4x)dx = x/8-sin4x/32 C
(7)=∫sin? x(1-pecado?x)? dsinx=∫(sinx)? 7-2(sinx)? 5 pecado? xdsinx=(sinx)? 8/8-(sinx)? 6/3 (seis)? 4/4 C
(10)=∫√(4-(x 1)?)dx reemplaza x=2sinu-1
=∫2co sud(2 sinu-1) = 2 ∫( cos2u 1)du = sin2u 2u C
=(x 1)√(4-(x 1)?)/2 2arcsin((x 1)/2) C
(11) ¿Intercambio? √x=u, dx=3u? Du (apellido)
=∫3u? sinudu=-3∫u? dcosu=-3u? cosu 3∫cosudu? =-3u? cosu 6∫udsinu
=-3u? cosu 6usinu-6∫sinudu=-3u? cosu 6usinu 6cosu C
(14) Sustituyendo x=2secu, =∫2tanu/2secud2secu=2∫tan? udu=2tanu-2u=√(x?-4)-2arccos(2/x) C
(17)=xln(x? 1)-∫xdln(x? 1)=xln(x ?1)-∫2x? /(x? 1)dx=xln(x? 1)-2x 2arctanx C
(22)=∫x/(x-1)(x? 1)= 1/2∫(1/ (x-1)-(x-1)/(x? 1))dx
= 1/2∫1/(x-1)dx-1/2∫x/(x? 1 )dx 1/2∫1/(x? 1)dx
=(1/2)ln x-1 |-(1/4)ln(x? 1) arctanx/2 C