1-ctg²a=(1+ctg²a)(sin²a-cos²a)
1+ctg²a=1/sin²a
sin²a-cos²a=sin²a-(1-sin²a)=2sin²a-1
(2sin²a-1)/sin²a=2-1/sin²a=2-(1+ctg²a)=1-ctg²a
1-ctg²a=1-ctg²a
что и требовалось доказать
<span>Sin(x+π/4)=1/2
x+</span>π/4=π/6+2πk U x+π/4=5π/6+2πk
x=-π/4+π/6+2πk U x=-π/4+5π/6+2πk
x=-π/12+2πk U x=7π/12+2πk,k∈z
Int[lnx]/x dx = int[lnx] d(lnx), т.к. d(lnx) = 1/x dx
int[lnx] d(lnx) = (lnx)^2/2
11n-7n+1-6n+8=-2n+9
-2*16+9=-32+9=-23