Properties — King's, Symmetry and Standard Results

King's property and the 2I technique:

• King's property: $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ (and $\int_b^c f(x)\,dx=\int_b^c f(b+c-x)\,dx$).
• The 2I move: let $I=\int_0^a f(x)\,dx$; write a second copy with $x\to a-x$; ADD them so the denominators match. Often $2I=\int_0^a 1\,dx = a$.
• Classic family: $\displaystyle\int_0^a \frac{f(x)}{f(x)+f(a-x)}\,dx = \frac{a}{2}$.
• Works on $\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx = \frac{\pi}{4}$ and the whole $\frac{a\cos x+b\sin x}{\sin x+\cos x}$ family.

Symmetry rules on $[-a,a]$:

• Odd $f(-x)=-f(x)$: $\displaystyle\int_{-a}^{a} f(x)\,dx = 0$.
• Even $f(-x)=f(x)$: $\displaystyle\int_{-a}^{a} f(x)\,dx = 2\int_0^{a} f(x)\,dx$.
• The $1+c^x$ property: for even $f$, $\displaystyle\int_{-a}^{a}\frac{f(x)}{1+c^{x}}\,dx = \frac12\int_{-a}^{a} f(x)\,dx = \int_0^a f(x)\,dx$.
• Also $\displaystyle\int_{-a}^{a} h(x)\,dx = \int_0^a [h(x)+h(-x)]\,dx$.
• Establishing oddness from a functional equation: if $f(x+y)=f(x)+f(y)$ for all $x,y$, put $x=y=0$ to get $f(0)=0$, then $y=-x$ to get $f(-x)=-f(x)$ — so $f$ is odd and $\int_{-a}^{a}f=0$ without knowing $f$. Likewise $f(x)f(-x)=1$-type conditions feed the $1+c^x$ property.
• Parity of the usual suspects: $x^{2k+1}$, $\sin x$, $\tan x$, $x\cos x$, $\ln\dfrac{1-x}{1+x}$, $\sin^3x=\tfrac{3\sin x-\sin3x}{4}$ are odd; $x^{2k}$, $\cos x$, $|x|$, $x\sin x$, $\cos^2x$ are even. A product is odd iff an odd number of its factors are odd.

Results and reductions to memorise:

• $\displaystyle\int_0^{\pi}\frac{dx}{1+\sin^2x} = \int_0^{\pi}\frac{dx}{1+\cos^2x} = \frac{\pi}{\sqrt2}$.
• $\sin^4x+\cos^4x = \dfrac{3+\cos 4x}{4}$, so $\int_0^{\pi}(\sin^4x+\cos^4x)\,dx=\tfrac{3\pi}{4}$.
• $1+\cos\theta = 2\cos^2\tfrac{\theta}{2}$, $1-\cos\theta=2\sin^2\tfrac{\theta}{2}$ (half-angle).
• Beta function: $\displaystyle\int_0^1 x^{m}(1-x)^{n}\,dx = \frac{m!\,n!}{(m+n+1)!}$.
• The METHOD behind $\pi/\sqrt2$, for the variants the memorised value does not cover ($1+2\tan^2x$, $a^2\sin^2x+b^2\cos^2x$, …): divide top and bottom by $\cos^2x$, use $\sec^2x=1+\tan^2x$, put $t=\tan x$ (so $\int_0^{\pi/2}\to\int_0^{\infty}$), and finish with $\int_0^\infty\dfrac{dt}{t^2+k^2}=\dfrac{\pi}{2k}$. Over $[0,\pi]$ split at $\tfrac\pi2$ (or use King's) and double.
• $\displaystyle\int_0^{\pi}\frac{dx}{1+\sin x}=2$ and $\displaystyle\int_0^{\pi/2}\frac{dx}{1+\sin x}=1$: multiply by $\dfrac{1-\sin x}{1-\sin x}$ to get $\sec^2x-\sec x\tan x$, or shift the half-angle $1+\sin x=\big(\sin\tfrac x2+\cos\tfrac x2\big)^2$. The $1+\cos$ sibling uses $2\cos^2\tfrac x2$ directly; $1+\sin$ needs one of these two moves.
• Trig preparation this chapter keeps assuming (full detail in the Differentiation toolkit, linked below): Weierstrass $t=\tan\tfrac x2$, $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$ for any rational function of $\sin x,\cos x$; the harmonic form $\sin x+\cos x=\sqrt2\sin(x+\tfrac\pi4)$, so $\dfrac{1}{(\sin x+\cos x)^2}=\tfrac12\csc^2(x+\tfrac\pi4)$; and $\sqrt{1+\sin\theta}=\big|\sin\tfrac\theta2+\cos\tfrac\theta2\big|$ (keep the modulus, resolve it on the interval).
• The $\sqrt{\tan x}+\sqrt{\cot x}$ finisher: after King's property gives $\int_0^{\pi/2}\dfrac{\sin x+\cos x}{\sqrt{\sin x\cos x}}\,dx$, put $u=\sin x-\cos x$: $du=(\cos x+\sin x)\,dx$ and $\sin x\cos x=\dfrac{1-u^2}{2}$, so the integral is $\sqrt2\int_{-1}^{1}\dfrac{du}{\sqrt{1-u^2}}=\sqrt2\,\pi$.

Simplification routes (then apply the limits):

• Trig factoring: $\tan^3x+\tan x = \tan x\,\sec^2x$, which integrates to $\tfrac{\tan^2x}{2}$.
• By-parts for products: $\int x\ln x\,dx$, $\int e^x\sin x\,dx = \tfrac{e^x(\sin x-\cos x)}{2}$.
• Substitution: $\int_0^{\pi/2}e^{\sin x}\cos x\,dx$ via $u=\sin x$.
• Is the whole integrand somebody's derivative? Ask before any technique: $x^x(1+\ln x)=\dfrac{d}{dx}x^x$, so $\int_1^2x^x(1+\ln x)\,dx=2^2-1^1=3$; $e^x(f+f')=\dfrac{d}{dx}(e^xf)$; $\dfrac{x\cos x-\sin x}{x^2}=\dfrac{d}{dx}\dfrac{\sin x}{x}$. The $f'/f\to\ln|f|$ case in the FTC block is the special case of this habit.

(For the full substitution / by-parts technique, see the Indefinite Integration notes — here just carry the limits through.)