Answer to Question #345560 in Calculus for Jason

Question #345560

Inregrate "\\int \\frac{\\cos x dx}{\\sin^2 x}"

Expert's answer

"\\int \\cfrac{\\cos x dx}{\\sin^2x} = [t = \\sin x, dt = \\cos x dx] = \\int \\cfrac{dt}{t^2} = -\\cfrac{1}{t}+C=-\\cfrac{1}{\\sin x}+C"

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