Properties of definite integrals

Some definite integrals defeat direct antidifferentiation but yield easily to clever properties. Mastering these is essential for board exams and JEE alike , every year, a problem appears whose three-line solution depends on the right property.

The eight standard properties

P1 , Reversal. $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$.

P2 , Same limits. $\int_a^a f(x)\,dx = 0$.

P3 , Variable renaming. $\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$. The variable inside is "dummy" , only the function and the limits matter.

P4 , King's rule. $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$. Substitute $u = a + b - x$.

P5 , Zero to $a$. $\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$. Special case of P4.

P6 , Splitting symmetric interval. $\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a - x)\,dx$.

P7 , Even/odd functions. $\int_{-a}^a f(x)\,dx = \begin{cases} 2 \int_0^a f(x)\,dx & f \text{ even} \\ 0 & f \text{ odd}\end{cases}$.

P8 , Periodic functions. If $f$ has period $T$, then $\int_a^{a+T} f(x)\,dx = \int_0^T f(x)\,dx$ for any $a$. And $\int_0^{nT} f(x)\,dx = n \int_0^T f(x)\,dx$.

The king's-rule trick

Property P5 is the workhorse. The technique: if direct integration looks ugly, replace $f(x)$ by $f(a-x)$ and add the two versions. The pieces often combine into a constant or a simpler function.

The classic example. Evaluate $I = \int_0^{\pi/2} \dfrac{\sin x}{\sin x + \cos x}\,dx$.

By P5, $I = \int_0^{\pi/2} \dfrac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)}\,dx = \int_0^{\pi/2} \dfrac{\cos x}{\cos x + \sin x}\,dx$. Adding the two versions: $2I = \int_0^{\pi/2} \dfrac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\pi/2} 1\,dx = \pi/2$. So $I = \pi/4$.

Even/odd shortcuts

A function is even if $f(-x) = f(x)$, odd if $f(-x) = -f(x)$. On a symmetric interval $[-a, a]$:

• Odd $\Rightarrow$ integral vanishes by cancellation.
• Even $\Rightarrow$ integral is twice the half-interval integral.

This makes integrals like $\int_{-1}^1 x^3 \cos x\,dx$ trivial: $x^3 \cos x$ is odd, so the answer is $0$.

Periodicity

For $\int_0^{n\pi} |\sin x|\,dx$: $|\sin x|$ has period $\pi$, and $\int_0^\pi \sin x\,dx = 2$. So $\int_0^{n\pi} |\sin x|\,dx = 2n$.

A useful family

A frequent JEE setup. If $f(a + b - x) + f(x) =$ constant or has a known relation to $f(x)$, then P4 with addition yields a closed form.

A special instance: when $f(a - x) = f(x)$ (the function is symmetric about $x = a/2$), then $\int_0^a x f(x)\,dx = \dfrac{a}{2}\int_0^a f(x)\,dx$. Proof: let $I = \int_0^a x f(x)\,dx$. Substitute $x \mapsto a - x$: $I = \int_0^a (a-x)f(a-x)\,dx = \int_0^a (a-x)f(x)\,dx = a \int_0^a f(x)\,dx - I$. Solve.

Worked examples

Example 1. $\int_{-\pi/2}^{\pi/2} \sin^7 x\,dx$.

$\sin^7 x$ is odd on a symmetric interval. Answer: $0$.

Example 2. $I = \int_0^{\pi/2} \log(\tan x)\,dx$.

By P5, $I = \int_0^{\pi/2} \log(\tan(\pi/2 - x))\,dx = \int_0^{\pi/2} \log(\cot x)\,dx = -\int_0^{\pi/2} \log(\tan x)\,dx = -I$. So $2I = 0$, giving $I = 0$.

Example 3. $I = \int_0^\pi \dfrac{x \sin x}{1 + \cos^2 x}\,dx$.

