Question Number 149809 by 0731619 last updated on 07/Aug/21

Answered by Ar Brandon last updated on 07/Aug/21

I=∫_0 ^1 ∫_0 ^(√(1−y^2 )) cos(x^2 +y^2 )dxdy { ((x=rcosθ),(0≤θ≤π)),((y=rsinθ),(0≤r≤(√(1−sin^2 θ))=cosθ)) :} I=∫_0 ^π ∫_0 ^(cosθ) rcos(r^2 )drdθ=(1/2)∫_0 ^π ∫_0 ^(cosθ) cos(r^2 )d(r^2 )dθ =(1/2)∫_0 ^π [sin(r^2 )]_0 ^(cosθ) dθ=(1/2)∫_0 ^π sin(cos^2 θ)dθ

Answered by Kamel last updated on 08/Aug/21

∫_0 ^(π/2) ∫_0 ^1 rsin(r^2 )drdθ=(π/4)sin(1)