Question

求解高数题？

画红线的两道题

Answer 1

利用不定积分的第二换元积分法的三角代换来求解。

Answer 2

(7)
let
x=2sinu
dx=2cosu du
∫ x^2/√(4-x^2) dx
=-∫ x d√(4-x^2)
=-x.√(4-x^2) +∫ √(4-x^2) dx
=-x.√(4-x^2) +4∫ (cosu)^2 du
=-x.√(4-x^2) +2∫ (1+cos2u) du
=-x.√(4-x^2) +2[u+(1/2)sin2u] +C
=-x.√(4-x^2) +2[arcsin(x/2) + (1/4)x.√(4-x^2) ] +C
= -(1/2)x.√(4-x^2) +2arcsin(x/2) +C
(8)
let
x=sinu
dx = cosu du
∫ dx/[ 1+√(1-x^2)]
=∫ [ 1-√(1-x^2)] /x^2 dx
=∫ [(1-cosu) /(sinu)^2 ] [ cosu du]
=∫ [(cosu-(cosu)^2 ] /(sinu)^2 du
=∫ [(cosu-1 + (sinu)^2 ] /(sinu)^2 du
=∫ [cosu/(sinu)^2 - (cscu)^2 + 1 ] du
=-1/sinu - cotu +u + C
=-1/x - √(1-x^2)/x + arcsinx + C