Aditya Gupta
Last Activity: 4 Years ago
differentiate both sides we get
f(x)sinxcosx= f’(x)/[2(b^2 – a^2)f(x)] where we have used chain rule to diff lnf(x).
let y= f(x) so that
dy/dx= 2(b^2 – a^2)*y^2sinxcosx
or dy/y^2= (b^2 – a^2)*2sinxcosx dx
integrate both sides
– 1/y= 2(b^2 – a^2)*sin^2x + c
or y= 1/[(a^2 – b^2)*sin^2x – c] where c is any constant.
if we let c= – b^2, then
y= 1/[(a^2sin^2x + b^2(1 – sin^2x)]
or y= f(x)= 1/(a^2sin^2x + b^2cos^2x)
KINDLY APPROVE :))