Question

Интеграл (1-cosx)/((x-sinx)^2)

Answer 1

$\int \frac{(1-cosx)dx}{(x-sinx)^2}=[\, t=x-sinx\; ,\; dt=(x-sinx)'dx=(1-cosx)dx\, ]=\\\\=\int \frac{dt}{t^2}=\int t^{-2}\, dt=\frac{t^{-1}}{-1}+C=-\frac{1}{x-sinx}+C$

Answer 2

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