求∫dx/3+(sinx)^2设u=tan(x/2),sinx=2u/1+u^2,dx=2du/1+u^2∫dx/3+(

gaojunqing2022-10-04 11:39:540条回答

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
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