2加sinx的平方分之1的微积分怎么求

Question

2加sinx的平方分之1的微积分怎么求

Answer 1

f1/(2+sin^2x)dx
=f1/(2+1/csc^2x)dx
=f 1/(2+1/(1+cot^2x))dx
=f (1+cot^2x)/(2+2cot^2x+1)dx
=f(1+cot^2x)/(3+2cot^2x)dx
=fcsc^2x/(3+2cot^2x)dx
=-f1/(3+2cot^2x)d(cotx)
设cotx=t
原式=-f1/(3+2t^2)dt=-1/3f1/(1+2t^2/3)dt
2t^2/3=tan^2u
(2/3)^(1/2) *t=tanu u=arctan[(根6t)/3]=arc
dt=(根6 sec^2u)/2du
原式=-1/3f1/sec^2u * (根6 sec^2u)/2du
=-1/3f(根6/2)du
=-根6/6u+C
=-根6/6arctan[(根6t)/3]+C
=-根6/6arctan[(根6cotx)/3]+C