`int sin2x.log cos xdx` is equal to
`1) cos^(2)x((1)/(2)+log cos x)+k`
`2) cos^(2)x*log cos x+k`
`3) cos^(2)x((1)/(2)-log cos x)+k`
`4) cos^(2)x+log cos x+k`

int log(sin x)*cos xdx

int(1)/(x cos^(2)(1+log x))dx

int cos(log x)dx is equal to (A)(x)/(2)(cos(log x)-sin(log x))+c(B)x(cos(log x)+sin(log x))+c(C)(x)/(2)(cos(log x)+sin(log x)+c(D)x(cos(log x)-sin(log x))+c

3((1)/(2)+log_(3)(cos x+sin x))-2^(log_(2)(cos x-sin x))=sqrt(2)

int sqrt(1+2cot x(cot x+cos ecx))dx is equal to (A) 2In((cos x)/(2))+c(B)2ln(s in(x)/(2))+c(C)(1)/(2)ln((cos x)/(2))+c(D)(1)/(2)ln(s in(x)/(2))+c

solve for x:log_(sin^(2)x)(2)+log_(cos^(2)x)(2)+2log_(sin^(2)x)(2)log_(cos^(2)x)(2)=0

The solution of the equation log _( cosx ^(2)) (3-2x) lt log _( cos x ^(2)) (2x -1) is:

int(cos4x-1)/(cot x-tan x)dx is equal to (A)(1)/(2)ln|sec2x|-(1)/(4)cos^(2)2x+c(B)(1)/(2)ln|sec2x|+(1)/(4)cos^(2)x+c(C)(1)/(2)ln|cos2x|-(1)/(4)cos^(2)2x+c(D)(1)/(2)ln|cos2x|+(1)/(4)cos^(2)x+c

int_(0)^((pi)/(2))cos2x log sin xdx

If f(x)sin x cos xdx=(1)/(2(b^(2)-a^(2)))ln f(x)+c, then f(x) is equal to (1)/(a^(2)sin^(2)x+b^(2)cos^(2)x)(b)(1)/(a^(2)sin^(2)x-b^(2)cos^(2)x+b^(2)cos^(2)x+b^(2)cos^(2)x)(d)(1)/(a^(2)cos^(2)x-b^(2)cos^(2)x)