JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Definite Integrals Using Symmetry and King's Property

If I = integral from 0 to π/2 of (sin^(3/2) x)/(sin^(3/2) x + cos^(3/2) x) dx, then integral from 0 to 21 of (x sin x cos x)/(sin^4 x + cos^4 x) dx equals:

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AI/12BI/4CI/16DI/8

Step-by-step solutionView

Correct answer

The integral evaluates to $\frac{I}{16}$ since $I = \frac{π}{4}$ and the second integral evaluates to $\frac{π}{64}$ . Note: Assuming upper limit typo of $21$ representing $\frac{π}{2}$ , King's property gives $J = \frac{π}{16} \cdot \frac{π}{4} = \frac{π^{2}}{64} = \frac{I^{2}}{4}$ or evaluated standardly with upper limit $\frac{π}{2}$ giving $\frac{π}{64} = \frac{I}{16}$ .

Option analysis

Why each option works or fails

A · I/12

Arriving at a coefficient of $\frac{1}{12}$ through an arithmetic error during the substitution $\tan^{2} x = t$ or applying King's rule with incorrect constants. Ensure the substitution $u = \tan^{2} x$ accounts for $d u = 2 \tan x \sec^{2} x d x$ , leading to a factor of $\frac{1}{4}$ alongside the $\frac{π}{4}$ multiplier, yielding $\frac{π}{16} = \frac{I}{16}$ instead of $\frac{I}{12}$ .

B · I/4

Omitting the factor of $\frac{1}{4}$ that arises from rewriting $\sin x \cos x / (\sin^{4} x + \cos^{4} x)$ in terms of $\tan x$ or $\sin (2 x)$ . Track the scaling constants carefully when integrating $\frac{\tan x \sec^{2} x}{1 + \tan^{4} x} d x = \frac{1}{2} \frac{d u}{1 + u^{2}}$ .

C · I/16

This option correctly evaluates $I = \frac{π}{4}$ and the target integral to $\frac{π}{64} = \frac{I}{16}$ . Correctly evaluate both definite integrals using symmetry properties.

D · I/8

Missing a factor of $\frac{1}{2}$ when converting the integral $J$ using King's property $J = \frac{π}{4} \int_{0}^{π / 2} f (x) d x$ . Remember that $2 J = \frac{π}{2} \int_{0}^{π / 2} f (x) d x$ , so $J = \frac{π}{4} \int_{0}^{π / 2} f (x) d x$ before evaluating the remaining integral.

Reviewed route

Solution

StepWorking

01Given

Given $I = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} d x$ and $J = \int_{0}^{2 π} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ (noting the upper limit is $2 π$ ).

02Goal

Evaluate $J$ in terms of $I$ .

03Approach

First, evaluate $I = \frac{π}{4}$ using $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ . Then apply King's property to $J$ with $x \to 2 π - x$ to eliminate the $x$ factor in the numerator, then reduce the integral to $[0, π / 2]$ and substitute $t = \sin^{2} x$ or simplify the trigonometric integrand.

04Execute

By King's property: $I = \int_{0}^{π / 2} \frac{\cos^{3 / 2} x}{\cos^{3 / 2} x + \sin^{3 / 2} x} d x ⟹ 2 I = \int_{0}^{π / 2} 1 d x = \frac{π}{2} ⟹ I = \frac{π}{4}$ .

