积分问题∫(1/sin^4x)d
∫(1/sin^4 x) dx请问这道题该怎么做呢
∫(1/sin^4 x)dx=∫[(sin^2 x+cos^2 x)/sin^4 x]dx
=∫(1/sin^2 x)dx+∫(cos^2 x/sin^4 x)dx
=∫(1/sin^2 x)dx+∫(cosx/sin^4 x)d(sinx)
=∫(1/sin^2 x)dx-(1/3)∫(cosx)d(1/sin^3 x)
=∫(1/sin^2 x)dx-(1/3)[(cosx/sin^3 x)-∫(1/sin^3 x)*(-sinx)]dx
=∫(1/sin^2 x)dx-(cosx/3sin^3 x)-(1/3)∫(1/sin^2 x...全部
∫(1/sin^4 x) dx请问这道题该怎么做呢
∫(1/sin^4 x)dx=∫[(sin^2 x+cos^2 x)/sin^4 x]dx
=∫(1/sin^2 x)dx+∫(cos^2 x/sin^4 x)dx
=∫(1/sin^2 x)dx+∫(cosx/sin^4 x)d(sinx)
=∫(1/sin^2 x)dx-(1/3)∫(cosx)d(1/sin^3 x)
=∫(1/sin^2 x)dx-(1/3)[(cosx/sin^3 x)-∫(1/sin^3 x)*(-sinx)]dx
=∫(1/sin^2 x)dx-(cosx/3sin^3 x)-(1/3)∫(1/sin^2 x)dx
=-(cosx/3sin^3 x)-(2/3)∫(1/sin^2 x)dx
=-(cosx/3sin^3 x)+(2/3)cotx+C。
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