letx^2=2sinu2xdx = 2cosu duxdx = cosu du∫ x/√(4-x^2) dx=∫ cosu du/ (2cosu)=(1/2)∫ du=(1/2)u + C=(1/2)arcsin(x^2/2) + C
1.先求f(x); 2.再求其导函数; 3.代入求不定积分; 4.具体步骤如下:
