求下面定积分我知道三角代换,但做不出结果：根号下4-x^2,根号下x^2+4,根号下x^2-4

Question

求下面定积分
我知道三角代换,但做不出结果：根号下4-x^2,根号下x^2+4,根号下x^2-4

Answer 1

∫√(4-x^2)dx=∫(2cosu)^2 du=∫(1+cos2u)d2u=2u+sin2u+C=arcsin(x/2)+(x/2)(√(4-x^2)+C
x=2sinu dx=2cosudu
∫√(x^2+4)dx=4∫secu^3du=2secutanu+2ln|secu+tanu|+C
=√(x^2+4)*(x/2)+2ln|x/2+√(x^2/4+1)|+C
x=2tanu dx=2secu^2du
∫secu^3du=∫secudtanu=secutanu-∫tanu^2secudu=secutanu-∫secu^3du+∫secudu
2∫secu^3du=secutanu+ln|secu+tanu|+C
∫√(x^2-4)dx=4∫tanudsecu=4secutanu-4∫secu^3du=2secutanu-2ln|secu+tanu|+C
=x√(x^2-4)-2ln|x/2+√(x^2/4-1)|+C
x=2secu