∫x^2(cosx)^2dx 求不定积分∫xcos^2dx
\u222bx^2cos(x/2)^2dx\u222bx^2cos(x/2)^2dx=(1/6)x³+(1/2)x²sinx+xcosx-sinx+C\u3002C\u4e3a\u5e38\u6570\u3002
\u89e3\u7b54\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u2235[cos(x/2)]²=(1+cosx)/2
\u2234\u539f\u5f0f=(1/2)\u222b(1+cosx)x²dx=(1/2)\u222bx²dx+(1/2)\u222bx²cosxdx\u3002
\u800c\u222bx²cosxdx=x²sinx-2\u222bxsinxdx=x²sinx+2xcosx-2sinx+c1\u3002
\u2234\u539f\u5f0f=(1/6)x³+(1/2)x²sinx+xcosx-sinx+C\u3002
\u6269\u5c55\u8d44\u6599\uff1a
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\u4e00\u4e2a\u51fd\u6570\uff0c\u53ef\u4ee5\u5b58\u5728\u4e0d\u5b9a\u79ef\u5206\uff0c\u800c\u4e0d\u5b58\u5728\u5b9a\u79ef\u5206\uff0c\u4e5f\u53ef\u4ee5\u5b58\u5728\u5b9a\u79ef\u5206\uff0c\u800c\u6ca1\u6709\u4e0d\u5b9a\u79ef\u5206\u3002\u8fde\u7eed\u51fd\u6570\uff0c\u4e00\u5b9a\u5b58\u5728\u5b9a\u79ef\u5206\u548c\u4e0d\u5b9a\u79ef\u5206\uff1b\u82e5\u5728\u6709\u9650\u533a\u95f4[a,b]\u4e0a\u53ea\u6709\u6709\u9650\u4e2a\u95f4\u65ad\u70b9\u4e14\u51fd\u6570\u6709\u754c\uff0c\u5219\u5b9a\u79ef\u5206\u5b58\u5728\uff1b\u82e5\u6709\u8df3\u8dc3\u3001\u53ef\u53bb\u3001\u65e0\u7a77\u95f4\u65ad\u70b9\uff0c\u5219\u539f\u51fd\u6570\u4e00\u5b9a\u4e0d\u5b58\u5728\uff0c\u5373\u4e0d\u5b9a\u79ef\u5206\u4e00\u5b9a\u4e0d\u5b58\u5728\u3002
\u534a\u89d2\u516c\u5f0f
sin^2(\u03b1/2)=(1-cos\u03b1)/2
cos^2(\u03b1/2)=(1+cos\u03b1)/2
tan^2(\u03b1/2)=(1-cos\u03b1)/(1+cos\u03b1)
tan(\u03b1/2)=sin\u03b1/(1+cos\u03b1)=(1-cos\u03b1)/sin\u03b1
\u5e38\u7528\u79ef\u5206\u516c\u5f0f\uff1a
1\uff09\u222b0dx=c
2\uff09\u222bx^udx=(x^(u+1))/(u+1)+c
3\uff09\u222b1/xdx=ln|x|+c
4\uff09\u222ba^xdx=(a^x)/lna+c
5\uff09\u222be^xdx=e^x+c
6\uff09\u222bsinxdx=-cosx+c
7\uff09\u222bcosxdx=sinx+c
8\uff09\u222b1/(cosx)^2dx=tanx+c
9\uff09\u222b1/(sinx)^2dx=-cotx+c
10\uff09\u222b1/\u221a\uff081-x^2) dx=arcsinx+c
\u7ed3\u679c\u662f
＝（1/2）∫x^2dx＋（1/2）∫x^2cos2xdx
＝（1/6）x^3＋（1/4）∫x^2d（sin2x）
＝（1/6）x^3＋（1/4）x^2sin2x－（1/4）∫sin2xd（x^2）
＝（1/6）x^3＋（1/4）x^2sin2x－（1/2）∫xsin2xdx
＝（1/6）x^3＋（1/4）x^2sin2x＋（1/4）∫xd（cos2x）
＝（1/6）x^3＋（1/4）x^2sin2x＋（1/4）xcos2x－（1/4）∫cos2xdx
＝（1/6）x^3＋（1/4）x^2sin2x＋（1/4）xcos2x－（1/8）sin2x＋C
绛旓細鍘熷紡锛濓紙1/2锛夆埆x^2锛1锛媍os2x锛dx 锛濓紙1/2锛夆埆x^2dx锛嬶紙1/2锛夆埆x^2cos2xdx 锛濓紙1/6锛墄^3锛嬶紙1/4锛夆埆x^2d锛坰in2x锛夛紳锛1/6锛墄^3锛嬶紙1/4锛墄^2sin2x锛嶏紙1/4锛夆埆sin2xd锛坸^2锛夛紳锛1/6锛墄^3锛嬶紙1/4锛墄^2sin2x锛嶏紙1/2锛鈭玿sin2xdx 锛濓紙1/6锛墄^3锛嬶紙1/4锛...
