Properties of Definite Integrals

The Toolkit: Properties P0 to P7

These properties evaluate definite integrals without finding an anti-derivative — often turning a hopeless-looking integral into a one-line answer.

P0 (dummy variable): $\displaystyle\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$ — the variable of integration is a placeholder; renaming it changes nothing. Proof: substitute $x = t$.

P1 (flip the limits): $\displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$. In particular $\displaystyle\int_a^a f(x)\,dx = 0$. Proof: by the Second FTC both sides are $\pm[F(b) - F(a)]$.

P2 (split the interval): $\displaystyle\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$. Proof: $[F(c) - F(a)] + [F(b) - F(c)] = F(b) - F(a)$. This is the property that handles modulus integrands: split at every point where the expression inside $|\cdot|$ changes sign.

P3 (reflect across the midpoint): $\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$. Proof: substitute $t = a + b - x$ (so $dt = -dx$, limits swap), then apply P1 and P0.

P4 (the King's rule — P3 with $a = 0$): $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$.

Key Point: P3/P4 replace the integrand by its mirror image without changing the value. The winning move is almost always add the two versions: $2I = \int_0^a [f(x) + f(a - x)]\,dx$, and the sum inside is often a constant or something elementary.

[JEE Tip] The King's rule is the single most-used property in JEE definite integrals. Whenever you see $x$ multiplying a symmetric trigonometric block on $[0, \pi]$ — like $\int_0^{\pi}\frac{x\sin x}{1+\cos^2 x}dx$ — apply P4, add, and the naked $x$ disappears, leaving $\frac{\pi}{2}\int_0^{\pi}(\text{trig part})$.

Doubling and Vanishing: P5, P6, P7

P5 (split $[0, 2a]$ and reflect): $\displaystyle\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a - x)\,dx$. Proof: split at $a$ by P2, then substitute $t = 2a - x$ in the second piece.

P6 (the $[0, 2a]$ shortcut):

$$\int_0^{2a} f(x)\,dx = \begin{cases} 2\displaystyle\int_0^a f(x)\,dx, & \text{if } f(2a - x) = f(x) \\[2mm] 0, & \text{if } f(2a - x) = -f(x) \end{cases}$$

P7 (even/odd symmetry on $[-a, a]$):

$$\int_{-a}^{a} f(x)\,dx = \begin{cases} 2\displaystyle\int_0^a f(x)\,dx, & \text{if } f \text{ is even: } f(-x) = f(x) \\[2mm] 0, & \text{if } f \text{ is odd: } f(-x) = -f(x) \end{cases}$$

Proof sketch of P7: split $\int_{-a}^{a}$ at $0$ by P2, substitute $t = -x$ in the left piece to get $\int_0^a f(-x)\,dx$, then use $f(-x) = \pm f(x)$.

Recognising even and odd quickly

1. Powers: even powers of $x$ are even; odd powers are odd. $\sin x$, $\tan x$, $x\cos x$, $x^2\sin x$ are odd; $\cos x$, $x\sin x$, $\sin^2 x$, $|x|$ are even.
2. Products: odd × odd = even; odd × even = odd; even × even = even — same rules as signs.
3. Mixed sums: split term by term. In $\int_{-a}^{a}(x^3 + x\cos x + \tan^5 x + 1)\,dx$, the first three terms are odd (each integrates to $0$) and only the constant survives.

Key Point: Before integrating anything on a symmetric interval $[-a, a]$ or $[0, 2a]$, spend ten seconds on a symmetry check — P6/P7 may finish the whole problem, or at least kill most of the terms.

[JEE Tip] JEE Main loves the sum-of-terms trap of Exercise 7.10 Q20: a long integrand on $[-\frac{\pi}{2}, \frac{\pi}{2}]$ where every term but one is odd. Also memorise the two famous results proved by these properties: $\int_0^{\pi/2}\log\sin x\,dx = -\frac{\pi}{2}\log 2$ and $\int_0^{\pi/4}\log(1 + \tan x)\,dx = \frac{\pi}{8}\log 2$ — both appear directly as questions.

Solved Examples

Example 1: Splitting a modulus

Evaluate $\displaystyle\int_{-1}^{2} \left|x^3 - x\right|dx$.

