# exam2_fa00_answer - Math 20B(Bender Second Exam November...

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Math 20B (Bender) Second Exam November 2000 1. (80 pts.) Evaluate the following integrals. Remember to show your work! (a) [7.5, #6] Z (sin x ) (cos(cos x )) dx Ans. Use u = cos x to get rid of the messy cosine stuff. Then du = - sin x dx and so we have R - cos u du = - sin u + C , which gives - sin(cos x ) + C . (b) [7.5, #10] Z t 2 1 - t 2 dt Ans. Set t = sin θ . Then dt = cos θ dθ and the integral becomes R sin 2 θ dθ = (1 / 2) R (1 - cos 2 θ ) , which is ( θ/ 2) - (sin 2 θ ) / 4 + C . You now must substi- tute back in for θ : sin - 1 t 2 + sin(2 sin - 1 t ) 4 + C. You can leave your answer in this form; however, you could simplify it using sin(2 θ ) = 2 sin θ cos θ = 2 t 1 - t 2 . (c) [7.5, #26] Z sin( t 1 / 2 ) dt Ans. Set t 1 / 2 = x to obtain dt = 2 x dx and R 2 x sin x dx . This integral can be done by parts with x = u and sin x dx = dv . We obtain - 2 x cos x + 2 R cos x dx . Integrating and substituting back gives - 2 t 1 / 2 cos t 1 / 2 + 2 sin t 1 / 2 + C . (d) [7.5, #44] Z 1 + e x 1 - e x dx Ans. The are various possibilities. Setting e x = u , we have dx = du/e u and the integral becomes Z 1 + u 1 - u du u = Z 2 1 - u + 1 u du, by partial fractions. Integrating and substituting back gives

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