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MATH
[1251]

Theorems from Reduction formulae

All theorems from Reduction formulae.

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Powers of xx times exe^x

For an integer n≥0n\ge0, choose compatible representatives InI_n of the antiderivatives of xnexx^n e^x. Then

In=xnex−nIn−1,n≥1.I_n=x^n e^x-n I_{n-1},\qquad n\ge 1.

Each application lowers the power of xx by one. Two antiderivatives can differ by a constant, so use the recurrence to construct one representative and add a single arbitrary constant when the reduction stops.

  1. [1]:Reduction formula
  2. [2]:Proof
  3. [3]:∫x3ex dx\int x^3 e^x \, dx

Sine and cosine powers on [0,π2]\bigl[0,\frac{\pi}{2}\bigr]

Let In=∫0π/2sin⁡nx dxI_n=\int_0^{\pi/2}\sin^n x \, dx and Jn=∫0π/2cos⁡nx dxJ_n=\int_0^{\pi/2}\cos^n x \, dx. Then

In=n−1nIn−2,Jn=n−1nJn−2,I_n=\frac{n-1}{n} I_{n-2},\qquad J_n=\frac{n-1}{n} J_{n-2},

for n≥2n\ge 2.

The recurrence stops at I0=π2I_0=\frac{\pi}{2} when nn is even and at I1=1I_1=1 when nn is odd and likewise for JnJ_n. In the proof, collect the reintroduced copy of InI_n before dividing by nn.

  1. [1]:Reduction formula
  2. [2]:Proof
  3. [3]:∫0π/2sin⁡5x dx\int_0^{\pi/2}\sin^5 x \, dx
  4. [4]:∫0π/2cos⁡7x dx\int_0^{\pi/2}\cos^7 x \, dx

Secant powers on [0,π4]\bigl[0,\frac{\pi}{4}\bigr]

Let In=∫0π/4sec⁡nx dxI_n=\int_0^{\pi/4}\sec^n x \, dx. Then

In=1n−1(2(n−2)/2+(n−2)In−2)I_n=\frac{1}{n-1}\bigl(2^{(n-2)/2}+(n-2)I_{n-2}\bigr)

for n≥2n\ge 2.

The base values are I0=∫0π/41 dx=π4I_0=\int_0^{\pi/4}1\,dx=\frac{\pi}{4} and I1=∫0π/4sec⁡x dx=ln⁡(1+2)I_1=\int_0^{\pi/4}\sec x\,dx=\ln(1+\sqrt{2}).

Without the limits, the same splitting sec⁡nx=sec⁡n−2xsec⁡2x\sec^n x=\sec^{n-2}x\sec^2 x produces the antiderivative of sec⁡3x\sec^3 x.

  1. [1]:Reduction formula
  2. [2]:∫sec⁡3x dx\int \sec^3 x \, dx
  3. [3]:Proof
  4. [4]:Secant and tangent recurrences on [0,π4]\bigl[0,\frac{\pi}{4}\bigr]

Tangent powers on [0,π4]\bigl[0,\frac{\pi}{4}\bigr]

Let Jn=∫0π/4tan⁡nx dxJ_n=\int_0^{\pi/4}\tan^n x \, dx. Then

Jn=1n−1−Jn−2,n≥2.J_n=\frac{1}{n-1}-J_{n-2},\qquad n\ge 2.

This recurrence needs no parts: the identity tan⁡2x=sec⁡2x−1\tan^2 x=\sec^2 x-1 turns the leftover into an elementary integral in u=tan⁡xu=\tan x.

Its base values are J0=π4J_0=\frac{\pi}{4} and J1=∫0π/4tan⁡x dx=ln⁡2J_1=\int_0^{\pi/4}\tan x\,dx=\ln\sqrt{2}.

  1. [1]:Reduction formula
  2. [2]:Proof
  3. [3]:Secant and tangent recurrences on [0,π4]\bigl[0,\frac{\pi}{4}\bigr]

Tangent powers times secant

Let Kn=∫0π/4tan⁡nθsec⁡θ dθK_n=\int_0^{\pi/4}\tan^n\theta\sec\theta \, d\theta. Then

Kn=1n(2−(n−1)Kn−2)K_n=\frac{1}{n}\bigl(\sqrt{2}-(n-1)K_{n-2}\bigr)

for n≥2n\ge 2.

Its base values are K0=∫0π/4sec⁡θ dθ=ln⁡(1+2)K_0=\int_0^{\pi/4}\sec\theta\,d\theta=\ln(1+\sqrt{2}) and K1=∫0π/4tan⁡θsec⁡θ dθ=2−1K_1=\int_0^{\pi/4}\tan\theta\sec\theta\,d\theta=\sqrt{2}-1.

  1. [1]:Reduction formula
  2. [2]:Proof
  3. [3]:Secant and tangent recurrences on [0,π4]\bigl[0,\frac{\pi}{4}\bigr]

On this page

Powers of xx times exe^xSine and cosine powers on [0,π2]\bigl[0,\frac{\pi}{2}\bigr]Secant powers on [0,π4]\bigl[0,\frac{\pi}{4}\bigr]Tangent powers on [0,π4]\bigl[0,\frac{\pi}{4}\bigr]Tangent powers times secant