Mohon bantu jawab ,masih belum tau bagaimana cara penyelesaiannya. nomor 1 - 2

integral
.
{ sec x = 1/cos x}
.
(π/2...π/4) ∫ cos x sec³ dx dx =
= (π/2..π/4) ∫ cos x (1/cos ³ x) dx
= (π/2..π/4) ∫ 1/cos² x dx
= (π/2 ..π/4) ∫ sec² x dx
= [tan x ](π/2 ...π/4)
= tan π/2 - tan π/4
= eror

2. ∫ ln² (x)  dx
dengan parsial ∫ u dv = uv - ∫v du
u = ln² (x)  --> du = (2 ln x)/x 
dv = dx --> v = ∫dx = x
∫ ln² (x) dx = x ln² x - ∫ x { 2 ln x )/ x) dx  + c
= x ln² x - ∫ 2 ln x dx + c
.
cari dulu ∫ 2 ln x dx , dgn parsial  ∫ f dg = fg - ∫ g df
f = 2 ln x , df = 2/x
dg = dx --> g = ∫ dx = x
∫2 ln x dx = 2x ln x - ∫ x (2/x) dx
∫ 2 ln x dx = 2(x). ln (x) - 2x
.
∫ ln² (x) dx = x ln² (x) - 2x ln x + 2x + c