Reduction formulae

Powers of $x$ times $e^x$

For an integer $n\ge0$ , choose compatible representatives $I_n$ of the antiderivatives of $x^n e^x$ . Then

I_n=x^n e^x-n I_{n-1},\qquad n\ge 1.

Each application lowers the power of $x$ 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.

Integrate by parts with $u=x^n$ and $\frac{dv}{dx}=e^x$ , so $du=n x^{n-1} \, dx$ and $v=e^x$ . Then

I_n=x^n e^x-n\int x^{n-1} e^x \, dx=x^n e^x-n I_{n-1}.

[1]:Powers of $x$ times $e^x$

$\int x^3 e^x \, dx$

The recurrence with $n=3$ , then $n=2$ , then $n=1$ and $I_0=\int e^x \, dx$ , gives

\begin{aligned} I_3&=x^3 e^x-3 I_2 \\ &=x^3 e^x-3\bigl(x^2 e^x-2 I_1\bigr) \\ &=x^3 e^x-3x^2 e^x+6 I_1 \\ &=x^3 e^x-3x^2 e^x+6\bigl(x e^x-I_0\bigr) \\ &=x^3 e^x-3x^2 e^x+6x e^x-6\int e^x \, dx \\ &=x^3 e^x-3x^2 e^x+6x e^x-6e^x+C. \end{aligned}

[exam]

[1]:Powers of $x$ times $e^x$

Reducing sine and cosine powers

On $\bigl[0,\frac{\pi}{2}\bigr]$ the boundary term from parts vanishes, so parts and a Pythagorean identity give a multiple of the integral two powers lower.

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

Let $I_n=\int_0^{\pi/2}\sin^n x \, dx$ and $J_n=\int_0^{\pi/2}\cos^n x \, dx$ . Then

I_n=\frac{n-1}{n} I_{n-2},\qquad J_n=\frac{n-1}{n} J_{n-2},

for $n\ge 2$ .

The recurrence stops at $I_0=\frac{\pi}{2}$ when $n$ is even and at $I_1=1$ when $n$ is odd and likewise for $J_n$ . In the proof, collect the reintroduced copy of $I_n$ before dividing by $n$ .

For $I_n$ , integrate by parts with $u=\sin^{n-1}x$ and $\frac{dv}{dx}=\sin x$ , so $du=(n-1)\sin^{n-2}x\cos x \, dx$ and $v=-\cos x$ . Then

I_n=\bigl[-\sin^{n-1}x\cos x\bigr]_0^{\pi/2}+(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x \, dx.

The boundary term is $0$ , since $\cos\frac{\pi}{2}=0$ and $\sin 0=0$ . Replace $\cos^2 x=1-\sin^2 x$ :

\begin{aligned} I_n&=(n-1)\int_0^{\pi/2}\sin^{n-2}x\bigl(1-\sin^2 x\bigr) \, dx \\ &=(n-1)\bigl(I_{n-2}-I_n\bigr). \end{aligned}

Hence $I_n=(n-1)I_{n-2}-(n-1)I_n$ , so $n I_n=(n-1)I_{n-2}$ and therefore

I_n=\frac{n-1}{n} I_{n-2},\qquad n\ge 2.

For $J_n$ , take $u=\cos^{n-1}x$ and $\frac{dv}{dx}=\cos x$ , so $du=-(n-1)\cos^{n-2}x\sin x \, dx$ and $v=\sin x$ . Then

J_n=\bigl[\cos^{n-1}x\sin x\bigr]_0^{\pi/2}+(n-1)\int_0^{\pi/2}\cos^{n-2}x\sin^2 x \, dx.

The boundary term is $0$ . Replace $\sin^2 x=1-\cos^2 x$ :

J_n=(n-1)\bigl(J_{n-2}-J_n\bigr).

The same rearrangement yields $J_n=\frac{n-1}{n} J_{n-2}$ for $n\ge 2$ .

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

$\int_0^{\pi/2}\sin^5 x \, dx$

I_5=\frac{4}{5} I_3=\frac{4}{5}\cdot\frac{2}{3} I_1=\frac{8}{15}\int_0^{\pi/2}\sin x \, dx=\frac{8}{15}\bigl[-\cos x\bigr]_0^{\pi/2}=\frac{8}{15}.

[exam]

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

$\int_0^{\pi/2}\cos^7 x \, dx$

J_7=\frac{6}{7} J_5=\frac{6}{7}\cdot\frac{4}{5} J_3=\frac{24}{35}\cdot\frac{2}{3} J_1=\frac{16}{35}\int_0^{\pi/2}\cos x \, dx=\frac{16}{35}\bigl[\sin x\bigr]_0^{\pi/2}=\frac{16}{35}.

