Question

Integral cos²x*dx/1-sinx

Answer 1

$\int \frac{cos^2x\cdot dx}{1-sinx} =\int \frac{(1-sin^2x)dx}{1-sinx} =\int \frac{(1-sinx)(1+sinx)dx}{1-sinx}=\\\\=\int (1+sinx)dx=x-cosx+C$

$P.S.\; \; sin^2x+cos^2x=1\; \; \to \; \; cos^2x=1-sin^2x\\\\a^2-b^2=(a-b)(a+b)$