M408D SPRING 2015 HOMEWORK #2 (1) - cheatham (sc36975) HW02...

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cheatham (sc36975) – HW02 – um – (53890)1Thisprint-outshouldhave18questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0 pointsEvaluate the integralkeywords:Stewart5e,indefiniteintegral,powers of sin, powers of cos, trig substitu-tion,00210.0 points
cheatham (sc36975) – HW02 – um – (53890)2which as the graph shows can in turn be writ-ten asI=π02 sin3(x)dx3π/2π2 sin3(x)dx .Sincesin2(x) = 1cos2(x),we thus see thatI=π03π/2π2 sin(x)(1cos2(x))dx .To evaluate these integrals, setu= cos(x).For thendu=sin(x)dx ,in which caseπ02 sin(x)(1cos2(x))dx=211(1u2)du= 211(1u2)du= 2u13u311=83,while3π/2π2 sin(x)(1cos2(x))dx=201(1u2)du=2u13u301=43.Consequently, the shaded region hasarea = 4.keywords:00310.0 pointsEvaluate the definite integralI=π/404 cosx2 sinxcos3xdx .1.I=522.I= 23.I=724.I= 45.I= 3correctExplanation:After division4 cosx2 sinxcos3x= 4 sec2x2 tanxsec2x= (42 tanx) sec2x .ThusI=π/40(42 tanx) sec2x dx .Letu= tanx; thendu= sec2x dxsoI=10(42u)du=4uu210.Consequently,I= 3.00410.0 pointsEvaluate the integralI= 22π/301 + cos(θ)dθ.Hint:use a double angle formula to express1 + cos(θ)in terms ofcos2(θ/2).1.I= 22
cheatham (sc36975) – HW02 – um – (53890)32.I= 463.I= 424.I= 45.I= 26.I= 26correctExplanation:By a double angle formula for cos(2x),cos(2x) = 2 cos2(x)1,so, takingx=θ/2, we see that1 + cos(θ) =2 cos2θ2=2 cosθ2when cos(θ/2)0, hence for 0θπ.ThusI= 222π/30cosθ2dθ= 42 sinθ22π/30.

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