Solution:

1. Find the sign pattern: $x^3 - x = x(x-1)(x+1)$ is $\geq 0$ on $[-1, 0]$, $\leq 0$ on $[0, 1]$, and $\geq 0$ on $[1, 2]$.
2. Split by P2: $I = \int_{-1}^{0}(x^3 - x)dx + \int_0^1 -(x^3 - x)dx + \int_1^2 (x^3 - x)dx$.
3. Integrate each piece with $F(x) = \frac{x^4}{4} - \frac{x^2}{2}$: pieces give $\frac14$, $\frac14$, and $\frac94$ respectively.

Final Answer: $\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{9}{4} = \dfrac{11}{4}$.

Takeaway: A modulus integral is really several ordinary integrals in disguise — the sign chart of the inside expression tells you where to cut.

Example 2: Even function shortcut

Evaluate $\displaystyle\int_{-\pi/4}^{\pi/4} \sin^2 x\,dx$.

Solution:

1. Symmetry check: $\sin^2(-x) = \sin^2 x$ — even. By P7(i), $I = 2\int_0^{\pi/4}\sin^2 x\,dx$.
2. Power-reduce: $2\int_0^{\pi/4}\frac{1 - \cos 2x}{2}dx = \int_0^{\pi/4}(1 - \cos 2x)\,dx$.
3. Evaluate: $\left[x - \frac{\sin 2x}{2}\right]_0^{\pi/4} = \frac{\pi}{4} - \frac{1}{2}$.

Final Answer: $\dfrac{\pi}{4} - \dfrac{1}{2}$.

Example 3: The King's rule classic

Evaluate $\displaystyle\int_0^{\pi} \frac{x\sin x}{1 + \cos^2 x}\,dx$.

Solution:

1. Apply P4 with $a = \pi$: $I = \int_0^{\pi}\frac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)}dx = \int_0^{\pi}\frac{(\pi - x)\sin x}{1 + \cos^2 x}dx$.
2. Add the two versions: $2I = \pi\int_0^{\pi}\frac{\sin x}{1 + \cos^2 x}dx$ — the $x$ vanished.
3. Substitute $t = \cos x$ ($dt = -\sin x\,dx$; limits $1 \to -1$): $2I = \pi\int_{-1}^{1}\frac{dt}{1 + t^2}$.
4. Even integrand (P7): $2I = 2\pi\int_0^1\frac{dt}{1+t^2} = 2\pi\cdot\frac{\pi}{4}$.

Final Answer: $I = \dfrac{\pi^2}{4}$.

Example 4: Odd function, zero work

Evaluate $\displaystyle\int_{-1}^{1} \sin^5 x\cos^4 x\,dx$.

Solution:

1. Symmetry check: $f(-x) = \sin^5(-x)\cos^4(-x) = -\sin^5 x\cos^4 x = -f(x)$ — odd.
2. Apply P7(ii): the integral over the symmetric interval $[-1, 1]$ is $0$.

Final Answer: $0$.

Example 5: The self-complementary trick

Evaluate $\displaystyle\int_0^{\pi/2} \frac{\sin^4 x}{\sin^4 x + \cos^4 x}\,dx$.

Solution:

1. Apply P4 with $a = \frac{\pi}{2}$: since $\sin\left(\frac{\pi}{2} - x\right) = \cos x$, the mirrored integral is $\int_0^{\pi/2}\frac{\cos^4 x}{\cos^4 x + \sin^4 x}dx$.
2. Add: $2I = \int_0^{\pi/2}\frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x}dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}$.

Final Answer: $I = \dfrac{\pi}{4}$.

Takeaway: Whenever numerator and denominator swap roles under $x \mapsto a - x$, adding the two copies gives $\int_0^a 1\,dx = a$, so $I = \frac{a}{2}$ — regardless of the powers involved.

Example 6: Mirror on a shifted interval

Evaluate $\displaystyle\int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}}$.

Solution:

1. Rewrite: $\frac{1}{1 + \sqrt{\tan x}} = \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}$.
2. Apply P3 with $a + b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$: the mirrored integrand is $\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}$.
3. Add: $2I = \int_{\pi/6}^{\pi/3} 1\,dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.

