# Using the pythagorean theorem we solve for the

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Using the Pythagorean theorem, we solve for the hypotenuseand getc=x2+16. So cosθ=4x2+16, and secθ=x2+164( for all values ofθwith-π2<θ<π2) .Therefore, the indefinite integral continues as-1Z1x2+16dx=-1ln|secθ+tanθ|+C=-1lnx2+164+x4+C=-1lnx2+16+x4+C=-1lnx2+16+x+1ln4+C=-1lnx2+16+x+CCorrect Answers:-1*ln(x+sqrt(xˆ2+16))+C+c16.(1 point)Evaluate the integralZ-6x4x2-2dxNote: Use an upper-case ”C” for the constant of integration.Solution:SOLUTIONLetx=2secθwhere 0θ<π2orπθ<3π2.Thendx=2secθtanθdθandx2-2=2sec2θ-2=2sec2θ-1=2tan2θ=2tanθ. ThusZ-6x4x2-2dx=-6Z14sec4θ2tanθ2secθtanθdθ=-6Z14sec3θdθ=-1.5Rcos3θdθ=-1.5Rcos2θcosθdθ=-1.5R(1-sin2θ)cosθdθ=-1.5R(1-u2)du[u=sinθ]=-1.5u-13u3+C=-1.5 sinθ-13sin3θ+CWe now need to go back to the variablex. From the originalsubstitution, we havesecθ=x2cosθ=2xsin2θ=1-cos2θ=1-2x2=x2-2x2so sinθ=x2-1x.ThusZ-6x4x2-2dx=-1.5 sinθ-13sin3θ+C=-1.5"x2-2x-13(x2-2)3/2x3#+CCorrect Answers:-6/4*(sqrt(xˆ2-2)/x-(xˆ2-2)ˆ(3/2)/(3*xˆ3))+C+c6
17.(1 point) Match each of the trigonometric expressionsbelow with the equivalent non-trigonometric function from thefollowing list. Enter the appropriate letter (A,B,C,D, or E) ineach blank.A. tan(sin-1(x/2))B. cos(sin-1(x/2))C.(1/2)sin(2sin-1(x/2))D. sin(tan-1(x/2))E. cos(tan-1(x/2))1.x4+x22.x4-x23.24+x24.x4p4-x25.4-x22Solution: 1. matches D.Letθ=tan-1(x/2). Then tanθ=x/2 If we interpretθas beingan angle in a right triangle, and label the side oppositeθasxandthe side adjacent toθas 2, then, by the Pythagorean Theorem,the hypotenuse is4+x2andsin(tan-1(x/2)) =sin(θ) =x4+x22. matches A,Letθ=sin-1(x/2). Then sinθ=x/2. If we interpretθas beingan angle in a right triangle, and label the side oppositeθasxand the hypotenuse as 2, then, by the Pythagorean Theorem, theadjacent side is4-x2andtan(sin-1(x/2)) =tan(θ) =x4-x23. matches E,Letθ=tan-1(x/2). Then tanθ=x/2 If we interpretθas beingan angle in a right triangle, and label the side oppositeθasxandthe side adjacent toθas 2, then, by the Pythagorean Theorem,the hypotenuse is4+x2andcos(tan-1(x/2)) =cos(θ) =24+x24. matches C,Letθ=sin-1(x/2). Then sinθ=x/2 and12sin(2sin-1(x/2)) =12sin(2θ) =12·2sin(θ)cos(θ) =sin(θ)p1-sin2θ=x2r1-x24=x4p4-x25. matches B,Letθ=sin-1(x/2). Then sinθ=x/2 andcos(sin-1(x/2)) =cos(θ) =p1-sin2θ=r1-x24=4-x22Correct Answers:DAECB18.(1 point) Evaluate the indefinite integral.