JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The value of $\frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx$ is

Choose the correct answer:

Explanation

Step 1: Integral ko $I$ maanein

I = \int_{0}^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx \quad \text{---(Equation 1)}

Step 2: Property use karein

Property: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$

Yahan $a+b-x = \frac{\pi}{2} - x$ .

Hume pata hai ki $\cos(\frac{\pi}{2} - x) = \sin x$ aur $\sin(\frac{\pi}{2} - x) = \cos x$ .

Naya integral banega:

I = \int_{0}^{\frac{\pi}{2}} \frac{(\sin x)^{2023}}{(\cos x)^{2023} + (\sin x)^{2023}} dx \quad \text{---(Equation 2)}

Step 3: Equation 1 aur 2 ko jodein

2I = \int_{0}^{\frac{\pi}{2}} \frac{(\cos x)^{2023} + (\sin x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx

2I = \int_{0}^{\frac{\pi}{2}} 1 \cdot dx

2I = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}

I = \frac{\pi}{4}

Step 4: Final value nikalna

Hume question mein $\frac{8}{\pi} \times I$ ki value puchi gayi hai:

\text{Value} = \frac{8}{\pi} \times \frac{\pi}{4}

\text{Value} = \frac{8}{4} = 2

Final Answer:

Iska sahi jawab 2 hai.