cos^2x求不定积分 cos^2x/2的不定积分？
cos^2X\u539f\u51fd\u6570\u662f\u4ec0\u4e48cos^2x\u7684\u539f\u51fd\u6570\u4e3a1/2x+1/4sin2x+C
\u89e3\uff1a\u4ee4F(x)\u4e3acos^2x\u7684\u539f\u51fd\u6570\u3002\u90a3\u4e48
F(x)=\u222bcos^2xdx
=\u222b(cos2x+1)/2dx
=1/2\u222bcos2xdx+1/2\u222b1dx
=1/4\u222bcos2xd(2x)+1/2\u222b1dx
=1/2x+1/4sin2x+C
\u6269\u5c55\u8d44\u6599\uff1a
1\u3001\u4e09\u89d2\u51fd\u6570\u57fa\u672c\u516c\u5f0f
\uff081\uff09\u4e09\u89d2\u51fd\u6570\u4e4b\u95f4\u53d8\u6362
tanx=sinx/cosx\u3001cotx=cosx/sinx\u3001secx=1/cosx\u3001cscx=1/sinx\u3001
tanx*cotx=1\u30011=(sinx)^2+(cosx)^2\u3002
\uff082\uff09\u4e8c\u500d\u89d2\u516c\u5f0f
cos2x=2cos²x-1=1-2sin²x\u3001cos²x=(cos2x+1)/2\u3001sin²x=(1-c0s2x)/2
2\u3001\u79ef\u5206\u516c\u5f0f\u6cd5\u6c42\u89e3\u4e0d\u5b9a\u79ef\u5206
\u76f4\u63a5\u5229\u7528\u79ef\u5206\u516c\u5f0f\u6c42\u51fa\u4e0d\u5b9a\u79ef\u5206\u3002
\u4f8b\uff1a\u222be^xdx=e^x\u3001\u222b1/xdx=ln|x|+C\u3001\u222bcosxdx=sinx+C\u3001\u222bsinxdx=-cosx+C
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u4e0d\u5b9a\u79ef\u5206
\u4f7f\u7528\u500d\u89d2\u516c\u5f0f
cos(x)^2/2=(1+cos2x)/4\uff0c\u7136\u540e\u79ef\u5206\u5f97\u51fa(x+sin2x/2)/4+C
∫cos^2xdx
=∫（1+cos2x）dx/2
=∫(1+cos2x)d2x/4
=(1/4)∫[d2x+cos2xd2x]
=(1/4){2x+sin2x+C1}
=x/2+(sin2x)/4+C
不定积分的公式
1、∫ a dx = ax + C，a和C都是常数
2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 a ≠ -1
3、∫ 1/x dx = ln|x| + C
4、∫ a^x dx = (1/lna)a^x + C，其中a > 0 且 a ≠ 1
5、∫ e^x dx = e^x + C
6、∫ cosx dx = sinx + C
7、∫ sinx dx = - cosx + C
8、∫ cotx dx = ln|sinx| + C = - ln|cscx| + C
∫cos^2xdx
=∫（1+cos2x）dx/2
=∫(1+cos2x)d2x/4
=(1/4)∫[d2x+cos2xd2x]
=(1/4){2x+sin2x+C1}
=x/2+(sin2x)/4+C
绛旓細鈭cos2xdx = 1/2 鈭玞os2xd(2x) = 1/2 sin2x+C锛屸埆cos²xdx = 鈭(1+cos2x)/2 dx = x/2 + 1/4 sin2x + C 銆
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绛旓細鎵浠os2x鐨勪笉瀹氱Н鍒嗘槸(1/2)sin2x+C銆
绛旓細cos2x鐨勪笉瀹氱Н鍒嗘槸(1/2)sin2x+C銆傗埆cos2xdx =(1/2)鈭玞os2xd2x =(1/2)sin2x+C 鎵浠os2x鐨勪笉瀹氱Н鍒嗘槸(1/2)sin2x+C銆傝В閲婏細鏍规嵁鐗涢】-鑾卞竷灏艰尐鍏紡锛岃澶氬嚱鏁扮殑瀹氱Н鍒嗙殑璁＄畻灏卞彲浠ョ畝渚垮湴閫氳繃姹備笉瀹氱Н鍒嗘潵杩涜銆傝繖閲岃娉ㄦ剰涓嶅畾绉垎涓庡畾绉垎涔嬮棿鐨勫叧绯伙細瀹氱Н鍒嗘槸涓涓暟锛岃屼笉瀹氱Н鍒嗘槸涓涓〃...
绛旓細瑙ｏ細cos^2x=(1+cos2x)/2锛屾墍浠モ埆cos^2x dx =鈭(1+cos2x)/2dx =x/2+sin2x/4+C锛孋涓绉垎甯告暟銆
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绛旓細鍥炵瓟锛氣埆cos^2xdx =鈭(1+cos2x)dx/2 =鈭(1+cos2x)d2x/4 =(1/4)鈭玔d2x+cos2xd2x] =(1/4){2x+sin2x+C1} =x/2+(sin2x)/4+C
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绛旓細cos^2x=(1+cos2x)/2 鈭粹埆 cos^2x dx=鈭 (1+cos2x)/2 dx =(1/4) 鈭 (1+cos2x) d2x =(2x+sin2x+C)/4
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