(sinx^2+1)/cox^4求不定积分
1/(sinx)^2\u7684\u4e0d\u5b9a\u79ef\u5206\u56e0\u4e3a\u5bfc\u6570(cotx)'=-csc²x=-1/sin²x
\u6240\u4ee5\u4e24\u8fb9\u53d6\u79ef\u5206\uff1a\u222b(cotx)'dx=\u222b(-1/sin²x) dx
cotx+C=-\u222b(1/sin²x)dx
\u6240\u4ee5\u222b(1/sin²x)dx=-cotx+C'
\u975e\u521d\u7b49\u79ef\u5206\uff0c\u8868\u793a\u4e3a\u4e00\u4e2a\u692d\u5706\u51fd\u6570\uff1a
=sqrt(1+sin(x))*sqrt(-2*sin(x)+2)*sqrt(-sin(x))*EllipticF(sqrt(1+sin(x)), (1/2)*sqrt(2))/(cos(x)*sqrt(sin(x)))
=∫[2-(cosx)^2] dx / [(cosx)^4]
=2∫dx/(cosx)^4 - ∫dx/(cosx)^2
=2∫(secx)^4dx-∫(secx)^2dx
=2∫(secx)^2d(tanx)-tanx
=2[∫(tanx)^2d(tanx) + ∫d(tanx)]-tanx
=2/3 (tanx)^3+tanx+c
=2/3sinx[1-(cosx)^2]/[(cosx)^3]+sinx/cosx+c
=(2/3)sinx / [(cosx)^2 + (1/3)(sinx/cosx)+c
绛旓細鈭(1/sin²x)dx=-cotx+C'鍥犱负瀵兼暟(cotx)'=-csc²x=-1/sin²x 鎵浠ヤ袱杈瑰彇绉垎锛氣埆(cotx)'dx=鈭(-1/sin²x) dx cotx+C=-鈭(1/sin²x)dx 鎵浠モ埆(1/sin²x)dx=-cotx+C'鍩烘湰浠嬬粛 绉垎鍙戝睍鐨勫姩鍔涙簮鑷疄闄呭簲鐢ㄤ腑鐨勯渶姹傘傚疄闄呮搷浣滀腑锛屾湁鏃跺欏彲浠ョ敤绮楃暐...
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