# prob5sol - Math 151 Spring 2008 Problem Set # 5 Solutions...

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Math 151 Spring 2008Problem Set # 5SolutionsTrigonometric and Hyperbolic Integrals and Substitutions1.We setu= sin (x)so thatdu= cos (x)dx:Zcos3(x) sin2(x)dx=Zcos2(x) sin2(x) cos (x)dx=Z¡1sin2(x)¢sin2(x) cos (x)dx=Z¡1u2¢u2du=Z¡u2u4¢du=13u315u5=13sin3(x)15sin5(x).2.We setu= cos (x)so thatdu=sin (x)dx:Zsin3(x) cos4(x)dx=Zsin2(x) cos4(x) sin (x)dx=Z¡1cos2(x)¢cos4(x) sin (x)dx=Z¡1u2¢u4du=Z¡u4u6¢du=15u5+17u7=15cos5(x) +17cos7(x).3.We setu= sinh (x)so thatdu= cosh (x)dx:Zcosh3(x) sinh2(x)dx=Zcosh2(x) sinh2(x) cosh (x)dx=Z¡sinh2(x) + 1¢sinh2(x) cosh (x)dx=Z¡u2+ 1¢u2du=Z¡u4+u2¢du=15u5+13u3=15sinh5(x) +13sinh3(x).1
4.We setu= cosh (x)so thatdu= sinh (x)dx:Zsinh3(x) cosh4(x)dx=Zsinh2(x) cosh4(x) sinh (x)dx=Z¡cosh2(x)1¢cosh4(x) sinh (x)dx=Z¡u21¢u4du=Z¡u6u4¢du=17u715u5=17cosh7(x)15cosh5(x).5.Zcos2(x)dx=Z12(1cos (2x))dx=x212Zcos (2x)dx=x212μ12sin (2x)=x214sin (2x)μ=x212sin (x) cos (x).6.We setu= cosh (x)anddv= cosh (x)dxso thatdu= sinh (x)dxandv=Zcosh (x)dx= sinh (x).Thus,Zcosh2(x)dx=Zcosh (x) cosh (x)dx=Zudv=uvZvdu=cosh (x) sinh (x)Zsinh (x) sinh (x)dx=cosh (x) sinh (x)Zsinh2(x)dx=cosh (x) sinh (x)Z¡cosh2(x)1¢dx=cosh (x) sinh (x) +xZcosh2(x)dx.2
Therefore,2Zcosh2(x) = cosh (x) sinh (x) +x,so thatZcosh2(x) =12cosh (x) sinh (x) +12x7.We setu= sinh (x)anddv= sinh (x)dxso thatdu= cosh (x)dxandv=Zsinh (x)dx= cosh (x).Thus,Zsinh2(x)dx=Zsinh (x) sinh (x)dx=Zudv=uvZvdu=sinh (x) cosh (x)Zcosh (x) cosh (x)dx=cosh (x) sinh (x)Zcosh2(x)dx=cosh (x) sinh (x)Z¡sinh2(x) + 1¢dx=cosh (x) sinh (x)xZsinh2(x)dx.Therefore,2Zsinh2(x) = cosh (x) sinh (x)x,so thatZsinh2(x) =12cosh (x) sinh (x)12x.3
8.We setx= 3 sin (u)so thatu= arcsin³x3´. dx= 3 cos (u)duandp9x2= 3 cos (u).Thus,Zp9x2dx=Z(3 cos (u)) (3 cos (u))du=9Zcos2(u)du=9μ12cos (u) sin (u) +12u=92cos (u) sin (u) +92u=92Ã9x23!³x3´+92arcsin³x3´=12xp9x2+92arcsin³x3´.9.We havex24+y29= 1y29= 1x24y2= 9μ1x24Therefore, the upper part of the ellipse is the graph ofy= 3r1x24=32p4x2,2x2.By symmetry, the area of the region inside the ellipse isZ2232p4x2dx=32Z22p4x2dxWe setx= 2 sin (u)so thatu= arcsin³x2´, dx= 2 cos (u)duandp4x2= 2 cos (u).It is practical to implemen t the definite integral version of the substitution rule:32Z22p4x2dx=32Zarcsin(1)arcsin(1)(2 cos (u)) (2 cos (u))du=6Zπ/2π/2cos2(u)du=6Ã12cos (u) sin (u) +12u¯¯¯¯π/2π/2!

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