Question

我大一，明天高数考试，今晚在复习，有一题做不来，求解··········大家都要期末了，帮帮忙

Answer 1

I=-1/2∫ (0,1) x√(1-x²) d(1-x² )
=-1/3∫ (0,1) x d(1-x² )^(3/2)
=-1/3[x(1-x² )^(3/2)|_0^1-∫ (0,1) (1-x²)^(3/2)dx]
=1/3∫ (0,1) (1-x²)^(3/2)dx(再令x=sint)
=1/3∫ (0,π/2) cos^4tdt=
=1/3∫ (0,π/2) (1+cos2t)^2/4dt
=1/12∫ (0,π/2) (1+2cos2t+cos^22t)dt
=1/12∫ (0,π/2) [1+2cos2t+(1+cos4t)/2]dt
=1/12[π/2+π/4]
=π/16

Answer 2

三角换元
I=∫ (0,1) x²√(1-x²) dx x=sina
=∫ (0,π/2) sin²a*cosa d(sina)
=∫ (0,π/2) sin²a*cos²a da
=∫ (0,π/2) 1/4*sin²2a da
=1/8 ∫ (0,π/2) (1-cos4a) da
=1/8 (a+1/4sin4a) |(0,π/2)
=π/16

Answer 3

设√（1-x²）=t，则dx=-t/√（1-t²）在自己算吧 电脑输入有点烦

Answer 4

亲，换元法，x=cost，即可

Answer 5

你试试用t=根号1-x^2代

Answer 6

大哥您肿么学的？我是高二的看一眼就会…