Question

计算定积分∫(-1,1) [(2+(x^2)*(sin^2011)*x)/√(4-x^2)dx]？

Answer 1

∫(-1,1) [(2+(x^2)*(sin^2011)*x)/√(4-x^2)dx]
=∫(-1,1) 2/√(4-x^2)dx+∫(-1,1)(x^2)*(sin^2011)*x)/√(4-x^2)dx]
=∫(-1,1) 2/√(4-x^2)dx
因为 (x^2)*(sinx)^2011/√(4-x^2) 是奇函数,而奇函数在对称区间是的定积分等于0,
令x=2sint,dx=2costdt,
=∫ (-1,1)2/√(4-x^2)dx
=2∫ (0,1)2/√(4-x^2)dx
=∫(0,1) 8cost/√(4-4(sint)^2)dt
=∫(0,1)4cost/costdt
=(0,1)4t=4
所以原积分=4.,1,∫(-1,1) [(2+(x^2)*(sin^2011)*x)/√(4-x^2)dx]
=∫(-1,1) /√(4-x^2)dx+∫(-1,1)(x^2)*(sin^2011)*x)/√(4-x^2)dx
因为(x^2)*(sin^2011)*x)/√(4-x^2)是(-1,1)上的奇函数
所以∫(-1,1)(x^2)*(sin^2011)*x)/√(4-x^2)dx=0,0,