∫1,-1(sinx+2)根号1-x^2dx
问题描述：
∫1,-1(sinx+2)根号1-x^2dx
答
答：
(-1→1) ∫ (sinx+2)√(1-x^2) dx
=(-1→1) ∫ sinx√(1-x^2)dx+ (-1→1) ∫ 2√(1-x^2) dx
=0+(-1→1) 2∫ √(1-x^2) dx,设x=sint
=(-π/2→π/2) 2∫ cost d(sint)
=(-π/2→π/2) ∫ (cos2t+1) dt
=(-π/2→π/2) [(1/2)sin2t+t]
=(0+π/2)-(0-π/2)
=π