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Integral Calculus: JEE Main Mathematics Question with Solution

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

No. For any power , because the transformation interchanges sine and cosine regardless of .

The value of

\frac{8}{\pi} \int_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx

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Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Let the given expression be $J = \frac{8}{\pi} I$ , where $I = \int_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx$ .

02goal

Evaluate $J = \frac{8}{\pi} I$ .

03approach

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

04execute

Using $x \to \frac{\pi}{2} - x$ , we get $\cos(\frac{\pi}{2} - x) = \sin x$ and $\sin(\frac{\pi}{2} - x) = \cos x$ . Thus: $I = \int_0^{\frac{\pi}{2}} \frac{(\sin x)^{2023}}{(\cos x)^{2023} + (\sin x)^{2023}} dx$ Adding the two forms of $I$ : $2I = \int_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023} + (\sin x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx = \int_0^{\frac{\pi}{2}} 1 dx = \frac{\pi}{2}$ Therefore, $I = \frac{\pi}{4}$ .

05execute

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

✓verify

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

Hints that build this answer step by step

Let $I = \int_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx$ . What equivalent integral is obtained by applying the property $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ ?
$I = \int_0^{\frac{\pi}{2}} \frac{(\sin x)^{2023}}{(\cos x)^{2023} + (\sin x)^{2023}} dx$
Adding the original expression for $I$ to the transformed expression yields $2I$ . What does the integrand simplify to?
$2I = \int_0^{\frac{\pi}{2}} 1 \, dx$
What is the final value of the expression $\frac{8}{\pi} I$ ?
$2$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $n$ , $\int_0^{\pi/2} \frac{\cos^n x}{\sin^n x + \cos^n x} dx = \frac{\pi}{4}$ because the transformation $x \to \frac{\pi}{2} - x$ interchanges sine and cosine regardless of $n$ .

Answer

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

Why each option works or fails

Step-by-step solution

given: Let the given expression be $J = \frac{8}{\pi} I$ , where $I = \int_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx$ .
goal: Evaluate $J = \frac{8}{\pi} I$ .
approach: Apply King's property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ with $a = 0$ and $b = \frac{\pi}{2}$ , then add the two representations of $I$ .
execute: Using $x \to \frac{\pi}{2} - x$ , we get $\cos(\frac{\pi}{2} - x) = \sin x$ and $\sin(\frac{\pi}{2} - x) = \cos x$ . Thus: $I = \int_0^{\frac{\pi}{2}} \frac{(\sin x)^{2023}}{(\cos x)^{2023} + (\sin x)^{2023}} dx$ Adding the two forms of $I$ : $2I = \int_0^{\frac{\pi}{2}} \frac{(\cos x)^{2023} + (\sin x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx = \int_0^{\frac{\pi}{2}} 1 dx = \frac{\pi}{2}$ Therefore, $I = \frac{\pi}{4}$ .
execute: Now multiply by the prefactor $\frac{8}{\pi}$ : $J = \frac{8}{\pi} I = \frac{8}{\pi} \times \frac{\pi}{4} = 2$
verify: By symmetry of $f(x) + f(\pi/2 - x) = 1$ over $[0, \pi/2]$ , the average value of the integrand is $\frac{1}{2}$ . The integral is $\frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}$ . Multiplying by $\frac{8}{\pi}$ gives $2$ .

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

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

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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