Almaz Khusnutdinov•Math enthusiast, love coding and mathematics
Published 9/20/2025
Integral Reductions
Problem 1.1
For In=∫0π/2sinnxdx,n∈N.(a) Derive a reduction formula relating In to In−2.(b) Compute I6.Answer: In=nn−1In−2(n≥2),I0=2π,I1=1;I6=325π.Solution: In=∫0π/2sinn−1xsinxdx.IBP: u=sinn−1x,dv=sinxdx⇒u′=(n−1)sinn−2xcosx,v=−cosx.In=[−sinn−1xcosx]0π/2+(n−1)∫0π/2sinn−2xcos2xdx=(n−1)∫0π/2sinn−2x(1−sin2x)dx=(n−1)(In−2−In).nIn=(n−1)In−2⇒In=nn−1In−2.Iterating: I6=65⋅43⋅21I0=325π.
Problem 1.2
Let Jn=∫cosnxdx,n≥2.(a) Derive a reduction formula for Jn.(b) Evaluate ∫cos4xdx.Answer: Jn=nsinxcosn−1x+nn−1Jn−2+C;∫cos4xdx=83x+4sin2x+32sin4x+C.Outline: IBP with u=cosn−1x,dv=cosxdx,cos2x=1−sin2x,solve for Jn.
Problem 1.3
Let In(a)=∫xneaxdx(a=0).(a) Derive a reduction formula for In(a).(b) Find I3(a).Answer: In(a)=axneax−anIn−1(a)+C;I3(a)=eax(ax3−a23x2+a36x−a46)+C.Outline: IBP with u=xn,dv=eaxdx,then recurse.
Problem 1.4
Let Kn=∫1elnnxdx.(a) Prove a reduction for Kn.(b) Compute K4.Answer: Kn=e−nKn−1,K0=e−1;K4=9e−24.Outline: IBP on [1,e]with u=lnnx,dv=dx,use [lnx]1e=1.
Problem 1.5
Let Sn=∫xnsin(ax)dx,Cn=∫xncos(ax)dx(a=0).(a) Derive coupled reductions for Sn and Cn.(b) Evaluate ∫x2sin(ax)dx.Answer: Sn=−axncos(ax)+anCn−1+C,Cn=axnsin(ax)−anSn−1+C;∫x2sin(ax)dx=−ax2cos(ax)+a22xsin(ax)+a32cos(ax)+C.Outline: IBP with u=xn,dv=sin(ax)dxand similarly for Cn.
Trigonometry
Problem 2.1
Let θ∈(0,π) and define E(θ)=sin(23π−θ)+cos(π−θ)−tan(2π+θ)+tan(23π−θ).(a) Reduce E(θ) to a function of sinθ,cosθ only.(b) Solve E(θ)=0 for θ∈(0,π).Answer: E(θ)=−2cosθ+2cotθ,and E(θ)=0⟺θ=2π.Solution: sin(23π−θ)=−cosθ,cos(π−θ)=−cosθ,tan(2π+θ)=−cotθ,tan(23π−θ)=cotθ.E(θ)=(−cosθ)+(−cosθ)−(−cotθ)+cotθ=−2cosθ+2cotθ.E(θ)=0⟺cotθ=cosθ⟺sinθcosθ=cosθ.Either cosθ=0⇒θ=2π;or sinθ=1⇒θ=2π.Hence the unique solution in (0,π) is θ=2π.(Domain note: E is defined on (0,π) since sinθ=0 there).
Problem 2.2
Prove the cofunction reductions for all θ∈R:sin(2π±θ)=cosθ,cos(2π±θ)=∓sinθ;sin(23π±θ)=∓cosθ,cos(23π±θ)=±sinθ.Answer: Identities hold.Outline: View (cosθ,sinθ) on the unit circle. Rotation by 2π sends (x,y)↦(−y,x),so (cos(2π±θ),sin(2π±θ))=(∓sinθ,cosθ).Rotation by 23π=−2π sends (x,y)↦(y,−x),yieldingthestatedsigns.Quadrant of 2π±θ or 23π±θ fixes the ± signs exactly.
Problem 2.3
Let θ∈R with sinθ=0,cosθ=0.(a) Simplify A(θ)=sin(2π+θ)+sin(π−θ)+cos(2π−θ)−cos(23π+θ)(a)+tan(π+θ)tan(23π−θ).(b) Solve A(θ)=0 for θ∈(0,2π).Answer: A(θ)=2sinθ+1;A(θ)=0⟺sinθ=−21⟺θ=67π,611π.(Used: sin(2π+θ)=sinθ,sin(π−θ)=sinθ,cos(2π−θ)=sinθ,cos(23π+θ)=sinθ,tan(π+θ)=tanθ,tan(23π−θ)=cotθ,tanθcotθ=1).
Problem 2.4
Prove the π and 2π reductions and periods: for all θ,sin(π±θ)=∓sinθ,cos(π±θ)=−cosθ,tan(π±θ)=tanθ;sin(2π±θ)=±sinθ,cos(2π±θ)=cosθ,tan(2π±θ)=tanθ.Deduce that sin,cos have period 2π and tan has period π.Answer: Identities and periods hold.Outline: On the unit circle, rotation by π maps (x,y)↦(−x,−y),giving sin(π±θ)=−sin(±θ)=∓sinθ,cos(π±θ)=−cosθ.Rotation by 2π is the identity, yielding the 2π formulas.Since adding π leaves tan=sin/cos unchanged, its period is π;sin,cos repeat after 2π.
Problem 2.5
Let θ∈R with sinθ=0.EvaluateR(θ)=cos(2π+θ)sin(2π−θ)sin(π−θ)cos(23π+θ)and specify the maximal domain.Answer: R(θ)=1 for sinθ=0; undefined when sinθ=0(θ∈πZ).Used: sin(π−θ)=sinθ,cos(23π+θ)=sinθ,cos(2π+θ)=−sinθ,sin(2π−θ)=−sinθ⇒R(θ)=(−sinθ)(−sinθ)sin2θ=1.
