求四道关于三角函数的不定积分 三角函数求不定积分，求助！
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∫ dx/sinx
=∫ cscx dx
=ln|cscx-cotx| + C
(2)
∫ dx/cosx
=∫ secx dx
=ln|secx+tanx| +C
(3)
∫ dx/(sinx)^2
=∫ (csx)^2 dx
=-cotx + C
(4)
∫ dx/(sinx)^3
=∫ (cscx)^3 dx
=-∫ cscx d(cotx)
=-cscx.cotx - ∫ (cotx)^2.cscx dx
=-cscx.cotx - ∫ [(cscx)^2 -1].cscx dx
2∫ (cscx)^3 dx =-cscx.cotx + ∫ cscx dx
∫ (cscx)^3 dx =(1/2)[-cscx.cotx + ln|cscx-cotx| ] + C
=>
∫ dx/(sinx)^3 =(1/2)[-cscx.cotx + ln|cscx-cotx| ] + C
前3个有公式
∫dx/sinx = ∫cscxdx = ln|cscx - cotx| + C
∫dx/cosx = ∫secxdx = ln|secx + tanx| + C
∫dx/(sinx)^2 = ∫(cscx)^2dx = -cotx + C
∫dx/(sinx)^3 = -∫sinxdx/(sinx)^4
= -∫dcosx/[1-(cosx)^2]^2 记 u = cosx
= -∫du/(1-u^2)^2
= (-1/4)∫[1/(1-u) + 1/(1+u) + 1(1-u)^2 + 1/(1+u)^2]du
= (-1/4)ln|(1+u)/(1-u)| + 1/(1-u) - 1/(1+u) + C
= (-1/4)ln|(1+cosx)/(1-cosx)| + 1/(1-cosx) - 1/(1+cosx) + C
不定积分结果不唯一求导验证应该能够提高凑微分的计算能力，
绛旓細(1)鈭 dx/sinx =鈭 cscx dx =ln|cscx-cotx| + C (2)鈭 dx/cosx =鈭 secx dx =ln|secx+tanx| +C (3)鈭 dx/(sinx)^2 =鈭 (csx)^2 dx =-cotx + C (4)鈭 dx/(sinx)^3 =鈭 (cscx)^3 dx =-鈭 cscx d(cotx)=-cscx.cotx - 鈭 (cotx)^2.cscx dx =-cscx.cotx - ...
绛旓細=1/4ln缁濆鍊紅an(x/2)+1/8{tan(x/2)}^2+C
绛旓細1. 鍩烘湰涓夎鍑芥暟鐨勪笉瀹氱Н鍒嗭細- sin(x)dx锛氣埆sin(x)dx = -cos(x) + C 锛堝叾涓瑿涓哄父鏁帮級- cos(x)dx锛氣埆cos(x)dx = sin(x) + C - tan(x)dx锛氣埆tan(x)dx = log_|tan(x)| + C - sec(x)dx锛氣埆sec(x)dx = ln|sec(x)| + C - csc...
绛旓細涓夎鍑芥暟绉垎鍏紡琛ㄤ负锛氾紙1锛鈭玸inxdx=-cosx+C锛涒埆cosxdx=sinx+C锛涳紙2锛夆埆tanxdx=ln|secx|+C锛涒埆cotxdx=ln|sinx|+C锛涒埆secxdx=ln|secx+tanx|+C锛涒埆cscxdx=ln|cscx_cotx|+C锛涳紙3锛夆埆sin_xdx=1/2x-1/4sin2x+C锛涒埆cos_xdx=1/2+1/4sin2x+C锛涒埆tan_xdx=tanx-x+C锛涒埆cot_xdx=-...
绛旓細鈭(sin2x/cos3x)dx =鈭(sinx/cos3x)d(-cosx) =鈭玸inx路d(-cosx)/cos3x =鈭玸inx路d[1/2cos2x] =sinx路[1/2cos2x]-鈭玔1/2cos2x]d(sinx) 鍒嗛儴绉垎 =?sinx路sec2x-?鈭玔1/(1-sin2x]d(sinx) =?sinx路sec2x-?鈭埆[1/(1-sinx)+1/(1+sinx)]卤d(1卤sinx) =?sinx路sec2x...
绛旓細绗竴绉嶆渶蹇細鈭 secx dx = 鈭 secx • (secx + tanx)/(secx + tanx) dx = 鈭 (secxtanx + sec²x)/(secx + tanx) dx = 鈭 d(secx + tanx)/(secx + tanx)= ln|secx + tanx| + C 绗簩绉嶏細鈭 secx dx = 鈭 1/cosx dx = 鈭 cosx/cos²x dx = 鈭 d...
绛旓細杩欓亾棰樺苟涓嶆槸鐗瑰埆绠鍗曪紝鍏蜂綋杩囩▼濡備笅锛氫互涓婏紝璇烽噰绾炽
绛旓細鍥炵瓟锛氱Н鍖栧拰宸叕寮 sin3xcos2x=(1/2)[sin5x-sinx] 鍐绉垎鍙緱鍒 -(1/10)cos5x+(1/2)cosx+C
绛旓細璁＄畻锛坱anx锛²涓嶅畾绉垎鐨勬柟娉曪細锛坱anx锛²=鈭玔(secx)^2-1]dx =鈭(secx)^2dx-x =tanx-x+c
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