int_0^1(sqrt(1-x)/(sqrt(1+x)))dx=?

প্রশ্ন: ∫₀¹√(1-x)/√(1+x) dx = ?

সমাধান:

ধরি, x = cos(2θ) 🤓

dx = -2sin(2θ) dθ

যখন x = 0, cos(2θ) = 0 => 2θ = π/2 => θ = π/4

যখন x = 1, cos(2θ) = 1 => 2θ = 0 => θ = 0

সুতরাং, ∫₀¹√(1-x)/√(1+x) dx = ∫_π/4⁰√(1-cos(2θ))/√(1+cos(2θ)) (-2sin(2θ)) dθ 😊

= ∫₀^π/4√(2sin²(θ))/√(2cos²(θ)) (2sin(2θ)) dθ 😎

= ∫₀^π/4 sin(θ)/cos(θ) * 2sin(θ)cos(θ) dθ 🤩

= ∫₀^π/4 2sin²(θ) dθ 😮

= ∫₀^π/4 (1 - cos(2θ)) dθ 😲

= [θ - (sin(2θ)/2)]₀^π/4 😵

= (π/4 - sin(π/2)/2) - (0 - sin(0)/2) 🤯

= π/4 - 1/2 😇

= (π - 2)/4 😲

এখন, প্রদত্ত উত্তর π/2 - 1 = (π-2)/2 🧐

∫₀^π/4 2sin²(θ) dθ = 2∫₀^π/4 sin²(θ) dθ = 2∫₀^π/4 (1-cos2θ)/2 dθ = ∫₀^π/4 (1-cos2θ) dθ = [θ - (sin2θ)/2]^π/4₀ = (π/4 - 1/2) - (0 - 0) = π/4 - 1/2 = (π-2)/4

আবার, ∫₀¹√(1-x)/√(1+x) dx = ∫₀¹(√(1-x)√(1-x))/(√(1+x)√(1-x)) dx = ∫₀¹(1-x)/√(1-x²) dx = ∫₀¹1/√(1-x²) dx - ∫₀¹x/√(1-x²) dx = [sin^-1x]¹₀ + 1/2∫₀¹1/√(1-x²) d(1-x²) = π/2 + [√(1-x²)]¹₀ = π/2 - 1 😊

সুতরাং, ∫₀¹√(1-x)/√(1+x) dx = π/2 - 1