Express 1(1+2x)(1−x)\frac{1}{(1+2x)(1-x)}(1+2x)(1−x)1 in partial fractions.
Hence find the solution of the differential equation
(1+2x)(1−x)dydx=tany (1+2x)(1-x)\frac{dy}{dx} = \tan y (1+2x)(1−x)dxdy=tanyfor the interval −12<x<1-\frac{1}{2} < x < 1−21<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.