求∫1/sin^2xcos^2x dx
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求∫1/sin^2xcos^2x dx
求∫1/sin^2xcos^2x dx
求∫1/sin^2xcos^2x dx
∫dx/(sinxcosx)^2
=∫4dx/(sin2x)^2
=2∫d2x/(sin2x)^2
=2∫(csc2x)^2 d2x
= -2cot2x+C
∫dx/(sinxcosx)^2
=∫4dx/(sin2x)^2
=2∫d2x/(sin2x)^2
=-2[∫[-(cos2x)^2-(sinx)^2]d2x/(sin2x)^2
=-2cos(2x)/sin2x+C
∫1/sin²xcos²xdx=4∫1/sin²2xdx=4∫csc²2xdx=2∫csc²2xd2x=-2cot2x+C.
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