Question

求定积分 见图片第7题 求详解

Answer 1

∫ (-2,2) (4x^2+x^3cosx)/(2+√4-x^2) dx
=∫ (-2,2) 4x^2/(2+√4-x^2) dx+∫ (-2,2)x^3cosx/(2+√4-x^2) dx
因为后面积分的函数x^3cosx/(2+√4-x^2)是奇函数，所以在对称区间[-2,2]中积分为0
=2∫ (0,2) 4x^2/(2+√4-x^2) dx 上下同时乘以(2-√4-x^2)
=2∫ (0,2) 4x^2(2-√4-x^2)/(2+√4-x^2)(2-√4-x^2) dx
=2∫ (0,2) 4x^2(2-√4-x^2)/x^2 dx
=8∫ (0,2) 2-√4-x^2 dx x=2sina 三角换元
=8∫(0,π/2) (2-2cosa) d2sina
=16∫(0,π/2) (2-2cosa)cosa da
=16∫(0,π/2) (2cosa-2cos^2a) da
=16∫(0,π/2) (2cosa-1-cos2a) da
=16(2sina-a-1/2sin2a) |(0,π/2)
=16(2-π/2)