求4个积分题目1.∫ ln(tanx)/(sinxcosx) dx pai/4≤x≤pai/32.∫ 1/ (1+2e^x -e^(-x)) dx 3.∫ (secxcos2x)/ (scex+sinx) dx4.∫ (2x-1)^1/2 / (2x+3) dx
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![求4个积分题目1.∫ ln(tanx)/(sinxcosx) dx pai/4≤x≤pai/32.∫ 1/ (1+2e^x -e^(-x)) dx 3.∫ (secxcos2x)/ (scex+sinx) dx4.∫ (2x-1)^1/2 / (2x+3) dx](/uploads/image/z/13963275-27-5.jpg?t=%E6%B1%824%E4%B8%AA%E7%A7%AF%E5%88%86%E9%A2%98%E7%9B%AE1.%E2%88%AB+ln%28tanx%29%2F%28sinxcosx%29+dx+pai%2F4%E2%89%A4x%E2%89%A4pai%2F32.%E2%88%AB+1%2F+%281%2B2e%5Ex+-e%5E%28-x%29%29+dx+3.%E2%88%AB+%28secxcos2x%29%2F+%28scex%2Bsinx%29+dx4.%E2%88%AB+%282x-1%29%5E1%2F2+%2F+%282x%2B3%29+dx)
求4个积分题目1.∫ ln(tanx)/(sinxcosx) dx pai/4≤x≤pai/32.∫ 1/ (1+2e^x -e^(-x)) dx 3.∫ (secxcos2x)/ (scex+sinx) dx4.∫ (2x-1)^1/2 / (2x+3) dx
求4个积分题目
1.∫ ln(tanx)/(sinxcosx) dx pai/4≤x≤pai/3
2.∫ 1/ (1+2e^x -e^(-x)) dx
3.∫ (secxcos2x)/ (scex+sinx) dx
4.∫ (2x-1)^1/2 / (2x+3) dx
求4个积分题目1.∫ ln(tanx)/(sinxcosx) dx pai/4≤x≤pai/32.∫ 1/ (1+2e^x -e^(-x)) dx 3.∫ (secxcos2x)/ (scex+sinx) dx4.∫ (2x-1)^1/2 / (2x+3) dx
1.∫ ln(tanx)/(sinxcosx) dx pai/4≤x≤pai/3
∫ ln(tanx)/(sinxcosx) dx =∫(ln(tanx)/tanx)(secx)^2 dx =∫ln(tanx)/tanx dtanx
=∫ln(tanx dln(tanx)+C
=1/2(ln(tanx))^2 +C
2.∫ 1/ (1+2e^x -e^(-x)) dx = ∫ e^x/ ( e^x+2e^2x -1) dx =(1/3)∫[1/(2e^x-1)+1/ ( e^x+1) ]dx
=(1/3)[∫1/(2e^x-1)dx+∫1/ ( e^x+1) dx]
=(1/3)[∫1/(2-e^(-x))d(2-e^(-x))-∫1/(1+e^(-x)d(1+e^(-x)]
=(1/3)[ln(2-e^(-x))-ln(1+e^(-x)]+C
=(1/3)[ln(2-e^(-x))/ln(1+e^(-x)]+C
3.∫ (secxcos2x)/ (scex+sinx) dx,该题目应该为∫ (secxcos2x)/ (secx+sinx) dx,
∫ (secxcos2x)/ (secx+sinx) dx=∫ cos2x/ (1+sinxcosx) dx=∫ 2cos2x/ (2+sin2x) dx
=∫ 1/ (2+sin2x) d(2+sin2x)
=ln(2+sin2x) +C
4.∫ (2x-1)^1/2 / (2x+3) dx
令(2x-1)^1/2=t,则x=(1+t^2)/2,dx=tdt,则
原式=∫[t/(4+t^2)]tdt=∫[1-4/(4+t^2)]dt=t-4∫[/(4+t^2)]tdt
=t-4×(1/2)arctan(t/2)+C
=(2x-1)^1/2-2arctan[((2x-1)^1/2)/2]+C