By P5 with $a = \pi$: $I = \int_0^\pi \dfrac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)}\,dx = \int_0^\pi \dfrac{(\pi - x)\sin x}{1 + \cos^2 x}\,dx$. Add: $2I = \int_0^\pi \dfrac{\pi \sin x}{1 + \cos^2 x}\,dx$. Substitute $u = \cos x$, $du = -\sin x\,dx$. Limits: $u(0) = 1$, $u(\pi) = -1$. So $2I = \pi \int_{-1}^1 \dfrac{du}{1 + u^2} = \pi \cdot 2 \tan^{-1}1 = \pi \cdot \dfrac{\pi}{2}$. Therefore $I = \pi^2/4$.

Example 4. $\int_0^{\pi/2} \dfrac{dx}{1 + \tan^3 x}$.

Let $f(x) = \dfrac{1}{1 + \tan^3 x}$. Then $f(\pi/2 - x) = \dfrac{1}{1 + \cot^3 x} = \dfrac{\tan^3 x}{1 + \tan^3 x}$. So $f(x) + f(\pi/2 - x) = 1$. Adding the two versions: $2I = \int_0^{\pi/2} 1\,dx = \pi/2$, so $I = \pi/4$.

Example 5. $\int_0^{2\pi} \cos^2 x\,dx$.

$\cos^2 x$ has period $\pi$. So the integral is $2 \int_0^\pi \cos^2 x\,dx = 2 \cdot \pi/2 = \pi$.

Example 6. $\int_0^1 \dfrac{\ln(1 + x)}{1 + x^2}\,dx$ (Putnam-style; substitute $x = \tan\theta$, $\theta \in [0, \pi/4]$).

After substitution: $\int_0^{\pi/4} \ln(1 + \tan\theta)\,d\theta$. By P5 (with $a = \pi/4$): replace $\theta$ by $\pi/4 - \theta$. We get $\ln(1 + \tan(\pi/4 - \theta)) = \ln\left(\dfrac{2}{1 + \tan\theta}\right) = \ln 2 - \ln(1 + \tan\theta)$. Adding: $2I = \int_0^{\pi/4}\ln 2\,d\theta = \dfrac{\pi \ln 2}{4}$. So $I = \dfrac{\pi \ln 2}{8}$.

Try it yourself

1. $\int_{-1}^1 (x^5 - x^3 + x)\,dx$
2. $\int_0^{\pi/2} \dfrac{\cos x}{\cos x + \sin x}\,dx$
3. $\int_0^{\pi} \dfrac{x\,dx}{1 + \sin x}$
4. $\int_0^a \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}}\,dx$
5. $\int_0^{\pi/4} \ln(1 + \tan x)\,dx$
6. $\int_0^{2\pi} |\cos x|\,dx$
7. $\int_{-\pi/4}^{\pi/4} \sin^2 x\,dx$
8. $\int_{-\pi/2}^{\pi/2} \dfrac{x^3 \cos x}{1 + x^2}\,dx$
9. $\int_0^\pi x \sin^3 x\,dx$
10. $\int_0^1 \dfrac{x(1-x)^n}{ }$ , wait, $\int_0^1 x(1-x)^n\,dx$. (Use $x \mapsto 1-x$.)
11. $\int_0^{\pi/2} \dfrac{\sin^n x}{\sin^n x + \cos^n x}\,dx$
12. $\int_{-1}^1 x^2 e^{x^2}\,dx$ (use parity)
13. $\int_0^{n\pi} |\sin x|\,dx$ where $n \in \mathbb{N}$
14. $\int_0^{\pi/2} \log(\sin x)\,dx$ (a famous one , use P5 and combine with $\log(\cos x)$ version)

Pitfalls and tricks

• Spot symmetry first. Always check if the integrand is odd/even on a symmetric interval , saves time.
• King's rule rewards persistence. Apply it, add the two versions, and look for cancellation or simplification.
• For periodic functions, integrate over one period and multiply by the number of periods.
• Don't forget to substitute the variable after using a property , it's just a rename, but mismatched variables cause confusion.
• Absolute values demand splitting. $\int_a^b |f(x)|\,dx$ requires locating sign changes of $f$ and splitting the interval.