Question:

If \( m, l, r, s, n \) are integers such that \( 9 >m >l >s >n >r >2 \) and \[ \int_{0}^{\frac{\pi}{2}} \sin^n x \cos^r x \, dx = 4 \int_{0}^{\frac{\pi}{2}} \sin^m x \cos^r x \, dx, \] \[ \int_{0}^{\frac{\pi}{2}} \sin^l x \cos^r x \, dx = 4 \int_{0}^{\frac{\pi}{2}} \sin^s x \cos^r x \, dx, \] \[ \int_{0}^{\frac{\pi}{2}} \sin^n x \cos^r x \, dx = 0, \] then the equation involving \( s, l, m, r \) is:

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For integrals involving trigonometric powers, use reduction formulas to establish relationships between different exponents.
Updated On: Mar 15, 2025
  • \( (s-2)(l-2) = mr \)
  • \( (s-2)(l+2) = rm + 5 \)
  • \( (s-2)(s+2) = ln - 3 \)
  • \( (l-2)(l+2) = ms - 5 \) 
     

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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Given Integrals 
We analyze the given integral equations and their recurrence relations: \[ \int_{0}^{\frac{\pi}{2}} \sin^n x \cos^r x \, dx = 4 \int_{0}^{\frac{\pi}{2}} \sin^m x \cos^r x \, dx. \] Using reduction formulas for trigonometric integrals: \[ I(n, r) = \frac{n-1}{r+1} I(n-2, r). \] Similarly, \[ I(l, r) = \frac{l-1}{r+1} I(l-2, r). \] This provides relationships between the powers of sine in the integrals. 

Step 2: Deriving the Relationship 
Given: \[ I(n, r) = 4 I(m, r), \quad I(l, r) = 4 I(s, r), \quad I(n, r) = 0. \] Using recurrence properties, \[ (n-1)(s+1) = 4(m-1)(r+1), \] \[ (l-1)(s+1) = 4(s-1)(r+1). \] Rearranging, \[ (s-2)(s+2) = ln - 3. \] 

Step 3: Conclusion 
Thus, the correct answer is: \[ \mathbf{(s-2)(s+2) = ln - 3}. \] 

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