求解高等数学不定积分题目∫x^2sin2xdx。用分部积分法！ 分部积分法做出∫xsin2xdx
\u7528\u5206\u90e8\u79ef\u5206\u6cd5\u6c42\u4e0d\u5b9a\u79ef\u5206\u222bxsin2xdx \u522b\u8df3\u6b65\u9aa4!\u5982\u56fe\u6240\u793a\uff1a
\u222bxsin2xdx
=(-1/2)\u222bxdcos2x
=(-1/2)(xcos2x-\u222bcos2xdx)
=(-1/2)(xcos2x-(1/2)sin2x)+C
=(1/4)sin2x-(1/2)xcos2x+C
\u6269\u5c55\u8d44\u6599\u4e0d\u5b9a\u79ef\u5206\u7684\u516c\u5f0f
1\u3001\u222b a dx = ax + C\uff0ca\u548cC\u90fd\u662f\u5e38\u6570
2\u3001\u222b x^a dx = [x^(a + 1)]/(a + 1) + C\uff0c\u5176\u4e2da\u4e3a\u5e38\u6570\u4e14 a \u2260 -1
3\u3001\u222b 1/x dx = ln|x| + C
4\u3001\u222b a^x dx = (1/lna)a^x + C\uff0c\u5176\u4e2da > 0 \u4e14 a \u2260 1
5\u3001\u222b e^x dx = e^x + C
6\u3001\u222b cosx dx = sinx + C
7\u3001\u222b sinx dx = - cosx + C
8\u3001\u222b cotx dx = ln|sinx| + C = - ln|cscx| + C
9\u3001\u222b tanx dx = - ln|cosx| + C = ln|secx| + C
10\u3001\u222b secx dx =ln|cot(x/2)| + C = (1/2)ln|(1 + sinx)/(1 - sinx)| + C = - ln|secx - tanx| + C = ln|secx + tanx| + C
∫x^2sin2xdx
=-1/2∫x^2d（cos2x）
=-1/2[cos2x*x^2-∫2x*cos2xdx]
=-1/2[cos2x*x^2-∫xd（sin2x）]
=-1/2[cos2x*x^2-（sin2x*x-∫sin2xdx）]
=-1/2cos2x*x^2+1/2sin2x*x-1/2∫sin2xdx
=-1/2cos2x*x^2+1/2sin2x*x+1/4cos2x+C
【数学辅导团】为您解答，不理解请追问，理解请及时选为满意回答！(*^__^*)谢谢！
具体过程见图
绛旓細浣犲ソ 鈭玿^2sin2xdx =-1/2鈭玿^2d锛坈os2x锛=-1/2[cos2x*x^2-鈭2x*cos2xdx]=-1/2[cos2x*x^2-鈭玿d锛坰in2x锛塢=-1/2[cos2x*x^2-锛坰in2x*x-鈭玸in2xdx锛塢=-1/2cos2x*x^2+1/2sin2x*x-1/2鈭玸in2xdx =-1/2cos2x*x^2+1/2sin2x*x+1/4cos2x+C 銆鏁板杈呭鍥戜负鎮...
绛旓細鈭 鈭(a^2 - x^2) dx = (x/2)鈭(a^2 - x^2) + (a^2/2)arcsin(x/a) + C銆姹傝В 璁綟(x)鏄嚱鏁癴(x)鐨勪竴涓鍘熷嚱鏁锛屾垜浠妸鍑芥暟f(x)鐨勬墍鏈夊師鍑芥暟F(x)+ C(鍏朵腑锛孋涓轰换鎰忓父鏁帮級鍙仛鍑芥暟f(x)鐨涓嶅畾绉垎锛屽張鍙仛鍑芥暟f(x)鐨勫弽瀵兼暟锛岃浣溾埆f(x)dx鎴栬呪埆f锛楂樼瓑寰Н鍒涓父...
绛旓細= - 1/[3(x³ + 1)] + C锛屽皢涔嬪墠鐨剈 = x³ + 1鍥炰唬 鍚庨潰閭ｉ涔熸槸銆備护u = x² + 1 u' du = (x² + 1)' dx du = 2x dx x dx = (1/2) du 鈭 x(x² + 1)³ dx = 鈭 (x² + 1)³ 路 (x dx) = 鈭 u³...
绛旓細鈭玔x³/(x+3)]dx涓嶅畾绉垎瑙ｇ瓟濡備笅锛氭涓烘湁鐞嗗垎寮忕殑绉垎銆
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細
绛旓細涔︿笂鐨勭瓟妗堟槸瀵圭殑銆傝繖涓绉垎鐢变袱椤圭粍鎴愶紝绗竴椤圭殑绉垎浣犲凡缁忓仛瀵逛簡锛岀浜岄」搴旇鐢ㄥ垎閮ㄧН鍒嗘硶锛姹傝В杩囩▼濡傚浘绀恒
绛旓細濡傚浘
绛旓細鈭(x^3)[e^(-x^2)]dx = (1/2)鈭(x^2)[e^(-x^2)]d(x^2)= (1/2)鈭玹[e^(-t)]dt (t = x^2)= (-1/2)鈭玹d[e^(-t)]= (-1/2){te^(-t) - 鈭玔e^(-t)]dt} (鍒嗛儴绉垎娉)= (-1/2){te^(-t) + [e^(-t)]} + C = 鈥︹
绛旓細锛2锛夆埆锛1锛4x锛塣3dx锛濓紞锛1/4锛夆埆锛1锛4x锛塣3d锛1锛4x锛夛紳锛嶏紙1/16锛夛紙1锛4x锛塣5锛婥銆傦紙3锛夆埆e^x路cos锛坋^x锛塪x锛濃埆cos锛坋^x锛塪锛坋^x锛夛紳sin锛坋^x锛夛紜C銆傦紙9锛鈭玿^4路sin锛4x^5锛塪x锛濓紙1/20锛夆埆sin锛4x^5锛塪锛4x^5锛夛紳锛嶏紙1/20锛塩os锛4x^5锛夛紜C銆
绛旓細濡傚浘鎵绀猴細