[exam]

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

Reducing secant and tangent powers

On $\bigl[0,\frac{\pi}{4}\bigr]$ the boundary term from parts does not vanish: $\tan\frac{\pi}{4}=1$ and $\sec\frac{\pi}{4}=\sqrt{2}$ leave a power of $2$ in the recurrence for $\sec^n x$ . Tangent powers need no parts at all.

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

Let $I_n=\int_0^{\pi/4}\sec^n x \, dx$ . Then

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

for $n\ge 2$ .

The base values are $I_0=\int_0^{\pi/4}1\,dx=\frac{\pi}{4}$ and $I_1=\int_0^{\pi/4}\sec x\,dx=\ln(1+\sqrt{2})$ .

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

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

Integrate by parts with $u=\sec^{n-2}x$ and $\frac{dv}{dx}=\sec^2 x$ , so

\frac{du}{dx}=(n-2)\sec^{n-2}x\tan x,\qquad v=\tan x.

Then

I_n=\bigl[\sec^{n-2}x\tan x\bigr]_0^{\pi/4}-(n-2)\int_0^{\pi/4}\sec^{n-2}x\tan^2 x \, dx.

The boundary term is $2^{(n-2)/2}$ , since $\sec\frac{\pi}{4}=\sqrt{2}$ and $\tan\frac{\pi}{4}=1$ , while the lower limit is $0$ . Replace $\tan^2 x=\sec^2 x-1$ :

\begin{aligned} I_n&=2^{(n-2)/2}-(n-2)\int_0^{\pi/4}\sec^{n-2}x\bigl(\sec^2 x-1\bigr) \, dx \\ &=2^{(n-2)/2}-(n-2)\bigl(I_n-I_{n-2}\bigr) \\ &=2^{(n-2)/2}-(n-2)I_n+(n-2)I_{n-2}. \end{aligned}

Hence $(n-1)I_n=2^{(n-2)/2}+(n-2)I_{n-2}$ and therefore

I_n=\frac{1}{n-1}\bigl(2^{(n-2)/2}+(n-2)I_{n-2}\bigr),\qquad n\ge 2.

[1]:Secant powers on $\bigl[0,\frac{\pi}{4}\bigr]$

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

Let $J_n=\int_0^{\pi/4}\tan^n x \, dx$ . Then

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

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

Its base values are $J_0=\frac{\pi}{4}$ and $J_1=\int_0^{\pi/4}\tan x\,dx=\ln\sqrt{2}$ .

Write $\tan^n x=\tan^{n-2}x\tan^2 x$ and replace $\tan^2 x=\sec^2 x-1$ :

J_n=\int_0^{\pi/4}\tan^{n-2}x\bigl(\sec^2 x-1\bigr) \, dx=\int_0^{\pi/4}\tan^{n-2}x\sec^2 x \, dx-J_{n-2}.

Let $u=\tan x$ , so $du=\sec^2 x \, dx$ . The limits change from $x=0$ to $x=\frac{\pi}{4}$ into $u=0$ to $u=1$ . Then

J_n=\int_0^1 u^{n-2} \, du-J_{n-2}=\biggl[\frac{u^{n-1}}{n-1}\biggr]_0^1-J_{n-2}=\frac{1}{n-1}-J_{n-2}

for $n\ge 2$ .

[1]:Tangent powers on $\bigl[0,\frac{\pi}{4}\bigr]$

The mixed integrand $\tan^n\theta\sec\theta$ on the same interval also drops by two, now with a factor $\sqrt{2}$ from the upper limit. Its integrals are written $K_n$ to keep them apart from the secant powers $I_n$ above.

Tangent powers times secant

Let $K_n=\int_0^{\pi/4}\tan^n\theta\sec\theta \, d\theta$ . Then

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

for $n\ge 2$ .

Its base values are $K_0=\int_0^{\pi/4}\sec\theta\,d\theta=\ln(1+\sqrt{2})$ and $K_1=\int_0^{\pi/4}\tan\theta\sec\theta\,d\theta=\sqrt{2}-1$ .

Integrate by parts with $u=\tan^{n-1}\theta$ and $\frac{dv}{d\theta}=\tan\theta\sec\theta$ , so $du=(n-1)\tan^{n-2}\theta\sec^2\theta \, d\theta$ and $v=\sec\theta$ . Then

K_n=\bigl[\tan^{n-1}\theta\sec\theta\bigr]_0^{\pi/4}-(n-1)\int_0^{\pi/4}\tan^{n-2}\theta\sec^3\theta \, d\theta.

The boundary term is $\sqrt{2}$ . Replace $\sec^2\theta=1+\tan^2\theta$ in the remaining integral:

\begin{aligned} K_n&=\sqrt{2}-(n-1)\int_0^{\pi/4}\tan^{n-2}\theta\bigl(1+\tan^2\theta\bigr)\sec\theta \, d\theta \\ &=\sqrt{2}-(n-1)\bigl(K_{n-2}+K_n\bigr). \end{aligned}

Collecting the $K_n$ terms gives $nK_n=\sqrt{2}-(n-1)K_{n-2}$ , so

K_n=\frac{1}{n}\bigl(\sqrt{2}-(n-1)K_{n-2}\bigr),\qquad n\ge 2.

