计算定积分∫[-2,2](2x-3)/((8-x^2))^1/2dx 求大神帮下忙
问题描述：
计算定积分∫[-2,2](2x-3)/((8-x^2))^1/2dx 求大神帮下忙
答
先求不定积分∫(2x-3)/√(8-x²)dx
=∫1/√(8-x²)dx²-3∫1/√(8-x²)dx
=-∫1/√(8-x²)d(8-x²)-(3/2√2)∫1/√(1-(x/2√2)²)dx
=-2√(8-x²)-3∫1/√(1-(x/2√2)²)d(x/2√2)
=-2√(8-x²)-3Arcsin(x/2√2)+C
-2到2积分
∴该定积分=-2√2-3π/4-(-2√2+3π/4)
=-3π/2