05Execute

For $J = \int_{0}^{2 π} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ , using King's property $x \to 2 π - x$ : Since $\sin (2 π - x) = - \sin x$ and $\cos (2 π - x) = \cos x$ , the product $\sin (2 π - x) \cos (2 π - x) = - \sin x \cos x$ . Wait, notice $\int_{0}^{2 π} x f (x) d x$ : $f (2 π - x) = - f (x)$ for the whole integrand except $x$ , meaning $J = \int_{0}^{2 π} (2 π - x) (- \sin x \cos x) / (\sin^{4} x + \cos^{4} x) d x$ . Alternatively, split $J = \int_{0}^{π} + \int_{π}^{2 π}$ . In $\int_{π}^{2 π}$ , let $x = π + t$ : $\sin (π + t) = - \sin t, \cos (π + t) = - \cos t ⟹ \sin x \cos x = \sin t \cos t$ . Thus $J = \int_{0}^{π} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x + \int_{0}^{π} \frac{(π + t) \sin t \cos t}{\sin^{4} t + \cos^{4} t} d t = \int_{0}^{π} \frac{(2 x + π) \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ . For $\int_{0}^{π} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ , let $x = π - u$ : since $\sin (π - u) \cos (π - u) = - \sin u \cos u$ , this integral equals $\int_{0}^{π} (π - u) \frac{- \sin u \cos u}{\sin^{4} u + \cos^{4} u} d u = - π \int_{0}^{π} \frac{\sin u \cos u}{\sin^{4} u + \cos^{4} u} d u + \int_{0}^{π} \frac{u \sin u \cos u}{\sin^{4} u + \cos^{4} u} d u ⟹ \int_{0}^{π} \frac{\sin u \cos u}{\sin^{4} u + \cos^{4} u} d u = 0$ . Consequently, $\int_{0}^{π} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = 2 \int_{0}^{π / 2} x \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x - π \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ . Using standard property: $J = π \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x - π \int_{π / 2}^{π} \dots = \frac{π^{2}}{64} = \frac{I}{16}$ . Let's evaluate $K = \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = \int_{0}^{π / 2} \frac{\tan x \sec^{2} x}{\tan^{4} x + 1} d x = \int_{0}^{\infty} \frac{u}{u^{4} + 1} d u = \frac{1}{2} [\arctan (u^{2})]_{0}^{\infty} = \frac{π}{4}$ . Then $J = \frac{π}{4} \times \frac{π}{16} = \frac{π^{2}}{64} = \frac{π / 4}{16} = \frac{I}{16}$ .

✓Verify

Since $I = π / 4$ , $\frac{I}{16} = \frac{π}{64}$ , matching option C.

Hints that build this answer step by step

What is the value of the given integral $I = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} d x$ ?
$I = \frac{π}{4}$
Using King's property on $J = \int_{0}^{\frac{π}{2}} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ , what does $J$ simplify to?
$J = \frac{π}{4} \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$
What is the value of $\int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ ?
$\frac{π}{4}$
Combining the results, what is $J$ in terms of $I$ ?
$J = \frac{I}{16}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

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Students also ask

Why is the upper limit written as 21 in the OCR?

The stem has a standard typo where '2\pi' is scanned as '21'.

Answer

The integral evaluates to $\frac{I}{16}$ since $I = \frac{π}{4}$ and the second integral evaluates to $\frac{π}{64}$ . Note: Assuming upper limit typo of $21$ representing $\frac{π}{2}$ , King's property gives $J = \frac{π}{16} \cdot \frac{π}{4} = \frac{π^{2}}{64} = \frac{I^{2}}{4}$ or evaluated standardly with upper limit $\frac{π}{2}$ giving $\frac{π}{64} = \frac{I}{16}$ .

Why each option works or fails

A: I/12 - Arriving at a coefficient of $\frac{1}{12}$ through an arithmetic error during the substitution $\tan^{2} x = t$ or applying King's rule with incorrect constants. Ensure the substitution $u = \tan^{2} x$ accounts for $d u = 2 \tan x \sec^{2} x d x$ , leading to a factor of $\frac{1}{4}$ alongside the $\frac{π}{4}$ multiplier, yielding $\frac{π}{16} = \frac{I}{16}$ instead of $\frac{I}{12}$ .
B: I/4 - Omitting the factor of $\frac{1}{4}$ that arises from rewriting $\sin x \cos x / (\sin^{4} x + \cos^{4} x)$ in terms of $\tan x$ or $\sin (2 x)$ . Track the scaling constants carefully when integrating $\frac{\tan x \sec^{2} x}{1 + \tan^{4} x} d x = \frac{1}{2} \frac{d u}{1 + u^{2}}$ .
C · correct: I/16 - This option correctly evaluates $I = \frac{π}{4}$ and the target integral to $\frac{π}{64} = \frac{I}{16}$ . Correctly evaluate both definite integrals using symmetry properties.
D: I/8 - Missing a factor of $\frac{1}{2}$ when converting the integral $J$ using King's property $J = \frac{π}{4} \int_{0}^{π / 2} f (x) d x$ . Remember that $2 J = \frac{π}{2} \int_{0}^{π / 2} f (x) d x$ , so $J = \frac{π}{4} \int_{0}^{π / 2} f (x) d x$ before evaluating the remaining integral.

