∫¹∫2 (cotx + sinx)dx
"\\int_{1}^{2}(cotx+sinx)dx=(\\ln sinx-cosx)""|_1^2=\\ln sin2-\\ln sin1-cos2+cos1\\approx1.03"
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment