Calc II Answers for Worksheet 7.2/7.3/7.1/7.4Worksheet 7.21. example2. .(a) sin2x =1cos 2x2;(b) cos2x =1+cos 2x2.3. Two ways to do it. Either way works.(a) sin 2x =2sinx cos x; or sin2x =1cos 2x2, cos2x =1+cos 2x2;(b)14sin22x;or14(1  cos22x);(c)x8132sin 4x + C.4. 05. .(a) u-sub, u =sinx;orcos x sin2x = cos x(1  cos2x) = cos x  cos3x;(b) u-sub.13sin3x + C.6. T.(a) Property of odd function. sin9x is an odd function, the integral interval is symmetric about the origin, so theintergral is 0;or(b) Evaluate the integral, let u = cos x,(the last equality is because the bounds of the ’new’ integral are the same)ˆ⇡⇡sin9xdx =ˆ⇡⇡(sin2x)4sin xdx = ˆcos ⇡cos(⇡)(1  u2)4du = ˆ11(1  u2)4du =0.Worksheet 7.31. example2. .(a) Trig Sub;(b) x = 3 tan ✓, dx =2cos2✓d✓;(c) hypotenuse=px2+9, adjacent=3,opposite= x(may not the only way);(d)19´cos ✓d✓3. .1(a) Trig Sub;(b) x =3sec✓, dx =3sec✓ tan ✓d✓;(c) hypotenuse= x, adjacent=3,opposite=px2 9(may not the only way);(d)13´sin2✓d✓;(e)16arc sec(x3) 12px29x2+ C;4.94⇡;5. F. In this case, a2=2)a =p2,thesubtitutionshouldbex =p2 tan ✓.Worksheet 7.11. example2. .(a) Integration by Parts;(b) u = x2, dv = e3xdx, du =2xdx, v = 13e3x.3. .(a) Integration by Parts;(b) u = x, dv = cos 3xdx, du = dx, v =13sin 3x;(c)13x sin 3x ´13sin 3xdx(d)13x sin 3x +19cos 3x + C(e) 294. ⇡2 45.exsin xexcos x2+ C6. TRUE. Apply Integration by Parts to the Left Hand Side and notice sin ⇡ =sin0=0to see Left Hand Side equalsthe Right Hand Side.Worksheet 7.41. example2. .(a) Partial Fractions;(b)Ax1+Bx3;(c) A = 12, B =12.3. .(a)Ax+Bx+Cx2+4;(b) A =14, B =34, C =1;(c)14ln |x| +38lnx2+4+12tan1(x2)+C4.116lnxx454⇥1x4+ C5. F, this integral can also be computed with U-Substitution.2Worksheet 7.71. example2. (a) 1344.79138;(b) 1603.73536(K = 307917.1892);(c)307917.189212n2or25659.76577n2;(d) n = 1602, 1603, 1604, ···.3. (a) 1.21895;(b) n = 10, 12, 14, ···(K =1516);4. T.By a picture, all the trapezoidal pieces line up to give one big trapezoid, since f (x) is linear, and this trapezoid isexactly the region whose signed area is´baf(x)dx.We can also use the error bound formula. f(x) is linear means that f(x)=ax + b for some constant a and b.Wehave f00(x)=0for all x. Thus we can take K =0, thus the error is bou nded by 0, i.e. there is no error.Worksheet 7.81. example2. (a) See Figure 1;(b) Shade the region between the gragh and x-axis (i.e. the line y=0) on 1  x<1;(c) It is bounded above and below on [1, 1).3. (a) See Figure 2;(b) Shade the region between the gragh and x-axis (i.e. the line y=0) on 0  x<1;(c)12ln(t2+ 1) !1,t !1. Thus the imp roper integral´10xx2+1diverges4. No, it is divergent.5. F. The integrand1x1blows up at 1, which is in [0, 2], so we have to treat´20as´10+´21, the sum of two improperintegral. Both integral are divergent (First integral equals 1 and the second one equals 1), thus the integral´20dxx1is divergent.30 2 4 6 8 10 12 14 16 18 20-0.2-0.100.10.20.30.40.50.6Figure 1: 7.8 2(a)0 2 4 6 8 10 12 14 16 18 2000.050.10.150.20.250.30.350.40.450.5Figure 2: 7.8