Draft:King's Property

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King's Property of Integration

In mathematics, specifically calculus, King's Property is a technique that can be implemented to solve certain integrals. Named by VK Bansal:, King's Property is simply a change of variables in an integral, where $x\mapsto a+b-x$. It takes the following form:

$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$

Geometrically, the property states that it does not matter if we integrate from left to right or from right to left. In other words, if the area under the curve is flipped horizontally, the area is unchanged.

Derivation

Take the definite integral: $\underset{a}{\overset{b}{\int }}f(x)dx$

Let $u(x)=a+b-x⟹du=-dx$.

Our limits of integration change accordingly

$u(a)=a+b-a=b$

$u(b)=a+b-b=a$

Following the substitution: $\underset{u(a)}{\overset{u(b)}{\int }}f(u)du=-\underset{a}{\overset{b}{\int }}f(a+b-x)dx$

We can generate another minus sign by flipping the bounds of integration leaving us with the end result of the King's Property:

$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$

Uses

King's Property can be used to evaluate definite integrals if the antiderivative of the integrand is non-elementary. The property can be especially helpful when dealing with integrals involving trigonometric functions or when Pi appears in the limits of integration. These two observations are by no means absolute, and there is no trick to determining if the property will be useful. An example is shown below.

Consider the integral: $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{1+(\tan x{)}^{\pi }}dx$

There is no obvious method to solving this integral. We cannot exploit any trigonometric identities, nor are there any useful substitutions to be made. However, we can use King's Property to solve:

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{1+(\tan (\frac{\pi }{2}-x){)}^{\pi }}dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{1+\frac{1}{(\tan x{)}^{\pi }}}dx$$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{(\tan x{)}^{\pi }}{(\tan x{)}^{\pi }+1}dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{(\tan x{)}^{\pi }+1-1}{(\tan x{)}^{\pi }+1}dx$

$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}1dx-\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{(\tan x{)}^{\pi }+1}dx=\frac{\pi }{2}-I$

$⟹I=\frac{\pi }{2}-I$

$⟹I=\frac{\pi }{4}$