Section 6.2MAT_266_ONLINE__Spg_2020.pdf - Larena Whitney-krug Assignment Section 6.2 due at 11:59pm MST Boerner MAT 266 ONLINE B Spring 2020 Answer(s

# Section 6.2MAT_266_ONLINE__Spg_2020.pdf - Larena...

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Larena Whitney-krugBoernerMAT266ONLINEBSpring2020Assignment Section6.2 due 03/22/2020 at 11:59pm MST1.(1 point) Evaluate the indefinite integral.Z8-2sinxcosxdx+CSolution:SOLUTIONWe simplify the integrand to getR8-2sinxcosxdx=R8cos(x)-2sin(x)cos(x)dx=R8sec(x)-2tan(x)dx=8ln sec(x)+tan(x)-2ln sec(x) +CWhere we consider it known that:Rsec(x)dx=ln sec(x)+tan(x)andRtan(x)dx=Rsin(x)cos(x)dx=R-1udu=-ln|u|=ln1u=ln sec(x).Answer(s) submitted:8ln(abs(sec(x)+tan(x)))+-2ln(abs(sec(x)))(correct)Correct Answers:8*ln(abs(sec(x)+tan(x)))+-2*ln(abs(sec(x)))2.(1 point) Evaluate the indefinite integral.Zsin3(11x)cos4(11x)dx+CSolution:SOLUTIONForRsin3(11x)cos4(11x)dx, we have an odd power of 3 forsin(11x). We separate one factor of sin(11x)and factor the re-maining sin2(11x)as sin2(11x) =1-cos2(11x):Rsin3(11x)cos4(11x)dx=Rsin2(11x)cos4(11x)sin(11x)dx=R(1-cos2(11x))cos4(11x)sin(11x)dxWemaynowusesubstitution:u=cos(11x),du=-11sin(11x)dx=⇒ -111du=sin(11x)dx.So continuing the above,Rsin3(11x)cos4(11x)dx=Rsin2(11x)cos4(11x)sin(11x)dx=R(1-cos2(11x))cos4(11x)sin(11x)dx=-111R(1-u2)u4du=-111R(u4-u6)du=-111hu55-u77i+C=-155cos5(11x)+177cos7(11x)+CAnswer(s) submitted:(-1/11)((cosˆ5(11x)/5)-(cosˆ7(11x)/7))(correct)Correct Answers:-cos(11 * x)**(5)/(5 *11)+ cos(11 * x)**(7)/(7 * 11)3.(1 point)Evaluate the indefinite integral.Z55cos3(2x)dxAnswer =+CSolution:SOLUTIONForR55cos3(2x)dx, we have an odd power of 3 for cos(2x).We separate one factor of cos(2x)and factor the remainingcos2(2x)as cos2(2x) =1-sin2(2x):R55cos3(2x)dx=55Rcos2(2x)sin(2x)dx=55R(1-sin2(2x))cos(2x)dxWemaynowusesubstitution:u=sin(2x),du=2cos(2x)dx=12du=cos(2x)dx.So continuing the above,R55cos3(2x)dx=55Rcos2(2x)sin(2x)dx=55R(1-sin2(2x))cos(2x)dx=552R(1-u2)du=552hu-u33i+C=552sin(2x)-556sin3(2x)+CAnswer(s) submitted:27.5 sin(2x)-(55/6)sinˆ3(2x)(correct)Correct Answers:55/2*sin(2*x) - 55/(3*2)*(sin(2*x))**31
4.(1 point)Evaluate the indefinite integral.Z11cos2(40x)dxAnswer =+CSolution:SOLUTIONForR11cos2(40x)dx, we have an even power of 2 forcos(40x).We use the half-angle identity cos2(A) =1+cos(2A)2to get:R11cos2(40x)dx=11R1+cos(80x)2dx=112R(1+cos(80x))dx=112hx+180sin(80x)i+C=112x+11160sin(80x)+CAnswer(s) submitted:(11/2)(x+(1/80)sin(80x))(correct)Correct Answers:11*x/2 + 11*sin(2*40*x)/(4*40)5.(1 point)Zπ/50sin4(5x)dx=Solution:SOLUTIONForZπ/50sin4(5x)dx, we have an even power of 4 for sin(5x).We separate this into sin4(5x) =sin2(5x)sin2(5x)and use thehalf-angle identity sin2(A) =1-cos(2A)2to get:Zπ/50sin4(5x)dx=Zπ/50sin2(5x)sin2(5x)dx=Zπ/501-cos(10x)21-cos(10x)2dx=14Zπ/50(1-2cos(10x)+cos2(10x))dxAnd now using the half-angle identity cos2(A) =1+cos(2A)2, wecontinue the above:Zπ/50sin4(5x)dx=14Zπ/50(1-2cos(10x)+cos2(10x))dx=14Zπ/501-2cos(10x)+1+cos(20x)2dx=18Zπ/50(3-4cos(10x)+cos(20x))dx=18h3x-410sin(10x)+120sin(20x)iπ/50=18h3(π5)-25sin(2π)+120sin(4π)i-h0-25sin(0)+=18h35π-0+0i-h0i=340πAnswer(s) submitted:(3/40)pi(correct)Correct Answers:0.235619449056.(1 point) Evaluate the integral.