Question #110060

∫¹∫2 (cotx + sinx)dx


1
Expert's answer
2020-04-30T19:49:07-0400

∫12(cotx+sinx)dx=(ln⁡sinx−cosx)\int_{1}^{2}(cotx+sinx)dx=(\ln sinx-cosx)∣12=ln⁡sin2−ln⁡sin1−cos2+cos1≈1.03|_1^2=\ln sin2-\ln sin1-cos2+cos1\approx1.03


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