![∫1[(sinx)^4(cosx)^4]dx求不定積分,第1張 ∫1[(sinx)^4(cosx)^4]dx求不定積分,第1張](data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==)
∫1/[(sinx)^4(cosx)^4]dx求不定積分網友廻答:
- ∫1/[(sinx)^4(cosx)^4]dx
=16∫1/[(2sinxcosx)^4]dx
=16∫1/(sin2x)^4dx
=16∫(csc2x)^4dx
=-8∫csc²2xdcot2x
=-8∫(cot²2x 1)dcot2x
=-8/3 cot³2x-8cot2x c
網友廻答:
- ∫ 1/[(sinx)^4(cosx)^4] dx
=16∫ 1/(2sinxcosx)^4 dx
=16∫ 1/(sin2x)^4 dx
=16∫ (csc2x)^4 dx
=-8∫ csc²2x d(cot2x)
=-8∫ (cot²2x 1) d(cot2x)
=-(8/3)cost³2x - 8cot2x C
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