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Concept 10 / 10Integral Calculus
Integrals of Piecewise and Special Functions, Reduction Formulas and Wallis' Formula
Definite integrals involving modulus, greatest integer and fractional part functions by splitting at break points, reduction formulas for powers of sine, cosine and tangent, Wallis' formula for integrals from 0 to pi/2, and the gamma-style result for sin^m cos^n.
Figure 1
Modulus integrands. ∫ab∣f(x)∣dx: find the zeros of f in [a,b], split there, and on each piece replace ∣f∣ by f or −f according to its sign. For ∫02∣x2−1∣dx: on [0,1], ∣x2−1∣=1−x2, contributing 32; on [1,2] it is x2−1, contributing 34. Total 2.
Greatest integer integrands. [f(x)] is constant between the points where f(x) crosses an integer. Split at those points, and on each piece integrate a constant. The figure shows ∫03[x]dx=0+1+2=3; in general ∫0n[x]dx=2n(n−1) for a positive integer n. For ∫02[x2]dx, the break points are x=1,2,3: the integral is 0⋅1+1(2−1)+2(3−2)+3(2−3)=5−2−3. For the fractional part, use {x}=x−[x], or its periodicity: ∫0n{x}dx=2n.
Reduction formulas express an integral with parameter n in terms of the same integral with a smaller parameter, usually via integration by parts. Let In=∫0π/2sinnxdx. Parts with u=sinn−1x and dv=sinxdx give
In=nn−1In−2,
with I0=2π and I1=1. The same holds for cosnx on [0,2π], since King's property swaps sin and cos. For tangent, Jn=∫0π/4tannxdx satisfies Jn+Jn−2=n−11, from writing tann=tann−2(sec2−1). For indefinite integrals: ∫sinnxdx=−nsinn−1xcosx+nn−1∫sinn−2xdx.
Wallis' formula. Iterating the reduction formula:
∫0π/2sinnxdx=∫0π/2cosnxdx=n(n−2)⋯(n−1)(n−3)⋯×K,
where the products continue down to 1 or 2, and K=2π if n is even, K=1 if n is odd. For example, ∫0π/2sin6xdx=6⋅4⋅25⋅3⋅1⋅2π=325π and ∫0π/2cos5xdx=5⋅3⋅14⋅2=158.
For mixed powers: ∫0π/2sinmxcosnxdx=(m+n)(m+n−2)⋯[(m−1)(m−3)⋯][(n−1)(n−3)⋯]×K, where K=2π if both m and n are even and K=1 otherwise. For instance, ∫0π/2sin4xcos2xdx=6⋅4⋅23⋅1⋅1⋅2π=32π.
Extending the interval uses symmetry: ∫0πsinnxdx=2∫0π/2sinnxdx, while ∫0πcosnxdx is 0 for odd n and 2∫0π/2cosnxdx for even n.
Key Takeaways:
- ∣f∣: split at zeros of f; [f(x)]: split where f(x) crosses integers.
- ∫0n[x]dx=2n(n−1); ∫0n{x}dx=2n.
- In=nn−1In−2 for ∫0π/2sinn or cosn.
- ∫0π/4(tann+tann−2)=n−11.
- Wallis: n(n−2)⋯(n−1)(n−3)⋯×(2π if n even, 1 if odd).
- Mixed: (m+n)!!(m−1)!!(n−1)!!×(2π if both even, else 1).
- Pitfall: applying Wallis' formula on intervals other than [0,2π] without adjusting by symmetry.
Brain-Teaser 1: Evaluate ∫−12(∣x∣+[x])dx.
Solution: Split at 0 and 1. ∫−12∣x∣dx=21+2=25. ∫−12[x]dx=(−1)(1)+0(1)+1(1)=0. Total 25. Each special function is broken at its own critical points, and the pieces are simple areas of triangles and rectangles.
Brain-Teaser 2: Evaluate ∫0πxsin4xcos2xdx.
Solution: By King's property, ∫0πxf(sinx,cos2x)dx=2π∫0πsin4xcos2xdx, valid because the integrand is unchanged under x→π−x. The function sin4xcos2x is symmetric about 2π, so ∫0π=2∫0π/2=2⋅32π=16π. The answer is 2π⋅16π=32π2. King's property removes the factor x, symmetry reduces the interval, and Wallis finishes it.
Brain-Teaser 3: If Jn=∫0π/4tannxdx, find J5, and show that Jn+Jn+2 decreases to 0.
Solution: J1=∫0π/4tanxdx=lnsec4π=21ln2. From Jn+Jn−2=n−11: J3=21−J1=21−21ln2, and J5=41−J3=−41+21ln2. Numerically J5≈0.0966, positive as it must be. And Jn+Jn+2=n+11, which decreases to 0 as n grows; since 0≤tanx≤1 on the interval, the powers shrink and so do the integrals. The alternating-sign formula for J5 shows why the reduction formula must be applied step by step from a known base case.
