sin^nxcos^nx在0到2/丌的定积分 sin^nx在0到2π积分，如果n为奇数，是否为0
cos^nx sinnx\u5728[0,\u03c0/2]\u5b9a\u79ef\u5206\u5c71\u4e1c\u5927\u5b66\u7ed3\u679c\u5982\u4e0b\uff1a
\u4e0d\u5b9a\u79ef\u5206\uff08Indefinite integral\uff09
\u5373\u5df2\u77e5\u5bfc\u6570\u6c42\u539f\u51fd\u6570\u3002\u82e5F\u2032(x)=f(x)\uff0c\u90a3\u4e48[F(x)+C]\u2032=f(x).(C\u2208R C\u4e3a\u5e38\u6570).\u4e5f\u5c31\u662f\u8bf4\uff0c\u628af(x)\u79ef\u5206\uff0c\u4e0d\u4e00\u5b9a\u80fd\u5f97\u5230F(x)\uff0c\u56e0\u4e3aF(x)+C\u7684\u5bfc\u6570\u4e5f\u662ff(x)\uff08C\u662f\u4efb\u610f\u5e38\u6570\uff09\u3002
\u6240\u4ee5f(x)\u79ef\u5206\u7684\u7ed3\u679c\u6709\u65e0\u6570\u4e2a\uff0c\u662f\u4e0d\u786e\u5b9a\u7684\u3002\u6211\u4eec\u4e00\u5f8b\u7528F(x)+C\u4ee3\u66ff\uff0c\u8fd9\u5c31\u79f0\u4e3a\u4e0d\u5b9a\u79ef\u5206\u3002\u5373\u5982\u679c\u4e00\u4e2a\u5bfc\u6570\u6709\u539f\u51fd\u6570\uff0c\u90a3\u4e48\u5b83\u5c31\u6709\u65e0\u9650\u591a\u4e2a\u539f\u51fd\u6570\u3002
\u5b9a\u79ef\u5206 \uff08definite integral\uff09
\u5b9a\u79ef\u5206\u5c31\u662f\u6c42\u51fd\u6570f(X)\u5728\u533a\u95f4[a,b]\u4e2d\u7684\u56fe\u50cf\u5305\u56f4\u7684\u9762\u79ef\u3002\u5373\u7531 y=0,x=a,x=b,y=f(X)\u6240\u56f4\u6210\u56fe\u5f62\u7684\u9762\u79ef\u3002\u8fd9\u4e2a\u56fe\u5f62\u79f0\u4e3a\u66f2\u8fb9\u68af\u5f62\uff0c\u7279\u4f8b\u662f\u66f2\u8fb9\u4e09\u89d2\u5f62\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u4e00\u822c\u5b9a\u7406
\u5b9a\u74061\uff1a\u8bbef(x)\u5728\u533a\u95f4[a,b]\u4e0a\u8fde\u7eed\uff0c\u5219f(x)\u5728[a,b]\u4e0a\u53ef\u79ef\u3002
\u5b9a\u74062\uff1a\u8bbef(x)\u533a\u95f4[a,b]\u4e0a\u6709\u754c\uff0c\u4e14\u53ea\u6709\u6709\u9650\u4e2a\u95f4\u65ad\u70b9\uff0c\u5219f(x)\u5728[a,b]\u4e0a\u53ef\u79ef\u3002
\u5b9a\u74063\uff1a\u8bbef(x)\u5728\u533a\u95f4[a,b]\u4e0a\u5355\u8c03\uff0c\u5219f(x)\u5728[a,b]\u4e0a\u53ef\u79ef\u3002
\u725b\u987f-\u83b1\u5e03\u5c3c\u8328\u516c\u5f0f
\u5b9a\u79ef\u5206\u4e0e\u4e0d\u5b9a\u79ef\u5206\u770b\u8d77\u6765\u98ce\u9a6c\u725b\u4e0d\u76f8\u53ca\uff0c\u4f46\u662f\u7531\u4e8e\u4e00\u4e2a\u6570\u5b66\u4e0a\u91cd\u8981\u7684\u7406\u8bba\u7684\u652f\u6491\uff0c\u4f7f\u5f97\u5b83\u4eec\u6709\u4e86\u672c\u8d28\u7684\u5bc6\u5207\u5173\u7cfb\u3002\u628a\u4e00\u4e2a\u56fe\u5f62\u65e0\u9650\u7ec6\u5206\u518d\u7d2f\u52a0\uff0c\u8fd9\u4f3c\u4e4e\u662f\u4e0d\u53ef\u80fd\u7684\u4e8b\u60c5\uff0c\u4f46\u662f\u7531\u4e8e\u8fd9\u4e2a\u7406\u8bba\uff0c\u53ef\u4ee5\u8f6c\u5316\u4e3a\u8ba1\u7b97\u79ef\u5206\u3002\u8fd9\u4e2a\u91cd\u8981\u7406\u8bba\u5c31\u662f\u5927\u540d\u9f0e\u9f0e\u7684\u725b\u987f-\u83b1\u5e03\u5c3c\u5179\u516c\u5f0f\uff0c\u5b83\u7684\u5185\u5bb9\u662f\uff1a
