Loading [MathJax]/jax/output/HTML-CSS/jax.js
AMPS | THARC | KE8QZC | SFW | TSW | WW
ORCID iD icon

Back to the class

Section 8.3 #4: sin3(3x)dx
Solution: Recall the Pythagorean identity cos2(θ)+sin2(θ)=1. We applying this rearranged to sin2(θ)=1cos2(θ) with θ=3x yields sin3(3x)=sin(3x)sin2(3x)=sin(3x)(1cos2(3x)). Therefore we may compute sin3(3x)dx=sin(3x)(1cos2(3x))dx=sin(3x)dxsin(3x)cos2(3x)dxu1=3x,u2=cos(3x)=13sin(u1)du1+13u22du2=13cos(u1)19u32+C=13cos(3x)19cos3(3x)+C. Section 8.3 #9: cos2(3x)dx
Solution: Recall the "reduction" identity cos2(θ)=1+cos(2θ)2. Apply that identity with θ=3x to see cos2(3x)=1+cos(6x)2. Therefore compute cos2(3x)dx=1+cos(6x)2dx=121dx+12cos(6x)dx=x2+112sin(6x)+C.

Section 8.4 #22: x236x2dx
Solution: Recall that the Pythagorean identity may be rearranged to say 1sin2(θ)=cos(θ). Also recall the "reduction" identity sin2(θ)=1cos(2θ)2. Let x=6sin(θ) so that dx=6cos(θ)dθ. Substituting this variable into the integral yields x236x2dx=36sin2(θ)3636sin2(θ)6cos(θ)dθ=3666sin2(θ)cos(θ)1sin2(θ)dθ=36sin2(θ)cos(θ)cos(θ)dθ=36sin2(θ)dθ=3621dθ362cos(2θ)dθ=18θ9sin(2θ)+C. Solving our assignment x=6sin(θ) for θ yields θ=arcsin(x6). Recall the double angle identity cos(2ψ)=2sin(ψ)cos(ψ). Applying this shows us that 9sin(2θ)=18sin(θ)cos(θ)=18(x6)cos(θ)=3cos(θ). io determine cos(θ) we must draw a triangle:

To find the side labelled "?", we use the Pythagorean theorem: ?2+x2=62. Solving for ? yields ?=36x2. Therefore cos(θ)=36x26. So, finally, we compute x236x2dx=18arcsin(x6)3xcos(θ)=18arcsin(x6)x236x2+C.

Section 8.5 #6: 29x21dx
Solution: The denominator is a difference of squares (i.e. a2b2=(ab)(a+b)) and so it factors: 9x21=(3x)21=(3x1)(3x+1). Therefore, we must apply partial fractions to 29x21=A3x1+B3x+1. Multiply by the common denominator to get 2=A(3x+1)+B(3x1). Combining like-terms on the right yields 2=(3A+3B)x+(AB). Noting that the "coefficient of x" on the left is zero, we equate coefficients to get the following system of two linear equations in two variables (A and B): {3A+3B=0(i)AB=2(ii). By (ii) we conclude that A=2+B. Plug this into (i) to get 3(2+B)+3B=0, or equivalently, 6B=6, i.e. B=1, and hence A=2+(1)=1. This gives us the partial fractions decomposition 29x21=13x113x+1. Therefore we may compute 29x21dx=13x1dx13x+1dxu=3x1,v=3x+1=131udu131vdv=13log(u)13log(v)+C=13log(3x1)13log(3x+1)+C.

Section 8.5 #9: x2+12x+12x34xdx
Solution: We may factor this denominator as x34x=x(x24)=x(x2)(x+2), so the partial fractions decomposition involves three terms (one for each linear factor): x2+12x+12x34x=Ax+Bx2+Cx+2. Multiplying by the common denominator x(x2)(x+2) yields x2+12x+12=A(x2)(x+2)+Bx(x+2)+Cx(x2)=A(x24)+B(x2+2x)+C(x22x). Combining like-terms yields x2+12x+12=(A+B+C)x2+(2B2C)x+(4A). Equating coefficients yields the following system of three linear equations in three variables: {A+B+C=1(i)2B2C=12(ii)4A=12(iii). From (iii), we may conclude that A=3. From (ii), we may conclude that BC=6, and hence B=6+C. Plugging A=3 and B=6+C into (i), we get 3+(6+C)+C=1, from which we may conclude C=1 and B=6+(1)=5, yielding the partial fractions decomposition x2+12x+12x34x=3x+5x21x+2. Therefore we may compute x2+12x+12x34xdx=3xdx+51x2dx1x+2dx=3log(x)+5log(x2)log(x+2)+C.

Section 8.5 #13: 4x2+2x1x3+x2dx
Solution: Factoring the denominator yields x3+x2=x2(x+1). Therefore we set up the partial fraction equation: 4x2+2x1x3+x2=Ax+Bx2+Cx+1. This yields a system of three equations in three variables and hence the following partial fractions decomposition: 4x2+2x1x3+x2=1x2+1x+1+3x. Therefore we compute 4x2+2x1x3+x2dx=x2dx+1x+1dx+31xdx=1x+log(x+1)+3log(x)+C.