设u=x^2 ; du=2xdx
dv=tgxdx ; v=-ln cosx
∫udv=uv-∫vdu
∫x^2tgx=-x^2.ln cosx+∫ln cosx.2xdx但还是不能用初等函数表示原函数是吧？再设u=2xdv=ln cosxdxdu=2dxv=∫ln cosxdx=∫ln cosxdcosx/sinx=cscx(cosx ln cosx-cosx)=ctgx(ln cosx-1)∫ln cosx.2xdx=2x.ctgx(ln cosx-1)-2∫ctgx(ln cosx)dx再设 u=ln cosx dv=ctgxdx v=ln sinx du=d(ln cosx)=-sinx/cosx.dx=-tgxdx∫ctgx(ln cosx)dx=ln cos.lnsinx+∫ln sinx.tgxdx无法用初等函表示.
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