1.∫(sinxcosx)/(1+sin^2x)dx2.设f'(sinx)=cos^2x 则f(x)=3.已知 f'(cosx)=sinx 则f(cosx)=4.∫dx/(e^x-1)^1/2=.....第三题的答案 最后给的是 1/2 (sinxcosx-x)+c
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![1.∫(sinxcosx)/(1+sin^2x)dx2.设f'(sinx)=cos^2x 则f(x)=3.已知 f'(cosx)=sinx 则f(cosx)=4.∫dx/(e^x-1)^1/2=.....第三题的答案 最后给的是 1/2 (sinxcosx-x)+c](/uploads/image/z/4354236-36-6.jpg?t=1.%E2%88%AB%EF%BC%88sinxcosx%EF%BC%89%2F%EF%BC%881%2Bsin%5E2x%EF%BC%89dx2.%E8%AE%BEf%27%EF%BC%88sinx%EF%BC%89%3Dcos%5E2x+%E5%88%99f%28x%29%3D3.%E5%B7%B2%E7%9F%A5+f%27%28cosx%29%3Dsinx+%E5%88%99f%28cosx%29%3D4.%E2%88%ABdx%2F%28e%5Ex-1%29%5E1%2F2%3D.....%E7%AC%AC%E4%B8%89%E9%A2%98%E7%9A%84%E7%AD%94%E6%A1%88+%E6%9C%80%E5%90%8E%E7%BB%99%E7%9A%84%E6%98%AF+1%2F2+%28sinxcosx-x%29%2Bc)
1.∫(sinxcosx)/(1+sin^2x)dx2.设f'(sinx)=cos^2x 则f(x)=3.已知 f'(cosx)=sinx 则f(cosx)=4.∫dx/(e^x-1)^1/2=.....第三题的答案 最后给的是 1/2 (sinxcosx-x)+c
1.∫(sinxcosx)/(1+sin^2x)dx
2.设f'(sinx)=cos^2x 则f(x)=
3.已知 f'(cosx)=sinx 则f(cosx)=
4.∫dx/(e^x-1)^1/2=
.....第三题的答案 最后给的是 1/2 (sinxcosx-x)+c
1.∫(sinxcosx)/(1+sin^2x)dx2.设f'(sinx)=cos^2x 则f(x)=3.已知 f'(cosx)=sinx 则f(cosx)=4.∫dx/(e^x-1)^1/2=.....第三题的答案 最后给的是 1/2 (sinxcosx-x)+c
(1)∫[(sinxcosx)/(1+sin²x)]dx,d(1+sin²x)=(2sinxcosx)dx
=∫[(sinxcosx)/(1+sin²x)*1/(2sinxcosx)]d(1+sin²x)
=(1/2)∫[1/(1+sin²x)]d(1+sin²x)
=(1/2)ln|1+sin²x|+C
(2)f'(sinx)=cos²x
f'[sin(arcsinx)]=cos²(arcsinx)
f'(x)=1-x²,两边积分
f(x)=∫dx-∫x²dx
=x-(1/3)x³+C
(3)f'(cosx)=sinx
f'(cosarccosx)=sinarccosx
f'(x)=√(1-x²),两边积分
f(x)=(1/2)[x√(1-x²)+arcsinx]+C
f(cosx)=(1/2)[cosx√(1-cos²x)+arcsincosx]+C
=(1/2)(sinxcosx+arcsincosx)+C
(4)∫[1/√(e^x-1)]dx,d(e^x)=(e^x)dx
=∫1/[e^x√(e^x-1)]d(e^x),d(e^x-1)=(e^x)dx
=∫{1/[(e^x)√(e^x-1)]}d(e^x-1),d√(e^x-1)=1/[2√(e^x-1)]d(e^x-1)
=2∫{[√(e^x-1)]/[(((√(e^x-1))²+1)√(e^x-1)]}d√(e^x-1)
=2∫{1/[(√(e^x-1))²+1]}d√(e^x-1)
=2arctan√(e^x-1)+C
换元不明显,
我建议你发到贴吧上去把