(sin(x)+cos(x)）分之一的求积分怎么算？ 1/（1+cosx)的积分怎么算？
\u6c42(cos/(sin x+cos x)\u7684\u4e0d\u5b9a\u79ef\u5206\u222bcosx/(sin x+cos x)dx=1/2x+1/2ln\u4e28cosx+sinx\u4e28+C\u3002C\u4e3a\u5e38\u6570\u3002
\u89e3\u7b54\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u222bcosx/(sin x+cos x)dx
=1/2\u222b[\uff08cosx+sinx\uff09+\uff08cosx-sinx\uff09]/\uff08cosx+sinx\uff09dx
=1/2\u222b1+\uff08cosx-sinx\uff09/\uff08cosx+sinx\uff09dx
=1/2x+1/2\u222b\uff08cosx-sinx\uff09/\uff08cosx+sinx\uff09dx
=1/2x+1/2\u222b1/\uff08cosx+sinx\uff09d\uff08cosx+sinx\uff09
=1/2x+1/2ln\u4e28cosx+sinx\u4e28+C
\u6269\u5c55\u8d44\u6599\uff1a
\u5206\u90e8\u79ef\u5206\uff1a
(uv)'=u'v+uv'
\u5f97\uff1au'v=(uv)'-uv'
\u4e24\u8fb9\u79ef\u5206\u5f97\uff1a\u222b u'v dx=\u222b (uv)' dx - \u222b uv' dx
\u5373\uff1a\u222b u'v dx = uv - \u222b uv' d,\u8fd9\u5c31\u662f\u5206\u90e8\u79ef\u5206\u516c\u5f0f
\u4e5f\u53ef\u7b80\u5199\u4e3a\uff1a\u222b v du = uv - \u222b u dv
\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
1/\uff081+cosx)\u7684\u79ef\u5206\u7b97\u6cd5\u5982\u4e0b\uff1a
1+cosx=2[cos(x/2)]^2
1/(1+cosx)=0.5[sec(x/2)]^2
\u222bdx/(1+cosx)
=\u222b0.5[sec(x/2)]^2dx
=\u222b[sec(x/2)]^2d0.5x
=\u222bdtan(x/2)
=tan(x/2)+c
\u6269\u5c55\u8d44\u6599\uff1a
\u4e8c\u500d\u89d2\u516c\u5f0f\u662f\u6570\u5b66\u4e09\u89d2\u51fd\u6570\u4e2d\u5e38\u7528\u7684\u4e00\u7ec4\u516c\u5f0f\uff0c\u901a\u8fc7\u89d2\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u7684\u4e00\u4e9b\u53d8\u6362\u5173\u7cfb\u6765\u8868\u793a\u5176\u4e8c\u500d\u89d22\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\uff0c\u4e8c\u500d\u89d2\u516c\u5f0f\u5305\u62ec\u6b63\u5f26\u4e8c\u500d\u89d2\u516c\u5f0f\u3001\u4f59\u5f26\u4e8c\u500d\u89d2\u516c\u5f0f\u4ee5\u53ca\u6b63\u5207\u4e8c\u500d\u89d2\u516c\u5f0f\u3002\u5728\u8ba1\u7b97\u4e2d\u53ef\u4ee5\u7528\u6765\u5316\u7b80\u8ba1\u7b97\u5f0f\u3001\u51cf\u5c11\u6c42\u4e09\u89d2\u51fd\u6570\u7684\u6b21\u6570\uff0c\u5728\u5de5\u7a0b\u4e2d\u4e5f\u6709\u5e7f\u6cdb\u7684\u8fd0\u7528\u3002
\u6b63\u5f26\u4e8c\u500d\u89d2\u516c\u5f0f\uff1asin2\u03b1 = 2cos\u03b1sin\u03b1
\u63a8\u5bfc\uff1asin2\u03b1 = sin(\u03b1+\u03b1) = sin\u03b1cos\u03b1 + cos\u03b1sin\u03b1= 2sin\u03b1cos\u03b1
\u4f59\u5f26\u4e8c\u500d\u89d2\u516c\u5f0f\uff1a\u4f59\u5f26\u4e8c\u500d\u89d2\u516c\u5f0f\u6709\u4e09\u7ec4\u8868\u793a\u5f62\u5f0f,\u4e09\u7ec4\u5f62\u5f0f\u7b49\u4ef7\uff1a
\uff081\uff09cos2\u03b1 = 2cos^2 \u03b1- 1
\uff082\uff09cos2\u03b1 = 1 − 2sin^2 \u03b1
\uff083\uff09cos2\u03b1 = cos^2 \u03b1 − sin^2 \u03b1
\u63a8\u5bfc\uff1acos2A = cos(A+A) = cosAcosA - sinAsinA = cos^2 A- sin^2 A = 2cos^2 A - 1=1 - 2sin^2 A
\u53c2\u8003\u8d44\u6599\uff1a\u767e\u5ea6\u767e\u79d1\u2014\u2014\u4e8c\u500d\u89d2\u516c\u5f0f
令t=x+∏/4
则等价于求1/sint的积分
dt/sint=dt*sint/sin^2t
=-dcost/(1-cos^2t)
再令cost=s
则等价于-ds/(1-s^2)
这是个初等积分,原函数为0.5[ln(s-1)-ln(s+1)]
最后把上述所有代换代回去,即得原答案
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