定积分∫(-2,2)xsin^4xdx怎么求?
这个不用具体求，这是奇函数在对称区间上得定积分，结果就是0，这是定积分基本性质之一绛旓細杩欎釜涓嶇敤鍏蜂綋姹傦紝杩欐槸濂囧嚱鏁板湪瀵圭О鍖洪棿涓婂緱瀹氱Н鍒锛岀粨鏋滃氨鏄0锛岃繖鏄畾绉垎鍩烘湰鎬ц川涔嬩竴
绛旓細瑙ｆ硶涓 鈭玿sin^2x dx =鈭玿(cos2x+1)/2dx =(1/4)鈭玿cos2xd2x =(1/4)鈭玿dsin2x =(1/4)xsin2x-(1/4)鈭玸in2xdx =(1/4)xsin2x-(1/8)鈭玸in2xd2x =(1/4)xsin2x+(1/8)cos2x [-蟺/2,蟺/2]=[(1/4)(蟺/2)*sin蟺+(1/8)cos蟺]-[(1/4)(-蟺/2)*sin(-蟺)+...
绛旓細鈭玿sin²x dx =鈭玿*(1-cos2x)/2 dx锛屽埄鐢ㄤ笁瑙掑嚱鏁版亽绛夊紡cos2x=1-2sin²x =(1/2)鈭x dx - (1/2)鈭玿cos2x dx =(1/2)(x²/2) - (1/2)(1/2)鈭玿cos2x d(2x)锛屽噾寰垎 =x²/4 - (1/4)鈭玿 d(sin2x)锛屽噾寰垎 =x²/4 - (1/4)[xsin2...
绛旓細xsinx^2=x*(1-cos2x)/2 鍏朵腑xcos2x锛岄噰鐢ㄥ垎閮ㄧН鍒嗘硶 x/2绉垎涓簒^2/4 渚嬪锛氳x=鈭2sint锛屽垯dx=鈭2costdt.(璇存槑锛氣埆(a,b)琛ㄧず浠巃鍒癰绉垎)鈭村師瀹氱Н鍒=鈭(0,蟺/2)[2cos²t]dt =鈭(0,蟺/2)[1+cos(2t)]dt =[t+1/2sin(2t)]|(0,蟺/2)=蟺/2+1/2sin蟺-0-1/...
绛旓細鈭xsin(x^2)dx =(1/2)鈭玸in(x^2) dx^2 =-(1/2)cos(x^2) + C 涓涓嚱鏁帮紝鍙互瀛樺湪涓瀹氱Н鍒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畾绉垎鍜屼笉瀹氱Н鍒嗐
绛旓細鈭玿sin^2xdx = (1/2)鈭x(1-cos2x)dx = (1/4)x^2 - (1/4)鈭玿dsin2x = (1/4)x^2 - (1/4)xsin2x + (1/4)鈭玸in2xdx = (1/4)x^2 - (1/4)xsin2x - (1/8)cos2x + C
绛旓細杩欐槸鐢ㄤ簡瀹氱Н鍒鍖洪棿鍐嶇幇鍏紡锛屽彲浠ヤ笉鏀瑰彉绉垎鍖哄煙鐨勬儏鍐典笅瀵硅绉嚱鏁拌繘琛屾敼閫狅紝鏂逛究绉垎璁＄畻銆傗埆(0,蟺)xsin²xdx =鈭(0,蟺)(蟺-x)sin²(蟺-x)dx =鈭(0,蟺)(蟺-x)sin²(x)dx =鈭(0,蟺)[蟺sin²(x)dx-鈭(0,蟺)xsin²xdx 鈭2鈭(0,蟺)xsin²xdx...
绛旓細鈭玿sin²xdx=(1/2)鈭玿(1-cos2x)dx=(1/2)鈭玿dx-(1/2)鈭 xcos2xdx =(1/4)x^2-(1/4)鈭 xd(sin2x)=(1/4)x^2-(1/4)xsin2x+(1/4)鈭 (sin2x)dx =(1/4)x^2-(1/4)xsin2x-(1/8)cos2x+C
绛旓細鐢ㄥ垎閮绉垎娉 鈭玿sin^2x dx =1/2鈭x(1-cos2x)dx=1/2(鈭玿dx -鈭玿cos2x dx)=1/2(1/2*x^2-1/2鈭玿 dsin2x)=1/4(x^2-xsin2x+鈭玸in2x dx )=1/4(x^2-xsin2x-1/2cos2x)+C
绛旓細"鍘熷紡=1/2*锛堚埆[x(1-cos2x)]dx)=1/2[鈭玿dx-1/2*鈭玿dsin2x]=1/2[(蟺--蟺)-1/2*(xsin2x-鈭sin2xdx)]=1/2*[2蟺-1/2*(0+cos2x/2)]=蟺"
