一道高数题（请各位用图片写出详细过程上传到网上，谢谢！）
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∫(π/3,π/4) x/sin²xdx
=-∫(π/3,π/4) xdcotx
=-(√3π/9-π/4)+∫(π/3,π/4) cotxdx
=-√3π/9+π/4+∫(π/3,π/4) cosx/sinx dx
=-√3π/9+π/4+∫(π/3,π/4) 1/sinx dsinx
=-√3π/9+π/4+ln(sin(π/3))-ln(sin(π/4))
=-√3π/9+π/4+ln(√3/2)-ln(√2/2)
=-√3π/9+π/4+ln(√6/2)
=-∫(π/3,π/4) xdcotx
=-(√3π/9-π/4)+∫(π/3,π/4) cotxdx
=-√3π/9+π/4+∫(π/3,π/4) cosx/sinx dx
=-√3π/9+π/4+∫(π/3,π/4) 1/sinx dsinx
=-√3π/9+π/4+ln(sin(π/3))-ln(sin(π/4))
=-√3π/9+π/4+ln(√3/2)-ln(√2/2)
=-√3π/9+π/4+ln(√6/2)
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