102-2007&2008-1-M20-Dec2007

102-2007&2008-1-M20-Dec2007 - Kuwait University...

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Unformatted text preview: Kuwait University Department of Mathematics & Computer Science Math 102 December 15, 2007 Calculus II ' Second Midterm Time: 75 minutes A. Evaluate the following integrals: [3 pts each] 1. f csc a: cot a: ln(sin at) da: 2. tan”2 :1: sec4 :6 d3: 3. [(205355 singxdx cosh 3: 4' f 1 + sin(sinh 3) (ix (x + 2')2 m ‘ \VE 6' fwd“; dz B. Determine whether the following integrals converge or diverge; if convergent, find the value. 7. [0° 6‘.” d3: [3 NS] 2 1 O. 09 . Integration by parts with u = ln(sin 3) and d1: = csca: cot x dx yields fescz cot$1n(sina:)d:r : — csca: ln(sinx] + [csca cotzdx which reduces to csc IL' ln(csc z) r- csca: + C. . Piece of cake: put as = tanm [tan3/2 1 sec“ coda: = ftanafl J: (1 + ten2 z) seczmda: = [913/2 + HT”) (11: . Use the identity 2 cost: (3056 = cos(a + b} + cos(a — b} and the half angle formula for sin2: . 1 1 c0333 511122: = E (30531: (1 —cos2x) = -2- cos 3m — i(cos 53:+ cosm) Then [ elsinSm—i '5 ls'n +0 — 6 511] 1'6 4 l I . Put 2 = sinha; and you’ll have f coshx dz“ f dz 1+sin(sinh:c) _ 1+sinz Weierstrass substitution reduces this to 2 2 z sinhx /(1+u)?d”"*1+u+0’ “Ami—tan 2 . Note that —:r2 m 455' = 4 i (m + 2)? Let :1: + 2 = 2 sin 9. The integral becomes (a: + 2)2 Wain:f4si1126d6=f2(1—cos23)d9 etc. — x . Put :6 = u” to get j W dx=/—12~—)du:4f[i— 1 ]du “us/hfi)‘ which becomes . Divergent! . Put x = sec2 9. Then 1 2 . fmdx=f2 cos 9d3:3+sm90036 which, in terms of 1', becomes 2 _ 1 + 4 S V‘ —1 lim sec‘lx/cT:+ m 341+ .12 1 2 O. 09 ...
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102-2007&2008-1-M20-Dec2007 - Kuwait University...

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