Final Answer: $I = \dfrac{\pi}{12}$.

Example 7: The famous log-sine integral

Evaluate $\displaystyle\int_0^{\pi/2} \log\sin x\,dx$.

Solution:

1. Apply P4: $I = \int_0^{\pi/2}\log\sin\left(\frac{\pi}{2} - x\right)dx = \int_0^{\pi/2}\log\cos x\,dx$.
2. Add: $2I = \int_0^{\pi/2}\log(\sin x\cos x)\,dx = \int_0^{\pi/2}\log\sin 2x\,dx - \int_0^{\pi/2}\log 2\,dx$ (using $\sin x\cos x = \frac{\sin 2x}{2}$).
3. Substitute $t = 2x$ in the first piece: $\frac12\int_0^{\pi}\log\sin t\,dt = \frac12\cdot 2\int_0^{\pi/2}\log\sin t\,dt = I$ (by P6, since $\sin(\pi - t) = \sin t$).
4. Solve the loop: $2I = I - \frac{\pi}{2}\log 2$, so $I = -\frac{\pi}{2}\log 2$.

Final Answer: $\displaystyle\int_0^{\pi/2}\log\sin x\,dx = -\dfrac{\pi}{2}\log 2$.

Example 8: Modulus with a simple kink

Evaluate $\displaystyle\int_{-5}^{5} |x + 2|\,dx$.

Solution:

1. Find the kink: $x + 2 = 0$ at $x = -2$; split there by P2.
2. Left piece: $\int_{-5}^{-2} -(x+2)\,dx = \left[-\frac{(x+2)^2}{2}\right]_{-5}^{-2} = 0 + \frac{9}{2}$.
3. Right piece: $\int_{-2}^{5}(x+2)\,dx = \left[\frac{(x+2)^2}{2}\right]_{-2}^{5} = \frac{49}{2}$.

Final Answer: $\dfrac{9}{2} + \dfrac{49}{2} = 29$.

Example 9: The log(1 + tan x) gem

Evaluate $\displaystyle\int_0^{\pi/4} \log(1 + \tan x)\,dx$.

Solution:

1. Apply P4 with $a = \frac{\pi}{4}$: $I = \int_0^{\pi/4}\log\left(1 + \tan\left(\frac{\pi}{4} - x\right)\right)dx$.
2. Simplify the inside: $\tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x}$, so $1 + \tan\left(\frac{\pi}{4} - x\right) = \frac{2}{1 + \tan x}$.
3. Add the two copies: $2I = \int_0^{\pi/4}\log\left[(1 + \tan x)\cdot\frac{2}{1 + \tan x}\right]dx = \int_0^{\pi/4}\log 2\,dx = \frac{\pi}{4}\log 2$.

Final Answer: $I = \dfrac{\pi}{8}\log 2$.

Example 10: Odd powers vanish, even powers don't

Evaluate (i) $\displaystyle\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx$ (ii) $\displaystyle\int_{-\pi/2}^{\pi/2} \sin^7 x\,dx$.

Solution:

1. (ii) first: $\sin^7 x$ is odd (odd power of an odd function), so by P7(ii) the integral is $0$ — no computation.
2. (i): $\sin^2 x$ is even, so $I = 2\int_0^{\pi/2}\sin^2 x\,dx = \int_0^{\pi/2}(1 - \cos 2x)\,dx$.
3. Evaluate: $\left[x - \frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{\pi}{2} - 0$.

Final Answer: (i) $\dfrac{\pi}{2}$; (ii) $0$.

Example 11: A beta-style polynomial

Evaluate $\displaystyle\int_0^1 x(1 - x)^n\,dx$.

Solution:

1. King's rule (P4) with $a = 1$: $I = \int_0^1 (1 - x)\left(1 - (1 - x)\right)^n dx = \int_0^1 (1 - x)x^n\,dx$.
2. Expand: $I = \int_0^1\left(x^n - x^{n+1}\right)dx = \frac{1}{n+1} - \frac{1}{n+2}$.

Final Answer: $\dfrac{1}{(n+1)(n+2)}$.

Takeaway: P4 converted the awkward $x(1-x)^n$ into plain powers of $x$ — reflecting the integrand is often easier than expanding a binomial to $n$ terms.