求不定积分：∫arcsinxdx 求∫arcsinx·arcsinxdx的不定积分
\u6c42\u222barcsinxdxarcsinx \u2018= 1/\u221a(1-x^2) \u8fd9\u4e2a\u662f\u5bfc\u6570\u91cc\u9762\u8981\u8bb0\u4f4f\u7684\u57fa\u672c\u516c\u5f0f\uff01\uff01\uff01
darcsinx \u2018= 1/\u221a(1-x^2)dx \u53ea\u662f\u5bfc\u6570\u516c\u5f0f\u7684\u5fae\u5206\u5f62\u5f0f\u3002
\u3000\u3000\u5199\u4e00\u5199\u5c31\u6709\uff1a\u6700\u540e\u4e00\u6b21\u5206\u90e8\u79ef\u5206\u540e\uff0c\u5f97\u5230\u79ef\u5206\u90e8\u5206\u662f
\u3000\u3000\u3000\u222b[\u221a(1-x²)]d(arcsinx-arccosx)
\u3000\u3000= \u222b[\u221a(1-x²)]{[1/\u221a(1-x²)]-[-1/\u221a(1-x²)]}dx
\u3000\u3000= \u2026\u2026
\u3000\u3000= 2x+C\u3002
令t=arcsinx
则 x=sint
则dx=costdt
∫tcostdt
=tsint-∫sintdt
=tsint+cost
=arcsinx*sin(aicsinx)+cos(arcsinx)+C
=xarcsinx+√[1-(sin(arcsinx))²]+C
=xarcsinx+√(1-x²)+C
楼上基本正确，题目有点小问题
绛旓細鍏蜂綋鍥炵瓟濡備笅锛氣埆 arcsinx dx =xarcsinx-鈭 x/鈭(1-x²) dx =xarcsinx-1/2鈭 1/鈭(1-x²) d(x²)=xarcsinx+鈭(1-x²) +C 涓嶅畾绉垎鐨勬剰涔夛細涓涓嚱鏁帮紝鍙互瀛樺湪涓嶅畾绉垎锛岃屼笉瀛樺湪瀹氱Н鍒嗭紝涔熷彲浠ュ瓨鍦ㄥ畾绉垎锛岃屾病鏈変笉瀹氱Н鍒嗐傝繛缁嚱鏁帮紝涓瀹氬瓨鍦ㄥ畾绉垎鍜屼笉瀹...
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細
绛旓細= 鈭 arcsinx d(x²/2)= (1/2)x²arcsinx - (1/2)鈭 x²/鈭(1 - x²) dx锛寈 = sinz = (1/2)x²arcsinx - (1/2)鈭 sin²z/|cosz| * (cosz dz)= (1/2)x²arcsinx - (1/2)鈭 (1 - cos2z)/2 dz = (1/2)x²...
绛旓細绛旓細鍦ㄨ绠鈭玜rcsinxd[鈭(1-x^2)]鍜屸埆arccosxd[鈭(1-x^2)]鏃讹紝鍒嗛儴绉垎銆佸父鏁1鍐嶇Н鍒嗚屽緱銆傚叿浣撴槸锛屸埆arcsinxd[鈭(1-x^2)]=arcsinx鈭(1-x^2)-鈭垰(1-x^2)/鈭(1-x^2)]dx=arcsinx鈭(1-x^2)-x锛屽悓鐞嗭紝鏈夆埆arccosxd[鈭(1-x^2)]=arccosx鈭(1-x^2)+x銆備緵鍙傝冦
绛旓細楂樿禐绛旀閿欒 閿欒 姝ｇ‘瑙ｇ瓟锛氭眰鈭玜rcsinx dx 锛氫护t = arcsinx 鍗硏 = sint 鍘熷紡 = 鈭玹dsint = tsint - 鈭玸intdt = tsint + cost +C 灏唗 = arcsinx 浠ｅ叆锛氬師寮 = xarcsinx + 鈭(1-x²) + C
绛旓細鈭玜rcsinxdarcsinx锛1/2arcsin²x锛婥.鎴栬呬护t锛漚rcsinx 鍒欏師寮忥紳鈭玹dt锛1/2t²锛婥锛1/2arcsin²x锛婥.
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細
绛旓細銆
绛旓細arcsinx鐨涓嶅畾绉垎姹傛硶锛氬埄鐢ㄥ垎閮ㄧН鍒嗘硶锛氬嵆鈭玼dv=uv-鈭玽du 鈭玜rcsinxdx=x路arcsinx-鈭玿d锛坅rcsinx锛=x路arcsinx-鈭玿/(1-x^2)^(1/2)dx =x路arcsinx+(1/2)鈭1/(1-x^2)^(1/2)d((1-x^2))=x路arcsinx+(1-x^2)^(1/2)+C =xarcsinx+鈭(1-x^2)+C銆傛牴鎹墰椤-鑾卞竷灏艰尐...
