请教一个积分题1/(cosx+sinx)的积分怎么求呢？只能用万能公式吗？
把cosx+sinx化成根号2sin（x+π/4）之后就可以求出来了。绛旓細鍋囪x=a^(1/2)tan(t)閭ｄ箞鍘熷紡=鈭(a+x^2)^(3/2)dx=鈭(a*(sec(t))^2)^(3/2)*d(a^(1/2)tan(t))=鈭玜^(3/2)*(sec(t))^3*d(a^(1/2)tan(t))=鈭玜^2*(sec(t))^5*dt =鈭玜^2*dt/(cos(t))^5 =鈭玜^2*d(sint)/(cos(t))^6 =a^2*鈭玠(sint)/(1-(...
绛旓細锛1锛変护x=tant锛屽垯dx=sec^2tdt 鍘熷紡=鈭玪n(tant+sect)/sect*sec^2tdt =鈭玪n(tant+sect)*sectdt =鈭玪n(tant+sect)d[ln(tant+sect)]=(1/2)*[ln(tant+sect)]^2+C =(1/2)*[ln(x+鈭(1+x^2)]^2+C锛屽叾涓瑿鏄换鎰忓父鏁 锛2锛夊師寮=鈭玡^(sin2x-2x)*sin^2xdx =(1/2)*鈭玡...
绛旓細鍚庨潰鐨绉垎绠楀嚭鏉ョ殑缁撴灉鏄涓涓鏁帮紒璁 a=鈭紙0鍒1锛塮(x)dx 閭ｄ箞f(x)=e^x+a 鈭(0鍒1)f(x)dx =鈭(0鍒1)e^x dx + [鈭(0鍒1)a dx]/e =e-1+a/e=a 鎵浠 e-1=a(1-1/e)=a(e-1)/e a/e=1 鎵浠 a=e f(x)=e^x+e/e=e^x+1 ...
绛旓細1锛寈e^x鏄痜(x)鐨涓涓鍘熷嚱鏁 锛屽嵆锛 鈭玣(x)dx=xe^x+C锛屾墍浠モ埆f(3x)dx=1/3*鈭玣(3x)d(3x)=1/3*3xe^(3x)+C=xe^(3x)+C 2锛宔^(-x^2)鏄痜(x)鐨勪竴涓師鍑芥暟锛屽嵆锛 鈭玣(x)dx=e^(-x^2)+C锛屸埆xf'(x)dx=鈭玿f'(x)dx=-1/2*鈭玣'(x)d(-x^2)=-1/2*e^(-x^...
绛旓細鈭玔1/5-4cos2x)]dx=(1/2)鈭玔1/5+(-4)cos2x]d2x.={(1/2)*2/鈭歔(5^2-(-4)^2]}*arctan{鈭歔(5-(-4))/5+(-4)]tan(2x/2}+C.=(1/3)*rarctan(3*tanx)+C.
绛旓細1锛夊洜涓猴細f鈥筹紙尉锛夛紴0锛屼笖f鈥诧紙尉锛=0锛屾墍浠ノ句负鍑芥暟f锛坸锛夌殑鏋佸皬鍊肩偣锛岀敱浜巉锛坸锛夆垐C[a锛屛綸锛屾墍浠ュ湪[a锛屛綸涓婃湁鏈澶у糵锛坱1锛夛紝鍚岀悊锛宖锛坸锛夆垐C[尉锛宐]锛屾墍浠ュ湪[尉锛宐]涓婃湁鏈澶у糵锛坱2锛夛紝涓嶅Θ璁緁锛坱1锛夆墹f锛坱2锛夛紟鍒欐湁f锛坱1锛夛紴f锛埼撅級锛
绛旓細鍦ㄥ井绉垎涓紝涓涓鍑芥暟f 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鍒嗐傘庝緥瀛愪竴銆 2鈭 x dx = x^2 +C 銆庝緥瀛愪簩銆 鈭 cosx dx = sinx +C 銆庝緥瀛愪笁銆忊埆 x^2 dx = (1/3)x^3...
绛旓細=銆0,1銆戔埆xe^x²dx+銆0,1銆戔埆ye^y²dy=銆0,1銆(1/2)鈭玡^x²d(x²)+銆0,1銆(1/2)鈭玡^y²d(y²)=(1/2)[e^x²+e^y²]銆0,1銆=(1/2)(2e-2)=e-1.銆愭敞鎰忓墠涓绉垎閲岋紝琚Н鍑芥暟鏄痚^x²锛岃岀Н鍒嗗彉閲忔槸y锛屽悗涓绉...
绛旓細鍙互杩欐牱鏉ュ仛鐨勶細鍒嗗瓙鍒嗘瘝鍚岄櫎浠^2,鍘熻绉嚱鏁板氨鍙互鍐欐垚(1+1/x^2)/(x^2+1/x^2),杩涗竴姝ュ彉涓涓嬪舰锛屽垎姣嶅彲浠ュ彉鎴(x-1/x)^2+2,鍒嗗瓙鍙互閫氳繃鍑戝井鍒嗗彉鎴恉(x-1/x),浜庢槸鍘熻绉嚱鏁板氨琚啓鎴 d(x-1/x)/[(x-1/x)^2+2]锛屾帴涓嬫潵鎴戠浉淇′綘灏变細鍋氫簡銆傚鏋滆繕涓嶄細璇疯拷闂 ...
绛旓細鎹㈠厓绉垎娉曞拰鍒嗛儴绉垎娉曠粨鍚 鍏堝垎閮ㄧН鍒嗭細鈭玔arcsin(e^x)]/e^x dx 锛濓紞鈭玔arcsin(e^x)] d[e^(-x)]锛濓紞e^(-x)脳arcsin(e^x)锛嬧埆e^(-x)脳1/鈭歔1-e^(2x)]脳e^xdx 锛濓紞e^(-x)脳arcsin(e^x)锛嬧埆1/鈭歔1-e^(2x)]dx 鍐嶆崲鍏冿細浠^x锛漵int锛屽垯 鈭1/鈭歔1-e^(2x)]...
