Why does integral from 0 to π/2 of x g(x) dx = π/4 integral from 0 to π/2 of g(x) dx hold?

Because g(π/2 - x) = g(x). Applying King's rule gives J = integral from 0 to π/2 of (π/2 - x) g(x) dx = π/2integral from 0 to π/2 of g(x) dx - J, so 2J = π/2integral from 0 to π/2 of g(x) dx.

JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Definite Integral of a Trigonometric Rational Function using King's Property and Symmetry

| 120/(π^3) integral from 0 to π of (x^2 sin x cos x)/(sin^4 x + cos^4 x) dx | is equal to hspace3cm.

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Correct answer

The value of the expression is 15.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

We are given the expression $E = ∣ \frac{120}{π^{3}} I ∣$ , where $I = \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ .

02Approach

Use the property $\int_{0}^{2 a} f (x) d x = \int_{0}^{a} (f (x) + f (2 a - x)) d x$ or split the integral from $0$ to $π / 2$ and $π / 2$ to $π$ , substitute $x = π - t$ , and then eliminate the $x$ factor using King's rule on the interval $[0, π / 2]$ .

03Execute

Splitting at $π / 2$ : since for $x \in [π / 2, π]$ , replacing $x$ by $π - u$ gives $\sin (π - u) \cos (π - u) = - \sin u \cos u$ and $\sin^{4} (π - u) + \cos^{4} (π - u) = \sin^{4} u + \cos^{4} u$ , we get $I = \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} (x^{2} - (π - x)^{2}) d x = \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} (2 π x - π^{2}) d x$ .

04Execute

Applying King's property $\int_{0}^{π / 2} x g (x) d x = \frac{π}{4} \int_{0}^{π / 2} g (x) d x$ for symmetric function $g (x) = \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x}$ , we have $2 π \int_{0}^{π / 2} x g (x) d x = 2 π \cdot \frac{π}{4} \int_{0}^{π / 2} g (x) d x = \frac{π^{2}}{2} \int_{0}^{π / 2} g (x) d x$ . Hence, $I = (\frac{π^{2}}{2} - π^{2}) \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = - \frac{π^{2}}{2} \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ .

05Execute

Rewrite the remaining integral: $\int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = \int_{0}^{π / 2} \frac{\frac{1}{2} \sin 2 x}{1 - 2 \sin^{2} x \cos^{2} x} d x = \int_{0}^{π / 2} \frac{\sin 2 x}{2 - \sin^{2} 2 x} d x = \int_{0}^{π / 2} \frac{\sin 2 x}{1 + \cos^{2} 2 x} d x$ . Let $t = \cos 2 x$ , so $d t = - 2 \sin 2 x d x$ . When $x = 0, t = 1$ ; when $x = π / 2, t = - 1$ . The integral becomes $\frac{1}{2} \int_{- 1}^{1} \frac{d t}{1 + t^{2}} = \frac{1}{2} [\arctan (t)]_{- 1}^{1} = \frac{1}{2} (\frac{π}{4} - (- \frac{π}{4})) = \frac{π}{4}$ . Thus, $I = - \frac{π^{2}}{2} \cdot \frac{π}{4} = - \frac{π^{3}}{8}$ .

06Execute

Substitute $I = - \frac{π^{3}}{8}$ into the required expression: $∣ \frac{120}{π^{3}} I ∣ = ∣ \frac{120}{π^{3}} (- \frac{π^{3}}{8}) ∣ = \frac{120}{8} = 15$ .

✓Verify

Check that the integral $I$ is negative since on $[0, π / 2]$ the integrand is positive, but on $[π / 2, π]$ $\sin x \cos x < 0$ and $x^{2}$ is larger, making the net value negative. Taking the absolute value yields a positive integer 15.

