Evaluate: `intcot^3xcos e c^2xdx`

Explanation

`cos y = xcos(a+y)`

Differentiating both sides w.r.t. `x`,

`-sinydy/dx = cos(a+y)(1)+x(-sin(a+y)dy/dx)`

`=>dy/dx(xsin(a+y) - siny) = cos(a+y)`

As, `cos y = xcos(a+y) => x = cosy/cos(a+y)`

`:. dy/dx(cosysin(a+y) - sinycos(a+y)) = cos^2(a+y)`

As `sinAcosB-cosAsinB = sin(A-B)`,

`:.dy/dx(sin(a+y-y)) = cos^2(a+y)`

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