Evaluate the following integral ∫_1^2√((x-1)/(2-x)) dx
∫_1^2√((x-1)/(2-x)) dxLet: 2-x=u^2→-dx=2udu ,x=1→u=1,x=2→u=0,x-1=1-u^2∴I=∫_0^1√(1-u^2 )/u.(-2udu)=2∫_0^1√(1-u^2 ) duLet:u=sinθ→du=cosθ dθ,u=0→θ=0,u=1→θ=π/2∴I=∫_0^(π⁄2)〖〖cos〗^2 θ〗 dθ=∫_0^(π⁄2)(1+cos2θ ) dθ=[θ+1/2 sin2θ ]_0^(π⁄2)=π/2
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∫_1^2√((x-1)/(2-x)) dx
need an explanation for this answer? contact us directly to get an explanation for this answerLet:
2-x=u^2→-dx=2udu ,x=1→u=1,x=2→u=0,x-1=1-u^2
∴I=∫_0^1√(1-u^2 )/u.(-2udu)=2∫_0^1√(1-u^2 ) du
Let:
u=sinθ→du=cosθ dθ,u=0→θ=0,u=1→θ=π/2
∴I=∫_0^(π⁄2)〖〖cos〗^2 θ〗 dθ=∫_0^(π⁄2)(1+cos2θ ) dθ=[θ+1/2 sin2θ ]_0^(π⁄2)=π/2