求∫1/sinxdx
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求∫1/sinxdx
求∫1/sinxdx
求∫1/sinxdx
∫1/sinxdx
= ∫sinx/sin²xdx
=-∫1/sin²xdcosx
=-∫1/(1-cos²x)dcosx
=-1/2∫1/(1-cosx)+1/(1+cosx)dcosx
=-1/2[ln(1+cosx)-ln(1-cosx)]+C
∫1/sinxdx
= ∫sinx/sin²xdx
=-∫1/sin²xdcosx
=-∫1/(1-cos²x)dcosx
=-1/2∫1/(1-cosx)+1/(1+cosx)dcosx
=-1/2[ln(1+cosx)-ln(1-cosx)]+C
==ln|(sinx/2)/(cosx/2)|+C=ln|tanx/2|+C
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