If f(x, y) = 0 be the solution of differential equation (2y cosec 2x + ln cot y)dx + (ln tan x - 2x cosec 2y)dy = 0 such that f(π4,π2)=0
f(x, y) is
(tan x)x(cot y)y
(tan x)y(cot y)x
(tan x)x2(cot y)y2
(tan x)y2(cot y)x2
f(x,y)=(tan x)y(cot y)x
If ∫∞0(5π4,2017π4)cos xxdx=π2,then∫∞01x(1−sin2x)32dx is equal to
If f(x)=0 is a quadratic equation such that f(−π)=f(π)=0 and f(π2)=−3π24, then limx→−πf(x)sin(sinx) is