If ∫(cosθ/(5+7sinθ-2cos²θ))dθ = A logₑ|B(θ)| + C, where C is a constant of integration, then B(θ)/A can be:
If ∫(cosθ/(5+7sinθ-2cos²θ))dθ = A logₑ|B(θ)| + C, where C is a constant of integration, then B(θ)/A can be:
Option 1 -
(5(sinθ+3))/(2sinθ+1)
Option 2 -
(5(2sinθ+1))/(sinθ+3)
Option 3 -
(2sinθ+1)/(5(sinθ+3))
Option 4 -
(sinθ+3)/(5(2sinθ+1))
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1 Answer
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Correct Option - 2
Detailed Solution:I = ∫ (cosθ / (2sin²θ + 7sinθ + 3) dθ
sinθ = t ⇒ cosθdθ = dt
= (1/2) ∫ (1 / (t² + (7/2)t + 3/2) dt
= (1/2) ∫ (1 / (t+7/4)² - (5/4)²) dt
= (1/5) ln | (2t+1)/ (t+3)| + c
= (1/5) ln | (2sinθ+1)/ (sinθ+3)| + C
So A = 1/5
B (θ) = 5 (2sinθ+1)/ (sinθ+3)
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