求这几道题的详细解题过程,感谢!
1个回答
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=-∫(cosx)^5dcosx=(-1/6)(cosx)^6
=∫x/2cos²xdx=∫x/2dtanx=(1/2)(xtanx-∫tanxdx)=xtanx/2+lncosx/2
=∫dt/√(ln(1-t))???题没错吧,换元或分部积分都出不来
=∫ln(t-1)/t²dt=-∫ln(t-1)d(1/t)=-ln(t-1)/t+∫1/tdln(t-1)=-ln(t-1)/t+∫1/t(t-1)dt=-ln(t-1)/t+∫1/(t-1)-1/tdt=-ln(t-1)/t+ln(t-1)-lnt=-ln(x+1)/(x+2)+ln(x+1)-ln(x+2)
=∫x/2cos²xdx=∫x/2dtanx=(1/2)(xtanx-∫tanxdx)=xtanx/2+lncosx/2
=∫dt/√(ln(1-t))???题没错吧,换元或分部积分都出不来
=∫ln(t-1)/t²dt=-∫ln(t-1)d(1/t)=-ln(t-1)/t+∫1/tdln(t-1)=-ln(t-1)/t+∫1/t(t-1)dt=-ln(t-1)/t+∫1/(t-1)-1/tdt=-ln(t-1)/t+ln(t-1)-lnt=-ln(x+1)/(x+2)+ln(x+1)-ln(x+2)
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