Definite integral properties — even/odd functions, King's property, Leibniz rule

Question

Evaluate: (a) $\int_{-\pi/2}^{\pi/2} \sin^3 x \, dx$ and (b) $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx$

Solution — Step by Step

$f(x) = \sin^3 x$ . Check: $f(-x) = \sin^3(-x) = -\sin^3 x = -f(x)$ .

Since $\sin^3 x$ is an odd function and the limits are symmetric about 0:

\int_{-a}^{a} f(x) \, dx = 0 \text{ when } f(-x) = -f(x)

\int_{-\pi/2}^{\pi/2} \sin^3 x \, dx = \mathbf{0}

No computation needed at all.

King’s property: $\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx$

Let $I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx$

Apply King’s property (replace $x$ with $\pi/2 - x$ ):

I = \int_0^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} \, dx = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} \, dx

Add both expressions:

2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}

I = \mathbf{\frac{\pi}{4}}

Why This Works

graph TD
    A["Definite Integral: Which property?"] --> B["Symmetric limits -a to a?"]
    B --> C["Is integrand odd? → Answer is 0"]
    B --> D["Is integrand even? → 2 × integral from 0 to a"]
    A --> E["Limits 0 to a?"]
    E --> F["f x + f a-x simplifies? → King's property"]
    A --> G["Limits 0 to 2a?"]
    G --> H["f 2a-x = f x? → 2 × integral from 0 to a"]
    G --> I["f 2a-x = -f x? → Answer is 0"]
    A --> J["Variable upper limit?"]
    J --> K["Leibniz rule: d/dx ∫ f t dt = f x"]

These properties are pattern-recognition shortcuts. The odd/even function property works because on symmetric intervals, the positive and negative areas cancel (odd) or double (even). King’s property works by reflecting the integration variable — the integral does not change, but the integrand transforms.

JEE Main asks 1-2 definite integral questions every paper. At least one of them can be solved using these properties without actually computing any antiderivative. The King’s property is especially powerful for integrals of the form $\int_0^{\pi/2} \frac{f(\sin x)}{f(\sin x) + f(\cos x)} dx$ — the answer is always $\pi/4$ .

Alternative Method

Walli’s formula for $\int_0^{\pi/2} \sin^n x \, dx$ :

If $n$ is even: $\frac{(n-1)(n-3)...1}{n(n-2)...2} \cdot \frac{\pi}{2}$
If $n$ is odd: $\frac{(n-1)(n-3)...2}{n(n-2)...1}$

Example: $\int_0^{\pi/2} \sin^4 x \, dx = \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$

This formula is a massive time-saver for JEE and appears almost every year.

Common Mistake

Applying King’s property with wrong substitution. The property says replace $x$ with $a + b - x$ where $a$ and $b$ are the limits. Students sometimes replace $x$ with $b - x$ or $a - x$ instead. For limits $[0, \pi/2]$ , the replacement is $x \to \pi/2 - x$ . For limits $[0, \pi]$ , it is $x \to \pi - x$ . Always use $a + b - x$ where $a$ is the lower limit and $b$ is the upper limit.

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

$\int_{-a}^{a} f(x) \, dx = 0$ if $f$ is odd; $= 2\int_0^a f(x) \, dx$ if $f$ is even

King’s property: $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$

Leibniz rule: $\frac{d}{dx}\int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$