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JEE Main 202324 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2023 · 24 January · Morning Shift · Mathematics · Definite Integration

The value of (8)/(π) ∫ _ 0 ^ (π)/(2) ( x ) ^ 2023 ( x) ^ 2023 + (...

The value of \dfrac{8}{\pi}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{{\left(\text{ \cos x }\right)}^{2023}}{{\left(\sin x\right)}^{2023} + {\left(\cos x\right)}^{2023}}dx is ______.

Correct answer

0

Step-by-step solution

Let

I = \dfrac{8}{\pi}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{{\left(\cos x\right)}^{2023}}{{\left(\sin x\right)}^{2023} + {\left(\cos x\right)}^{2023}}dx . . . \left(1\right)</p><p>Using</p><p>Using{∫}_{0}^{a}f ( x ) dx = {∫}_{0}^{a}f ( a - x ) dx</p><p></p><p>I = \dfrac{8}{\pi}{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{{\left(\sin x\right)}^{2023}}{{\left(\sin x\right)}^{2023} + {\left(\cos x\right)}^{2023}}dx . . . \left(2\right)

Adding \left(1\right)and(2)and \left(2\right), we get

2I=8π∫0π21dx2I = \dfrac{8}{\pi}{∫}_{0}^{\dfrac{\pi}{2}}1dx

⇒2I=8π[x]0π2⇒ 2I = \dfrac{8}{\pi}{\left[x\right]}_{0}^{\dfrac{\pi}{2}}

⇒I=2⇒ I = 2

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