Economics 3364A/B Lecture 8: Calc Quiz 6

• School: Western University
• Department: Economics
• Course: Economics 3364A/B
• Professor: holdi

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Then dv = v dx = (ln x) dx = 1 x ln x. 1 x dx; and v = 2 when x = e2, v = 1 when. 1 = 2 2 2 1 = 2 2 2. Solution: noticing that (cot x) = csc2 x, we use the substitute v = cot x. Then, dv = v dx = (cot x) dx = csc2 x dx, i. e. , csc2 x dx = dv. Therefore r csc2 x cot x dx = r cot x csc2 x dx = R v( dv) = r v dv = v. 1 4x2 cos 1(2x) dx = z 1 u 1. = cos 1 u + c = cos 1(e x) + c: evaluate the limit limx 0. 0 sin(t2) dt x6 one sees that this is an 0 with the fundamental theorem of calculus i and chaine rule) to obtain. 0 type indeterminated form, and hence one can apply l. h"s rule (combined.