求1/{(1+5x^2)(1+x^2)^1/2}的定积分,x的范围是（0,1/3^1/2）,

题目

求1/{(1+5x^2)(1+x^2)^1/2}的定积分,x的范围是（0,1/3^1/2）,

答案

原式=∫(0,1/√3)dx/[(1+5x²)√(1+x²)] (∫(0,1/√3)表示从0到1/√3积分)
=∫(0,π/6)secαdα/(1+5tan²α) (做变换x=tanα)
=∫(0,π/6)cosαdα/(1+4sin²α)
=∫(0,π/6)d(sinα)/(1+4sin²α)
=1/2∫(0,1)dx/(1+x²) (做变换2sinα=x)
=1/2[arctanx]|(0,1)
=(1/2)(π/4)
=π/8.