\u5982\u679cf(x)\u662f[a,b]\u4e0a\u7684\u8fde\u7eed\u51fd\u6570\uff0c\u5e76\u4e14\u6709F\u2032(x)=f(x)\uff0c\u90a3\u4e48\uff1a
\u7528\u6587\u5b57\u8868\u8ff0\u4e3a\uff1a\u4e00\u4e2a\u5b9a\u79ef\u5206\u5f0f\u7684\u503c\uff0c\u5c31\u662f\u539f\u51fd\u6570\u5728\u4e0a\u9650\u7684\u503c\u4e0e\u539f\u51fd\u6570\u5728\u4e0b\u9650\u7684\u503c\u7684\u5dee\u3002
\u6b63\u56e0\u4e3a\u8fd9\u4e2a\u7406\u8bba\uff0c\u63ed\u793a\u4e86\u79ef\u5206\u4e0e\u9ece\u66fc\u79ef\u5206\u672c\u8d28\u7684\u8054\u7cfb\uff0c\u53ef\u89c1\u5176\u5728\u5fae\u79ef\u5206\u5b66\u4ee5\u81f3\u66f4\u9ad8\u7b49\u7684\u6570\u5b66\u4e0a\u7684\u91cd\u8981\u5730\u4f4d\uff0c\u56e0\u6b64\uff0c\u725b\u987f-\u83b1\u5e03\u5c3c\u5179\u516c\u5f0f\u4e5f\u88ab\u79f0\u4f5c\u5fae\u79ef\u5206\u57fa\u672c\u5b9a\u7406\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u5b9a\u79ef\u5206
0\u52302\u03c0\u662f\u4e00\u4e2a\u5468\u671f\uff0c\u5982\u679cn\u4e3a\u5947\u6570\uff0c\u79ef\u5206\u4e3a\u96f6\u3002
绛旓細璁 p=sin(nx),q=(cosx)^n 鍒檖'=ncos(nx),q'=cos(x+n蟺/2)鈭磞'=p'q+pq'=ncos(nx)路(cosx)^n+sin(nx)路cos(x+n蟺/2)
绛旓細鐩存帴鐢ㄥ叕寮99鍗冲彲锛岀瓟妗堝鍥炬墍绀 涔熷彲浠ョ敤鍏紡95鍜96
绛旓細涓ら閮界粰浣犲仛涓嬶細鈥斺斺斺埆 xsin(nx) dx = (- 1/n)鈭 x d[cos(nx)]= (- x/n)cos(nx) + (1/n)鈭 cos(nx) dx = (- x/n)cos(nx) + (1/n^2)sin(nx) + C 鈭 鈭(- 蟺鈫0) xsin(nx) dx = (- x/n)cos(nx) + (1/n^2)sin(nx)_(x = 0) - (- x/n)...
绛旓細=鈭(-蟺/2,蟺/2) cos^ntdt =2*鈭(0,蟺/2) cos^ntdt 鈭(0,蟺) sin^nxdx-2*鈭(0,蟺/2) sin^nxdx =2*鈭(0,蟺/2) cos^nxdx-2*鈭(0,蟺/2) sin^nxdx =2*鈭(0,蟺/2) [cos^nx-sin^nx]dx 浠=蟺/2-t锛宒x=-dt 鈭(0,蟺) sin^nxdx-2*鈭(0,蟺/2) sin^n...
绛旓細=(sinx+cosx)[1-(sinx+cosx)^2/2-1/2]=(1/2)(sinx+cosx)-(3/2)(sinx+cosx)^3 鏈鍛ㄦ湡2蟺 n>3鏃讹紝n濂囨暟f(x)=(sinx+cosx)^n-g(x) g(x)= (i=1,n-1)鈭慳i (sinx)^i cos^(n-i)鍥犱负sinx+cosx=鈭2sin(x+蟺/4) 鏈灏忔鍛ㄦ湡2蟺 鎵浠(x) 鏈灏忔鍛ㄦ湡2蟺 n...
绛旓細=sin0/cos0=0/1=0
绛旓細绠鍗曡绠椾竴涓嬪嵆鍙紝绛旀濡傚浘鎵绀
绛旓細n=1 f(x)=sinx+cosx=鈭2sin(x+蟺/4) 鍛ㄦ湡2蟺n=2 f(x)=1 涓嶅瓨鍦ㄦ渶灏忔鍛ㄦ湡n=3 f(x)=(sinx)^3+(cosx)^3=(sinx+cosx)(sinx^2+cosx^2-sinxcosx)=(sinx+cosx)(1-sinxcosx)=(sinx+cosx)[1-(sinx+cosx)^2/2-1/2]=(1/2)(sinx...
绛旓細璇佹槑锛氭牴鎹畾绉垎鐨勫嚑浣曟剰涔夌煡鏇茬嚎y=sinnx鍦銆0,蟺/n銆曚笂鐨勯潰绉氨鏄嚱鏁皔=sinnx鍦ㄣ0,蟺/n銆曚笂鐨勫畾绉垎锛屽嵆 s=sinnx鍦ㄣ0,蟺/n銆曚笂鐨勫畾绉垎 =-(1/n)cosnx| [0,蟺/n]=-(1/n)cos[n*(蟺/n)]+(1/n)cos0 =-(1/n)cos蟺+1/n =1/n+1/n =2/n 鏁厃=sinnx鍦ㄣ0,蟺/n...
绛旓細cos^nx鐨勭Н鍒嗙殑涓鑸舰寮忎负鈭玞os?(x)dx銆傚浜庣Н鍒嗏埆cos?(x)dx锛屽叾涓璶鏄竴涓鏁存暟锛岃冭檻浠ヤ笅涓ょ鎯呭喌锛1銆佸綋n=1鏃讹細鈭玞os(x)dx=sin(x)+C锛屽叾涓瑿鏄Н鍒嗗父鏁般2銆佸綋n鈮1鏃讹細娌℃湁绠鍗曠殑闂紡琛ㄨ揪寮忔潵琛ㄧず鈭玞os?(x)dx銆傚綋n鏄鏁版椂锛屽彲浠ヤ娇鐢ㄧ壒娈婂嚱鏁帮紙濡傝秴鍑犱綍鍑芥暟鎴栬礉濉炲皵鍑芥暟锛夋潵琛ㄧず绉垎...
