12028-0130670227_ismSec7 - Section 8.7 C08S07.001 C08S07.002 x2 1 dx = 4x 5 1 dx = arctan(x 2 C(x 2)2 1 2x 5 dx = x2 4x 5 2x 4 1 x2 4x 5 x2 4x 5 dx =

12028-0130670227_ismSec7 - Section 8.7 C08S07.001...

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Section 8.7C08S07.001:1x2+ 4x+ 5dx=1(x+ 2)2+ 1dx= arctan(x+ 2) +C.C08S07.002:2x+ 5x2+ 4x+ 5dx=2x+ 4x2+ 4x+ 5+1x2+ 4x+ 5dx= ln(x2+4x+5)+arctan(x+2)+C(the antiderivative of the second fraction was computed in the solution of Problem 1).C08S07.003:53xx2+ 4x+ 5dx=322x+ 4223x2+ 4x+ 5dx=32ln(x2+ 4x+ 5) + 111x2+ 4x+ 5dx=32ln(x2+ 4x+ 5) + 11arctan(x+ 2) +C(see Problem 1).C08S07.004:We will obtainx2+ 4x+ 5 = (x+ 2)2+ 1 = 1 + tan2θ= sec2θif we letx=2 + tanθ. Ifso, thendx= sec2θ dθ,x+ 1 =1 + tanθ, and tanθ=x+ 2. ThusI=1 + tanθsec4θsec2θ dθ=(1 + tanθ)cos2θ dθ=(cos2θ+ sinθcosθ)=sinθcosθ1 + cos2θ2=12sin2θ12θ14sin2θ+C=12sin2θ12θ12sinθcosθ+C.A reference triangle with acute angleθ, opposite sidex+ 2, and adjacent side 1 has hypotenuse of lengthx2+ 4x+ 5. ThereforeI=12·(x+ 2)2x2+ 4x+ 512arctan(x+ 2)12·x+ 2x2+ 4x+ 5+C=12·x2+ 4x+ 4x2x2+ 4x+ 512arctan(x+ 2) +C=x2+ 3x+ 22(x2+ 4x+ 5)12arctan(x+ 2) +C.C08S07.005:32xx2=(x2+2x3) =(x2+2x+14) = 4(x+1)244sin2θ= 4cos2θifwe (and we do) letx+ 1 = 2sinθ. Thenx=1 + 2sinθ,dx= 2cosθ dθ, and sinθ=12(x+ 1). So132xx2dx=12cosθ·2cosθ dθ=1=θ+C= arcsinx+ 12+C.C08S07.006:The same substitution as in the solution of Problem 5 yieldsJ=x+ 332xx2dx=2 + 2sinθ2cosθ·2cosθ dθ= 2θ2cosθ+C.A reference triangle with acute angleθ, opposite sidex+ 1, and hypotenuse 2 has adjacent side of length32xx2. ThereforeJ= 2arcsinx+ 1232xx2+C.C08S07.007:32xx2=(x2+2x3) =(x2+2x+14) = 4(x+1)244sin2θ= 4cos2θifwe (and we do) letx+ 1 = 2sinθ. Thenx=1 + 2sinθ,dx= 2cosθ dθ, and sinθ=12(x+ 1). So1
x32xx2dx=(1 + 2sinθ)·(2cosθ)·(2cosθ)= 4(2cos2θsinθcos2θ)= 42cos2θsinθ1 + cos2θ2= 423cos2θ12θ12sinθcosθ+C=13(2cosθ)33θ2sinθcosθ+C=13(32xx2)3/22arcsinx+ 12x+ 1232xx2+C=2arcsinx+ 1213(32xx2) +12(x+ 1)32xx2+C=2arcsinx+ 12162(32xx2) + 3(x+ 1)32xx2+C=2arcsinx+ 1216(2x24x+ 6 + 3x+x)32xx2+C=2arcsinx+ 12+16(2x2+x9)32xx2+C.C08S07.008:4x2+ 4x3 = 4x2+ 4x+ 14 = (2x+ 1)24 = 4sec2θ4 = 4tan2θif we (and we do)let 2secθ= 2x+ 1. Thusx= secθ12, secθ=x+12, anddx= secθtanθ dθ. ThereforeK=14x2+ 4x3dx=14tan2θ·secθtanθ dθ=14secθtanθ=14cscθ dθ=14ln|cscθcotθ|+C.

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