Express 1(1+4x)(1−x)\frac{1}{(1+4x)(1-x)}(1+4x)(1−x)1 in partial fractions.
Hence find the solution of the differential equation
(1+4x)(1−x)dydx=tany (1+4x)(1-x)\frac{dy}{dx} = \tan y (1+4x)(1−x)dxdy=tanyfor the interval −14<x<1-\frac{1}{4} < x < 1−41<x<1, given that y=π2y = \frac{\pi}{2}y=2π when x=0x = 0x=0.
Give your answer in the form sinny=f(x)\sin^n y = f(x)sinny=f(x) where nnn is an integer to be found.