Definite Integrals — Properties & Standard Results

Property 1 — Dummy Variable

${\displaystyle {\int }_a^{b}f(x)dx={\int }_a^{b}f(t)dt}$

The variable of integration is a dummy — it doesn't affect the value. Useful for comparing or adding two integrals that differ only in variable name.

Property 2 — Reversal of Limits

${\displaystyle {\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx}$

Swapping limits changes the sign. In particular, ${\displaystyle {\int }_a^{a}f(x)dx=0}$.

Property 3 — Additivity (Splitting the Interval)

${\displaystyle {\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx}$

The point $c$ need not lie between $a$ and $b$. This is the key tool for integrating functions with absolute values — split at the point where the expression inside $|\cdot |$ changes sign.

Example: ${\displaystyle {\int }_0^2|x-1|dx}$

$|x-1|=\{\begin{array}{ll}1-x& 0\le x<1\\ x-1& 1\le x\le 2\end{array}$

$={\displaystyle {\int }_0^1(1-x)dx+{\int }_1^2(x-1)dx={[x-\frac{x^2}{2}]}_0^1+{[\frac{x^2}{2}-x]}_1^2=\frac{1}{2}+\frac{1}{2}=1}$

Property 4 — King Property ★ Most Important

${\displaystyle {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx}$

Replace $x$ with $a+b-x$. The integral value is unchanged. For the symmetric case $a=0$:

${\displaystyle {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx}$

Classic application: ${\displaystyle {\int }_0^{\pi /2}\frac{\sin x}{\sin x+\cos x}dx}$

Let $I={\displaystyle {\int }_0^{\pi /2}\frac{\sin x}{\sin x+\cos x}dx}$. By King ($x\to \frac{\pi }{2}-x$):

$I={\displaystyle {\int }_0^{\pi /2}\frac{\cos x}{\cos x+\sin x}dx}$

Adding: $2I={\displaystyle {\int }_0^{\pi /2}\frac{\sin x+\cos x}{\sin x+\cos x}dx={\int }_0^{\pi /2}1dx=\frac{\pi }{2}}$. Therefore $I={\displaystyle \frac{\pi }{4}}$.

Property 5 — Odd and Even Functions

${\displaystyle {\int }_{-a}^{a}f(x)dx=\{\begin{array}{ll}2{\displaystyle {\int }_0^{a}f(x)dx}& \text{if }f\text{ is even }(f(-x)=f(x))\\ 0& \text{if }f\text{ is odd }(f(-x)=-f(x))\end{array}}$

Examples:

${\displaystyle {\int }_{-1}^1{x}^{3}dx=0}$ (odd function)

${\displaystyle {\int }_{-\pi }^{\pi }\sin xdx=0}$ (odd function)

${\displaystyle {\int }_{-1}^1{x}^{2}dx=2{\int }_0^1{x}^{2}dx={\displaystyle \frac{2}{3}}}$ (even function)

Property 6 — Periodicity

If $f(x)$ is periodic with period $T$ (i.e., $f(x+T)=f(x)$), then:

${\displaystyle {\int }_0^{nT}f(x)dx=n{\int }_0^{T}f(x)dx}$

${\displaystyle {\int }_a^{a+T}f(x)dx={\int }_0^{T}f(x)dx}$ (independent of $a$)

Example: ${\displaystyle {\int }_0^{10\pi }|\sin x|dx}$

$|\sin x|$ has period $\pi$. ${\displaystyle {\int }_0^{\pi }|\sin x|dx=2}$. So ${\displaystyle {\int }_0^{10\pi }|\sin x|dx=10\times 2=20}$.

Property 7 — Symmetry about $x=a/2$

${\displaystyle {\int }_0^{2a}f(x)dx=\{\begin{array}{ll}2{\displaystyle {\int }_0^{a}f(x)dx}& \text{if }f(2a-x)=f(x)\\ 0& \text{if }f(2a-x)=-f(x)\end{array}}$

Wallis' formula gives a direct result for ${\displaystyle {\int }_0^{\pi /2}{\sin }^{n}xdx={\int }_0^{\pi /2}{\cos }^{n}xdx}$:

${\displaystyle {\int }_0^{\pi /2}{\sin }^{n}xdx=\{\begin{array}{ll}{\displaystyle \frac{(n-1)(n-3)\cdots 2}{n(n-2)\cdots 3}}& n\text{ odd}\\ {\displaystyle \frac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}}\cdot {\displaystyle \frac{\pi }{2}}& n\text{ even}\end{array}}$

Wallis Values — Quick Reference

Key pattern: Even $n$ → answer has $\pi$; Odd $n$ → rational number only.

Note: ${\displaystyle {\int }_0^{\pi /2}{\sin }^{n}xdx={\int }_0^{\pi /2}{\cos }^{n}xdx}$ — they are always equal.

Integration of Absolute Value Functions

Use Property 3 (splitting) at points where the argument of $|\cdot |$ is zero:

${\displaystyle {\int }_a^b|f(x)|dx}$: find zeros of $f(x)$ in $[a,b]$, split there, and flip sign where $f(x)<0$.

Useful Substitution in Definite Integrals

When limits change with substitution: If $x=g(t)$, then $dx=g^{\prime }(t)dt$ and the limits transform as $a\to g^{-1}(a)$, $b\to g^{-1}(b)$.

Key result by substitution:

${\displaystyle {\int }_0^{\pi }xf(\sin x)dx=\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)dx}$

Proof: By King ($x\to \pi -x$): LHS becomes ${\displaystyle {\int }_0^{\pi }(\pi -x)f(\sin x)dx}$. Adding both: $2I=\pi {\displaystyle {\int }_0^{\pi }f(\sin x)dx}$.

The $\ln$ and Trig Definite Integral

${\displaystyle {\int }_0^{\pi /2}\ln (\sin x)dx={\int }_0^{\pi /2}\ln (\cos x)dx=-\frac{\pi }{2}\ln 2}$

This is a standard JEE Advanced result — the proof uses the King Property followed by the product-to-sum log identity.