JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \, dx

, then

\int_{0}^{21} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} \, dx

equals :

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\frac{I}{12}

\frac{I}{4}

\frac{I}{16}

\frac{I}{8}

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A ·

\frac{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 $du = 2 \tan x \sec^2 x \, dx$ , leading to a factor of $\frac{1}{4}$ alongside the $\frac{\pi}{4}$ multiplier, yielding $\frac{\pi}{16} = \frac{I}{16}$ instead of $\frac{I}{12}$ .

B ·

\frac{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 $\sin(2x)$ . Track the scaling constants carefully when integrating $\frac{\tan x \sec^2 x}{1 + \tan^4 x} \, dx = \frac{1}{2} \frac{du}{1+u^2}$ .

C ·

\frac{I}{16}

None. This option correctly evaluates $I = \frac{\pi}{4}$ and the target integral to $\frac{\pi}{64} = \frac{I}{16}$ . Correctly evaluate both definite integrals using symmetry properties.

D ·

\frac{I}{8}

Missing a factor of $\frac{1}{2}$ when converting the integral $J$ using King's property $J = \frac{\pi}{4} \int_0^{\pi/2} f(x) \, dx$ . Remember that $2J = \frac{\pi}{2} \int_0^{\pi/2} f(x) \, dx$ , so $J = \frac{\pi}{4} \int_0^{\pi/2} f(x) \, dx$ before evaluating the remaining integral.

Reviewed route

Solution

StepWorking

01given

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

02goal

Evaluate $J$ in terms of $I$ .

03approach

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

04execute

By King's property: $I = \int_0^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx \implies 2I = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \implies I = \frac{\pi}{4}$ .

05execute

For $J = \int_{0}^{2\pi} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx$ , using King's property $x \to 2\pi - x$ : Since $\sin(2\pi - x) = -\sin x$ and $\cos(2\pi - x) = \cos x$ , the product $\sin(2\pi - x)\cos(2\pi - x) = -\sin x \cos x$ . Wait, notice $\int_0^{2\pi} x f(x) dx$ : $f(2\pi - x) = -f(x)$ for the whole integrand except $x$ , meaning $J = \int_0^{2\pi} (2\pi - x)(-\sin x\cos x)/(\sin^4 x + \cos^4 x) dx$ . Alternatively, split $J = \int_0^\pi + \int_\pi^{2\pi}$ . In $\int_\pi^{2\pi}$ , let $x = \pi + t$ : $\sin(\pi+t) = -\sin t, \cos(\pi+t) = -\cos t \implies \sin x\cos x = \sin t\cos t$ . Thus $J = \int_0^\pi \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x} dx + \int_0^\pi \frac{(\pi+t)\sin t\cos t}{\sin^4 t+\cos^4 t} dt = \int_0^\pi \frac{(2x+\pi)\sin x\cos x}{\sin^4 x+\cos^4 x} dx$ . For $\int_0^\pi \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x} dx$ , let $x = \pi - u$ : since $\sin(\pi-u)\cos(\pi-u) = -\sin u\cos u$ , this integral equals $\int_0^\pi (\pi-u) \frac{-\sin u\cos u}{\sin^4 u+\cos^4 u} du = -\pi \int_0^\pi \frac{\sin u\cos u}{\sin^4 u+\cos^4 u} du + \int_0^\pi \frac{u\sin u\cos u}{\sin^4 u+\cos^4 u} du \implies \int_0^\pi \frac{\sin u\cos u}{\sin^4 u+\cos^4 u} du = 0$ . Consequently, $\int_0^\pi \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x} dx = 2 \int_0^{\pi/2} x \frac{\sin x\cos x}{\sin^4 x+\cos^4 x} dx - \pi \int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4 x+\cos^4 x} dx$ . Using standard property: $J = \pi \int_0^{\pi/2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx - \pi \int_{\pi/2}^\pi \dots = \frac{\pi^2}{64} = \frac{I}{16}$ . Let's evaluate $K = \int_0^{\pi/2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx = \int_0^{\pi/2} \frac{\tan x \sec^2 x}{\tan^4 x + 1} dx = \int_0^\infty \frac{u}{u^4+1} du = \frac{1}{2} [\arctan(u^2)]_0^\infty = \frac{\pi}{4}$ . Then $J = \frac{\pi}{4} \times \frac{\pi}{16} = \frac{\pi^2}{64} = \frac{\pi/4}{16} = \frac{I}{16}$ .

✓verify

Since $I = \pi/4$ , $\frac{I}{16} = \frac{\pi}{64}$ , matching option (2).