绛旓細绛旓細鈭2xcosx^2dx鈮犫埆sinx^2dx^2 鍥犱负锛屸埆2xcosx^2dx = 鈭玞osx^2dx^2 = sinx^2+C 鑰 鈭玸inx^2dx^2 =-cosx^2+C 娉ㄦ剰锛 sinx^2涓巗in²x 鐨勫尯鍒 鎵浠ワ紝瀹冧滑涓嶇浉绛
绛旓細鈭2xcosx鈭2dx =鈭玿(cos2x+1)dx =x^2/2+(1/2)鈭玿dsin2x =x^2/2+(1/2)xdsin2x-(1/2)鈭玸in2xdx =x^2/2+(1/2)xdsin2x+(1/4)鈭玠cos2x =x^2/2+(1/2)xdsin2x+(1/4)dcos2x+c
绛旓細鈭玿cosx^2dx=(1/2)鈭玞osx^2dx^2=(1/2)sinx^2+C锛涘湪寰Н鍒嗕腑锛屼竴涓嚱鏁癴 鐨勪笉瀹氱Н鍒嗭紝鎴栧師鍑芥暟锛屾垨鍙嶅鏁帮紝鏄竴涓鏁扮瓑浜巉 鐨勫嚱鏁 F 锛屽嵆F 鈥 = f銆備笉瀹氱Н鍒嗗拰瀹氱Н鍒嗛棿鐨勫叧绯荤敱寰Н鍒嗗熀鏈畾鐞嗙‘瀹氥傚叾涓璅鏄痜鐨勪笉瀹氱Н鍒嗐
绛旓細鈭玿(cosx)^2dx=鈭玿cos^2xdx =鈭玿(1+cos2x/2)dx =1/4x^2+1/2鈭玿cos2xdx =1/4x^2+1/4鈭玿d(sin2x)=1/4x^2+1/4xsin2x-1/4鈭玸in2xdx =1/4x^2+1/4xsin2x+1/8cos2x+C 璇存槑锛欳鏄父鏁 涓嶅彲绉嚱鏁 铏界劧寰堝鍑芥暟閮藉彲閫氳繃濡備笂鐨勫悇绉嶆墜娈佃绠楀叾涓嶅畾绉垎锛屼絾杩欏苟涓嶆剰鍛崇潃鎵鏈...
绛旓細x^2*(cosx)^2鐨勭Н鍒嗕负1/6 x³+1/4x² *sin2x+1/4cos2x-1/8sin2x+C 瑙ｏ細 鈭紙 x² cos²x锛塪x= 鈭紙 x² 锛坈os2x+1锛/2锛dx =1/2鈭锛 x² cos2x+x²锛塪x =1/2鈭玿²dx+1/2鈭 x² cos2xdx =1/6 x³+1/2...
绛旓細cosx=1-x^2/2!+x^4/4!+鈥+(-1)^m脳x^(2m)/(2m)!+鈥﹂偅涔坈os(x^2)=1-x^4/2!+x^8/4!+鈥+(-1)^m脳x^(4m)/(2m)!+鈥﹂偅涔 鈭玞os(x^2)dx=鈭玔1-x^4/2!+x^8/4!+鈥+(-1)^m脳x^(4m)/(2m)!+鈥dx =鈭(-1)^m脳x^(2m+1)/{[4m+1]脳(4m)!} (m浠0...
绛旓細鈭玿cosx^2dx =(1/2)鈭玞osx^2dx^2 =(1/2)sinx^2+C 璇佹槑锛氬鏋渇(x)鍦ㄥ尯闂碔涓婃湁鍘熷嚱鏁帮紝鍗虫湁涓涓嚱鏁癋(x)浣垮浠绘剰x鈭圛锛岄兘鏈塅'(x)=f(x)锛岄偅涔堝浠讳綍甯告暟鏄剧劧涔熸湁[F(x)+C]'=f(x).鍗冲浠讳綍甯告暟C锛屽嚱鏁癋(x)+C涔熸槸f(x)鐨勫師鍑芥暟銆傝繖璇存槑濡傛灉f(x)鏈変竴涓師鍑芥暟,閭ｄ箞f(x)灏...
绛旓細鏄(cosx)^2鍚楋紵鈭紙cosx锛塣2dx=(1/2)鈭锛1+cos2x)dx =(1/2)鈭玠x+(1/4)鈭(cos2x)d(2x)=x/2+(sin2x)/4+C.
绛旓細鈭(cosx)^2dx=x/2 + sin2x /4+c銆俢涓虹Н鍒嗗父鏁般傝繃绋嬪涓嬶細y=(cosx)^2 =(1+cos2x)/2 瀵瑰叾绉垎锛氣埆(cosx)^2dx =鈭(1+cos2x)/2dx = 1/2 鈭紙1+cos2x锛塪x = 1/2 銆 x + 1/2 sin2x 銆= x/2 + sin2x /4+c
