求∫﹙sin²x+cosx﹚dx的不定积分
∫﹙sin²x+cosx﹚dx=∫[(1-cos2x)/2+cosx]dx
=(1/2) ∫ (1-cos2x)dx+∫ cosxdx
=(1/2) [x-(sin2x)/2]+sinx+C
=x/2-(sin2x)/4+sinx+C
绛旓細濡傚浘鎵绀恒
绛旓細=(1/2)x-(1/4)sin2x+C(C涓哄父鏁帮級鍒嗛儴绉垎锛(uv)'=u'v+uv'寰楋細u'v=(uv)'-uv'涓よ竟绉垎寰楋細鈭玼'v dx=鈭紙uv)' dx -鈭玼v' dx銆傚嵆锛氣埆u'v dx = uv -鈭玼v' d,杩欏氨鏄垎閮ㄧН鍒嗗叕寮忋備篃鍙畝鍐欎负锛氣埆v du = uv -鈭玼 dv銆傚垎閮ㄧН鍒嗘硶鐨勫疄璐細灏嗘墍姹傜Н鍒嗗寲涓轰袱涓Н鍒嗕箣宸紝...
绛旓細璁＄畻杩囩▼濡備笅锛氣埆锛0锝炏/2锛sin蠁cos鈭3蠁d蠁 =-鈭蓟0锝炏/2锛絚os鈭3蠁dcos蠁 =-(1/4)cos^4蠁(0锝炏/2)=1/4
绛旓細sin^3=sin*(1-cos^2)=sin-sin*cos^2.鎵浠in^3sx=(sin-sin*cos^2)dx=-cosx+cos^2dcos=-cos+1/3cos^3+c
绛旓細浠も垰x锛漷 鈭玸in鈭歺dx 锛2鈭玹sintdt 锛濓紞2鈭玹dcost 锛濓紞2tcost锛2鈭玞ostdt 锛濓紞2tcost锛2sint锛婥 锛濓紞2鈭歺cos鈭歺锛2sin鈭歺锛婥
绛旓細= (1/2)鈭玸inxdx + (1/2)鈭玞os2x sinxdx = -(1/2)cosx + (1/4)sin2x 鎺ヤ笅鏉ユ垜浠妸姣忎竴姝ョ殑缁撴灉鍚堣捣鏉ワ紝寰楀埌锛鈭(sinx)^6 dx = x - 3/4sinx + 3/32sin2x - 1/32sin3x + 1/192sin4x - 1/192sin5x + C 鍏朵腑C鏄竴涓父鏁帮紝鍙互鏄换鎰忓疄鏁般傝繖灏辨槸鎵姹傜殑閫氳В銆
绛旓細杩囩▼濡備笅锛鈭玸in^2xdx sin^2x=(1-cos2x)/2 鈭玸in^2xdx =1/2鈭1dx-1/2鈭玞os2xdx =x/2-1/4鈭玞os2xd2x =x/2-sin2x/4+C
绛旓細鐢ㄧ浜岀被鎹㈠厓娉曪紝灏嗙Н鍒嗗紡涓殑cosx鍙樺埌dx鍚庨潰鍙樻垚sinx锛屽嵆锛岀Н鍒嗗彿sinx^2dsinx 锛堝洜涓锛坰inx锛'=cosx)杩欐牱锛岀敤绗竴绫绘崲鍏冩硶锛屽皢sinx=t锛屽垯绉垎寮忓彉鎴愮Н鍒嗗彿t^2dt 杩欎釜灏辨槸鍩烘湰鐨勭Н鍒嗗紡浜嗭紝骞傜殑绉垎锛屾牴鎹(x^m)'=(m-1)x^(m-1)鏈鍚庣Н鍒嗗緱鍒1/3t^3 鍗1/3sinx^3 涓嶇煡鍥炵瓟婊℃剰鍚楋紝鐧惧害...
绛旓細濡傚浘
绛旓細鈭玸in^2xdx鐨勪笉瀹氱Н鍒嗘槸x/2-sin2x/4+C銆傗埆sin^2xdxsin^2x=(1-cos2x)/2鍒欌埆sin^2xdx =1/2鈭1dx-1/2鈭玞os2xdx =x/2-1/4鈭玞os2xd2x =x/2-sin2x/4+C 鎵浠モ埆sin^2xdx鐨勪笉瀹氱Н鍒嗘槸x/2-sin2x/4+C銆
