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Answer to Question #110060 in Calculus for Nimra
Question #110060
∫¹∫2 (cotx + sinx)dx
Expert's answer
"\\int_{1}^{2}(cotx+sinx)dx=(\\ln sinx-cosx)""|_1^2=\\ln sin2-\\ln sin1-cos2+cos1\\approx1.03"
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