Hints that build this answer step by step

Using King's property $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ with $f (x) = \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x}$ , how does $I = \int_{0}^{π} x^{2} f (x) d x$ relate to $\int_{0}^{π} x f (x) d x$ ?
$I = π \int_{0}^{π} x f (x) d x$
What is the value of the base integral $K = \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ ?
$\frac{π}{4}$
What is the value of $\int_{0}^{π} x f (x) d x$ in terms of $K$ ?
$- \frac{π^{2}}{8}$
Evaluating the original expression $∣ \frac{120}{π^{3}} I ∣$ , what is the final numerical value?
$15$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why does

\int_{0}^{π / 2} x g (x) d x = \frac{π}{4} \int_{0}^{π / 2} g (x) d x

hold?

Because $g (π / 2 - x) = g (x)$ . Applying King's rule gives $J = \int_{0}^{π / 2} (π / 2 - x) g (x) d x = \frac{π}{2} \int_{0}^{π / 2} g (x) d x - J$ , so $2 J = \frac{π}{2} \int_{0}^{π / 2} g (x) d x$ .

Answer

Accepted answer: 15

The value of the expression is 15.

Why each option works or fails

Step-by-step solution

Given: We are given the expression $E = ∣ \frac{120}{π^{3}} I ∣$ , where $I = \int_{0}^{π} \frac{x^{2} \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ .
Approach: Use the property $\int_{0}^{2 a} f (x) d x = \int_{0}^{a} (f (x) + f (2 a - x)) d x$ or split the integral from $0$ to $π / 2$ and $π / 2$ to $π$ , substitute $x = π - t$ , and then eliminate the $x$ factor using King's rule on the interval $[0, π / 2]$ .
Execute: Splitting at $π / 2$ : since for $x \in [π / 2, π]$ , replacing $x$ by $π - u$ gives $\sin (π - u) \cos (π - u) = - \sin u \cos u$ and $\sin^{4} (π - u) + \cos^{4} (π - u) = \sin^{4} u + \cos^{4} u$ , we get $I = \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} (x^{2} - (π - x)^{2}) d x = \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} (2 π x - π^{2}) d x$ .
Execute: Applying King's property $\int_{0}^{π / 2} x g (x) d x = \frac{π}{4} \int_{0}^{π / 2} g (x) d x$ for symmetric function $g (x) = \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x}$ , we have $2 π \int_{0}^{π / 2} x g (x) d x = 2 π \cdot \frac{π}{4} \int_{0}^{π / 2} g (x) d x = \frac{π^{2}}{2} \int_{0}^{π / 2} g (x) d x$ . Hence, $I = (\frac{π^{2}}{2} - π^{2}) \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = - \frac{π^{2}}{2} \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ .
Execute: Rewrite the remaining integral: $\int_{0}^{π / 2} x g (x) d x = \frac{π}{4} \int_{0}^{π / 2} g (x) d x$ . Let $t = \cos 2 x$ , so $d t = - 2 \sin 2 x d x$ . When $x = 0, t = 1$ ; when $x = π / 2, t = - 1$ . The integral becomes $\frac{1}{2} \int_{- 1}^{1} \frac{d t}{1 + t^{2}} = \frac{1}{2} [\arctan (t)]_{- 1}^{1} = \frac{1}{2} (\frac{π}{4} - (- \frac{π}{4})) = \frac{π}{4}$ . Thus, $I = - \frac{π^{2}}{2} \cdot \frac{π}{4} = - \frac{π^{3}}{8}$ .
Execute: Substitute $I = - \frac{π^{3}}{8}$ into the required expression: $∣ \frac{120}{π^{3}} I ∣ = ∣ \frac{120}{π^{3}} (- \frac{π^{3}}{8}) ∣ = \frac{120}{8} = 15$ .
Verify: Check that the integral $I$ is negative since on $[0, π / 2]$ the integrand is positive, but on $[π / 2, π]$ $\sin x \cos x < 0$ and $x^{2}$ is larger, making the net value negative. Taking the absolute value yields a positive integer 15.

Definite Integral of a Trigonometric Rational Function using King's Property and Symmetry

What feels right?

The value of the expression is 15.

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

06Execute

✓Verify

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