Last Week: Two Techniques of IntegrationLast week we talked about two tools for integrating difficultfunctions:IIntegration by Substitution (Change of Variable)Zx=bx=af(u(x))dx=Zu=u(b)u=u(a)f(u)dxdudu=Zu=u(b)u=u(a)f(u)x0(u)du=Zu=u(b)u=u(a)f(u)u0(x)duIIntegration by PartsZudv=uv-ZvduZbaudv=uvba-Zbavdu
Remember the following “tips” we talked about:IIf you see a compound function, think about trying asubstitution. Eg.,Z21e3xdx=Z63eu13du=13Z63euduIIf you’re integrating the product of a polynomial times afunction that anti-differentiates nicely (likeexor sin(x) orcos(x)), try integration by parts. Eg.,Zxexdx=xex-Zexdx=xex-ex+CIIf you have a polynomial times ln(x), ignore the last tip—differentiate the ln(x) and anti-differentiate the polynomial.Eg.,Zx2ln(x)=x33ln(x)-Zx331xdx=13x3ln(x)-1Zx2dx3
More TipsWe saw some instances where it was not obvious which tool(s) touse.ITrigonometric Identities: Eg.,IZtan(x)dx=Zsin(x)cos(x)dxIZsin2(x)dx=Z12-cos(2x)2dxIZsin3(x)dx=Z(1-cos2(x)) sin(x)dxwhich lead to a natural substitution.
More TipsIThe “·1” Trick , eg.Ztan-1(x)dx=Ztan-1(x)·1dx=xtan-1(x)-Zx·11 +x2dxby integration by parts, and the integralRx1+x2dxon the rightcan be solved with the substitutionu= 1 +x2.ICyclic Integration by Parts, eg.Zsin(x) cos(x)dx=sin2(x)-Zcos(x) sin(x)dx2Zsin(x) cos(x)dx=sin2(x) +CZsin(x) cos(x)dx=12sin2(x) +C
Get answer to your question and much more
Substitutions will not help you here! (At least, not directly.)This integral requires the method of “partial fractions”.
Trigonometric SubstitutionsThe name of the game is the following trigonometric identities:I1-sin2(θ) = cos2(θ)I1 + tan2(θ) =cos2(θ)cos2(θ)+sin2(θ)cos2(θ)= sec2(θ)Isec2(θ)-1 = tan2(θ)Notice that in each case, the right hand side of the identity is aperfect square.
Trigonometric Substitution—R11+x2dxRemember the identity1 + tan2(θ) = sec2(θ)Maybe we could use the substitutionx= tan(θ) to turn thedenominator 1 +x2into sec2(θ)? This will turn out to be usefulbecausedx= sec2(θ)dθ. Watch:Z11 +x2dx=Z11 + tan2(θ)·sec2(θ)dθ=Z1sec2(θ)sec2(θ)dθ=Zdθ=θ+C= tan-1(x) +C
Trigonometric Substitution—R10√1-x2dxWe already met this integral, and we know it gives the area ofone-quarter of the unit circle. This time, remember the identity1-sin2(θ)=cos2(θ)q1-sin2(θ)=|cos(θ)|So perhaps it makes sense to use the change of variablesx= sin(θ)dx= cos(θ)dθThis givesZx=1x=0p1-x2dx=Zθ=π/2θ=0q1-sin2(θ)·(cos(θ)dθ)=Zθ=π/2θ=0|cos(θ)| ·cos(θ)dθ=Zθ=π/2θ=0cos2(θ)dθsince cos(θ)≥0 for 0≤θ≤π/2.
Trigonometric Substitution—R10√1-x2dxSo a trigonometric substitutionx= sin(θ) shows thatZx=1x=0p1-x2