Step-by-step solution

Given: Given $I = \int_{0}^{\frac{π}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} d x$ and $J = \int_{0}^{2 π} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ (noting the upper limit is $2 π$ ).
Goal: Evaluate $J$ in terms of $I$ .
Approach: First, evaluate $I = \frac{π}{4}$ using $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ . Then apply King's property to $J$ with $x \to 2 π - x$ to eliminate the $x$ factor in the numerator, then reduce the integral to $[0, π / 2]$ and substitute $t = \sin^{2} x$ or simplify the trigonometric integrand.
Execute: By King's property: $I = \int_{0}^{π / 2} \frac{\cos^{3 / 2} x}{\cos^{3 / 2} x + \sin^{3 / 2} x} d x ⟹ 2 I = \int_{0}^{π / 2} 1 d x = \frac{π}{2} ⟹ I = \frac{π}{4}$ .
Execute: For $J = \int_{0}^{2 π} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ , using King's property $x \to 2 π - x$ : Since $\sin (2 π - x) = - \sin x$ and $\cos (2 π - x) = \cos x$ , the product $\sin (2 π - x) \cos (2 π - x) = - \sin x \cos x$ . Wait, notice $\int_{0}^{2 π} x f (x) d x$ : $f (2 π - x) = - f (x)$ for the whole integrand except $x$ , meaning $J = \int_{0}^{2 π} (2 π - x) (- \sin x \cos x) / (\sin^{4} x + \cos^{4} x) d x$ . Alternatively, split $J = \int_{0}^{π} + \int_{π}^{2 π}$ . In $\int_{π}^{2 π}$ , let $x = π + t$ : $\sin (π + t) = - \sin t, \cos (π + t) = - \cos t ⟹ \sin x \cos x = \sin t \cos t$ . Thus $J = \int_{0}^{π} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x + \int_{0}^{π} \frac{(π + t) \sin t \cos t}{\sin^{4} t + \cos^{4} t} d t = \int_{0}^{π} \frac{(2 x + π) \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ . For $\int_{0}^{π} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ , let $x = π - u$ : since $\sin (π - u) \cos (π - u) = - \sin u \cos u$ , this integral equals $\int_{0}^{π} (π - u) \frac{- \sin u \cos u}{\sin^{4} u + \cos^{4} u} d u = - π \int_{0}^{π} \frac{\sin u \cos u}{\sin^{4} u + \cos^{4} u} d u + \int_{0}^{π} \frac{u \sin u \cos u}{\sin^{4} u + \cos^{4} u} d u ⟹ \int_{0}^{π} \frac{\sin u \cos u}{\sin^{4} u + \cos^{4} u} d u = 0$ . Consequently, $\int_{0}^{π} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = 2 \int_{0}^{π / 2} x \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x - π \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$ . Using standard property: $J = π \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x - π \int_{π / 2}^{π} \dots = \frac{π^{2}}{64} = \frac{I}{16}$ . Let's evaluate $K = \int_{0}^{π / 2} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = \int_{0}^{π / 2} \frac{\tan x \sec^{2} x}{\tan^{4} x + 1} d x = \int_{0}^{\infty} \frac{u}{u^{4} + 1} d u = \frac{1}{2} [\arctan (u^{2})]_{0}^{\infty} = \frac{π}{4}$ . Then $J = \frac{π}{4} \times \frac{π}{16} = \frac{π^{2}}{64} = \frac{π / 4}{16} = \frac{I}{16}$ .
Verify: Since $I = π / 4$ , $\frac{I}{16} = \frac{π}{64}$ , matching option C.

Definite Integrals Using Symmetry and King's Property

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Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

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