Question

求∫dx/3+(sinx)^2

设u=tan(x/2),sinx=2u/1+u^2,dx=2du/1+u^2
∫dx/3+(sinx)^2=2∫(1+u^2)du/3u^4+10u^2+3=1/2∫du/(3u^2+1)+1/2∫du/(u^2+3)=1/2ln[3tan(x/2)^2+1]+1/2ln[tan(x/2)^2+1]+C
这道题解的有错吗？

Answer 1

最后一步跌倒
u=tan(x/2),sinx=2u/1+u^2,du=[1+[tan(x/2)]^2]/2dx=((1+u^2)/2)dx
dx=2du/(1+u^2)
3+(sinx)^2=(3+3u^4+6u^2+4u^2)/(1+u^2)^2
=(3u^4+10u^2+3)/(1+u^2)^2
∫dx/[3+(sinx)^2]=2∫(1+u^2)du/(3u^4+10u^2+3)
=2∫(1+u^2)du/[(3u^2+1)(u^2+3)]
=(1/2)∫[(3u^2+1)+(u^2+3)]du/[(3u^2+1)(u^2+3)]
=(1/2)∫du/(3u^2+1)+(1/2)∫du/(u^2+3)
=(1/(2√3)∫d(√3u)/[(√3u)^2+1] +(√3/6)∫d(u/√3)/[(u/√3)^2+1]
=(1/(2√3))arctan(√3u+1)+(√3/6)arctan(u/√3)
=(1/(2√3))arctan[√3tan(x/2)+1]+(√3/6)arctan[tan(x/2)/√3]+C

追问

感谢，我发现错了，但答案是1/(2√3)arctan(2tanx)/√3+C,我知道是另一种方法做的，请问你能帮我证明，这两种方法结果都是对的吗

追答

刚才我也出现失误
我的答案应为=(1/(2√3))arctan[√3tan(x/2)]+(1/(2√3))arctan[tan(x/2)/√3]+C
tana=√3t
tanb=t/√3
tan(a+b)=4t/(√3-√3t)=(2/√3)2t/(1-t^2)
t=tan(x/2)
两个答案一致

追问

哇，你太有才了

Answer 2

your answer here