[1]:Tangent powers times secant

Secant and tangent recurrences on $\bigl[0,\frac{\pi}{4}\bigr]$

The base value $I_2=\int_0^{\pi/4}\sec^2 x \, dx=\bigl[\tan x\bigr]_0^{\pi/4}=1$ gives

\int_0^{\pi/4}\sec^4 x \, dx=I_4=\frac{1}{3}\bigl(2^{(4-2)/2}+2I_2\bigr)=\frac{1}{3}(2+2)=\frac{4}{3}.

The base value $J_2=\int_0^{\pi/4}\tan^2 x \, dx=\int_0^{\pi/4}(\sec^2 x-1)\, dx=\bigl[\tan x-x\bigr]_0^{\pi/4}=1-\frac{\pi}{4}$ gives

\int_0^{\pi/4}\tan^4 x \, dx=J_4=\frac{1}{3}-J_2=\frac{1}{3}-\bigl(1-\frac{\pi}{4}\bigr)=\frac{\pi}{4}-\frac{2}{3}.

For mixed powers, $K_1=\int_0^{\pi/4}\tan\theta\sec\theta \, d\theta=\bigl[\sec\theta\bigr]_0^{\pi/4}=\sqrt{2}-1$ , so

\int_0^{\pi/4}\tan^3\theta\sec\theta \, d\theta=K_3=\frac{1}{3}\bigl(\sqrt{2}-2K_1\bigr)=\frac{1}{3}\bigl(\sqrt{2}-2(\sqrt{2}-1)\bigr)=\frac{2-\sqrt{2}}{3}.

[routine]

Reducing two powers together

When both exponents vary, each reduction drops one index by two, as in the one-parameter trigonometric recurrences. Reduce whichever index leads to a known base integral.

[DEFINITION]

Reduction formula

A reduction formula expresses an integral $I_n$ in terms of $I_{n-k}$ for a smaller index, typically by integration by parts.

[PROPOSITION]

A two-parameter recurrence

Let $I_{m,n}=\int_0^{\pi/2}\cos^m x\sin^n x \, dx$ . Then

I_{m,n}=\frac{m-1}{m+n} I_{m-2,n}

if $m\ge 2$ and

I_{m,n}=\frac{n-1}{m+n} I_{m,n-2}

if $n\ge 2$ .

The case $m=0$ recovers the sine-power recurrence; the case $n=0$ recovers the cosine-power recurrence. The base values needed to stop the reduction are $I_{0,0}=\frac{\pi}{2}$ , $I_{0,1}=1$ , $I_{1,0}=1$ and $I_{1,1}=\frac{1}{2}$ .

[1]:Reduction formula
[2]:Proof
[3]: $\int_0^{\pi/2}\cos^5 x\sin^4 x \, dx$

Assume $m\ge 2$ . Integrate by parts with $u=\cos^{m-1}x$ and $dv=\cos x\sin^n x \, dx$ , so $du=-(m-1)\cos^{m-2}x\sin x \, dx$ and $v=\frac{\sin^{n+1}x}{n+1}$ . Then

I_{m,n}=\biggl[\frac{\cos^{m-1}x\sin^{n+1}x}{n+1}\biggr]_0^{\pi/2}+\frac{m-1}{n+1}\int_0^{\pi/2}\cos^{m-2}x\sin^{n+2}x \, dx.

The boundary term is $0$ . Replace $\sin^2 x=1-\cos^2 x$ :

\begin{aligned} I_{m,n}&=\frac{m-1}{n+1}\int_0^{\pi/2}\cos^{m-2}x\sin^n x\bigl(1-\cos^2 x\bigr) \, dx \\ &=\frac{m-1}{n+1}\bigl(I_{m-2,n}-I_{m,n}\bigr). \end{aligned}

Hence $\frac{m+n}{n+1} I_{m,n}=\frac{m-1}{n+1} I_{m-2,n}$ and therefore

I_{m,n}=\frac{m-1}{m+n} I_{m-2,n}.

For $n\ge 2$ reverse the roles. Take $u=\sin^{n-1}x$ and $dv=\sin x\cos^m x\,dx$ , so $v=-\frac{\cos^{m+1}x}{m+1}$ . The boundary term vanishes and $\cos^2x=1-\sin^2x$ gives

I_{m,n}=\frac{n-1}{m+1}\bigl(I_{m,n-2}-I_{m,n}\bigr).

Rearranging this equation yields $I_{m,n}=\frac{n-1}{m+n}I_{m,n-2}$ .

[1]:A two-parameter recurrence