8/3int_(pi/2)^0(3cosθ)^2\ dθ 

প্রশ্ন: ∫_π/2⁰ (8/3) (3cosθ)² dθ

সমাধান:

প্রথমে, ইন্টিগ্র‍্যালটিকে সরল করা যাক:

∫_π/2⁰ (8/3) (9cos²θ) dθ = ∫_π/2⁰ 24cos²θ dθ

আমরা জানি, cos²θ = (1 + cos2θ)/2

সুতরাং, ∫_π/2⁰ 24((1 + cos2θ)/2) dθ = ∫_π/2⁰ 12(1 + cos2θ) dθ

= 12 ∫_π/2⁰ (1 + cos2θ) dθ

= 12 [θ + (sin2θ)/2] _π/2⁰

= 12 [(0 + (sin0)/2) - (π/2 + (sinπ)/2)]

= 12 [0 - (π/2 + 0)]

= 12 (-π/2)

= -6π

অতএব, ∫_π/2⁰ (8/3) (3cosθ)² dθ = -6π

উত্তর: -6π

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