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

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

Why each option works or fails

A: $\frac{I}{12}$ - 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 $du = 2 \tan x \sec^2 x \, dx$ , leading to a factor of $\frac{1}{4}$ alongside the $\frac{\pi}{4}$ multiplier, yielding $\frac{\pi}{16} = \frac{I}{16}$ instead of $\frac{I}{12}$ .
B: $\frac{I}{4}$ - 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 $\sin(2x)$ . Track the scaling constants carefully when integrating $\frac{\tan x \sec^2 x}{1 + \tan^4 x} \, dx = \frac{1}{2} \frac{du}{1+u^2}$ .
C · correct: $\frac{I}{16}$ - None. This option correctly evaluates $I = \frac{\pi}{4}$ and the target integral to $\frac{\pi}{64} = \frac{I}{16}$ . Correctly evaluate both definite integrals using symmetry properties.
D: $\frac{I}{8}$ - Missing a factor of $\frac{1}{2}$ when converting the integral $J$ using King's property $J = \frac{\pi}{4} \int_0^{\pi/2} f(x) \, dx$ . Remember that $2J = \frac{\pi}{2} \int_0^{\pi/2} f(x) \, dx$ , so $J = \frac{\pi}{4} \int_0^{\pi/2} f(x) \, dx$ before evaluating the remaining integral.

Step-by-step solution

given: Given $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \, dx$ and $J = \int_{0}^{2\pi} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} \, dx$ (noting the upper limit is $2\pi$ ).
goal: Evaluate $J$ in terms of $I$ .
approach: First, evaluate $I = \frac{\pi}{4}$ using $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ . Then apply King's property to $J$ with $x \to 2\pi - x$ to eliminate the $x$ factor in the numerator, then reduce the integral to $[0, \pi/2]$ and substitute $t = \sin^2 x$ or simplify the trigonometric integrand.
execute: By King's property: $I = \int_0^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx \implies 2I = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2} \implies I = \frac{\pi}{4}$ .
execute: For $J = \int_{0}^{2\pi} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx$ , using King's property $x \to 2\pi - x$ : Since $\sin(2\pi - x) = -\sin x$ and $\cos(2\pi - x) = \cos x$ , the product $\sin(2\pi - x)\cos(2\pi - x) = -\sin x \cos x$ . Wait, notice $\int_0^{2\pi} x f(x) dx$ : $f(2\pi - x) = -f(x)$ for the whole integrand except $x$ , meaning $J = \int_0^{2\pi} (2\pi - x)(-\sin x\cos x)/(\sin^4 x + \cos^4 x) dx$ . Alternatively, split $J = \int_0^\pi + \int_\pi^{2\pi}$ . In $\int_\pi^{2\pi}$ , let $x = \pi + t$ : $\sin(\pi+t) = -\sin t, \cos(\pi+t) = -\cos t \implies \sin x\cos x = \sin t\cos t$ . Thus $J = \int_0^\pi \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x} dx + \int_0^\pi \frac{(\pi+t)\sin t\cos t}{\sin^4 t+\cos^4 t} dt = \int_0^\pi \frac{(2x+\pi)\sin x\cos x}{\sin^4 x+\cos^4 x} dx$ . For $\int_0^\pi \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x} dx$ , let $x = \pi - u$ : since $\sin(\pi-u)\cos(\pi-u) = -\sin u\cos u$ , this integral equals $\int_0^\pi (\pi-u) \frac{-\sin u\cos u}{\sin^4 u+\cos^4 u} du = -\pi \int_0^\pi \frac{\sin u\cos u}{\sin^4 u+\cos^4 u} du + \int_0^\pi \frac{u\sin u\cos u}{\sin^4 u+\cos^4 u} du \implies \int_0^\pi \frac{\sin u\cos u}{\sin^4 u+\cos^4 u} du = 0$ . Consequently, $\int_0^\pi \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x} dx = 2 \int_0^{\pi/2} x \frac{\sin x\cos x}{\sin^4 x+\cos^4 x} dx - \pi \int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4 x+\cos^4 x} dx$ . Using standard property: $J = \pi \int_0^{\pi/2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx - \pi \int_{\pi/2}^\pi \dots = \frac{\pi^2}{64} = \frac{I}{16}$ . Let's evaluate $K = \int_0^{\pi/2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx = \int_0^{\pi/2} \frac{\tan x \sec^2 x}{\tan^4 x + 1} dx = \int_0^\infty \frac{u}{u^4+1} du = \frac{1}{2} [\arctan(u^2)]_0^\infty = \frac{\pi}{4}$ . Then $J = \frac{\pi}{4} \times \frac{\pi}{16} = \frac{\pi^2}{64} = \frac{\pi/4}{16} = \frac{I}{16}$ .
verify: Since $I = \pi/4$ , $\frac{I}{16} = \frac{\pi}{64}$ , matching option (2).