JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectNumericalPrevious-year question

Definite Integrals Using King's Property

Q: Does the power 2023 affect the final value of the integral?

No. For any power n, integral from 0 to π/2 of (cos^n x)/(sin^n x + cos^n x) dx = π/4 because the transformation x → π/2 - x interchanges sine and cosine regardless of n.

The value of 8/π integral from 0 to π/2 of ((cos x)^2023)/((sin x)^2023 + (cos x)^2023) dx is

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

Step-by-step solutionView

Correct answer

The integral evaluates to $\frac{π}{4}$ , so multiplying by $\frac{8}{π}$ gives an answer of $2$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Let the given expression be $J = \frac{8}{π} I$ , where $I = \int_{0}^{\frac{π}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} d x$ .

02Goal

Evaluate $J = \frac{8}{π} I$ .

03Approach

Apply King's property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ with $a = 0$ and $b = \frac{π}{2}$ , then add the two representations of $I$ .

04Execute

Using $x \to \frac{π}{2} - x$ , we get $\cos (\frac{π}{2} - x) = \sin x$ and $\sin (\frac{π}{2} - x) = \cos x$ . Thus: $I = \int_{0}^{\frac{π}{2}} \frac{(\sin x)^{2023}}{(\cos x)^{2023} + (\sin x)^{2023}} d x$ Adding the two forms of $I$ : $2 I = \int_{0}^{\frac{π}{2}} \frac{(\cos x)^{2023} + (\sin x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} d x = \int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}$ Therefore, $I = \frac{π}{4}$ .

05Execute

Now multiply by the prefactor $\frac{8}{π}$ : $J = \frac{8}{π} I = \frac{8}{π} \times \frac{π}{4} = 2$

✓Verify

By symmetry of $f (x) + f (π / 2 - x) = 1$ over $[0, π / 2]$ , the average value of the integrand is $\frac{1}{2}$ . The integral is $\frac{1}{2} \times \frac{π}{2} = \frac{π}{4}$ . Multiplying by $\frac{8}{π}$ gives $2$ .

Hints that build this answer step by step

Let $I = \int_{0}^{\frac{π}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} d x$ . What equivalent integral is obtained by applying the property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ ?
$I = \int_{0}^{\frac{π}{2}} \frac{(\sin x)^{2023}}{(\cos x)^{2023} + (\sin x)^{2023}} d x$
Adding the original expression for $I$ to the transformed expression yields $2 I$ . What does the integrand simplify to?
$2 I = \int_{0}^{\frac{π}{2}} 1 d x$
What is the final value of the expression $\frac{8}{π} I$ ?
$2$

✓ 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: 9 September 2026

Quick checks

Students also ask

Does the power 2023 affect the final value of the integral?

No. For any power $n$ , $\int_{0}^{π / 2} \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x} d x = \frac{π}{4}$ because the transformation $x \to \frac{π}{2} - x$ interchanges sine and cosine regardless of $n$ .

Answer

The integral evaluates to $\frac{π}{4}$ , so multiplying by $\frac{8}{π}$ gives an answer of $2$ .

Why each option works or fails

Step-by-step solution

Given: Let the given expression be $J = \frac{8}{π} I$ , where $I = \int_{0}^{\frac{π}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} d x$ .
Goal: Evaluate $J = \frac{8}{π} I$ .
Approach: Apply King's property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ with $a = 0$ and $b = \frac{π}{2}$ , then add the two representations of $I$ .
Execute: Using $x \to \frac{π}{2} - x$ , we get $\cos (\frac{π}{2} - x) = \sin x$ and $\sin (\frac{π}{2} - x) = \cos x$ . Thus: $I = \int_{0}^{\frac{π}{2}} \frac{(\sin x)^{2023}}{(\cos x)^{2023} + (\sin x)^{2023}} d x$ Adding the two forms of $I$ : $2 I = \int_{0}^{\frac{π}{2}} \frac{(\cos x)^{2023} + (\sin x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} d x = \int_{0}^{\frac{π}{2}} 1 d x = \frac{π}{2}$ Therefore, $I = \frac{π}{4}$ .
Execute: Now multiply by the prefactor $\frac{8}{π}$ : $J = \frac{8}{π} I = \frac{8}{π} \times \frac{π}{4} = 2$
Verify: By symmetry of $f (x) + f (π / 2 - x) = 1$ over $[0, π / 2]$ , the average value of the integrand is $\frac{1}{2}$ . The integral is $\frac{1}{2} \times \frac{π}{2} = \frac{π}{4}$ . Multiplying by $\frac{8}{π}$ gives $2$ .

Definite Integrals Using King's Property

What feels right?

The integral evaluates to $\frac{π}{4}$ , so multiplying by $\frac{8}{π}$ gives an answer of $2$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

We think you should solve this next.

Let $A$ be the area bounded by the curve $y = x ∣ x - 3 ∣$ , the x-axis and the ordinates…

The same chapter, from another paper — Integral Calculus, 2023

Let a tangent to the curve $y^{2} = 24 x$ meet the curve $x y = 2$ at the points $A$ a…

Students also ask

Definite Integrals Using King's Property

The integral evaluates to π44π​, so multiplying by 8ππ8​ gives an answer of 22.

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

Let AA be the area bounded by the curve y=x∣x−3∣y=x∣x−3∣, the x-axis and the ordinates…

The same chapter, from another paper — Integral Calculus, 2023

Let a tangent to the curve y2=24xy2=24x meet the curve xy=2xy=2 at the points AA a…

Students also ask

The integral evaluates to $\frac{π}{4}$ , so multiplying by $\frac{8}{π}$ gives an answer of $2$ .

Let $A$ be the area bounded by the curve $y = x ∣ x - 3 ∣$ , the x-axis and the ordinates…

Let a tangent to the curve $y^{2} = 24 x$ meet the curve $x y = 2$ at the points $A$ a…