sin(x)*cos(x)不定积分问题 求(cos/(sin x+cos x)的不定积分
sin(x).cos(x)\u7684\u5012\u6570\u7684\u4e0d\u5b9a\u79ef\u5206\u600e\u4e48\u7b97
\u6bd4\u8f83\u9ebb\u70e6\uff0c\u5e0c\u671b\u9a8c\u7b97\u4e00\u4e0b
\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
你那两个答案都是sin(x)*cos(x)的原函数，而且只要差个常数都是它的原函数。不过写成+C的形式才是真正正确的。
sinxcosx=(1/2)sin2x，则：∫[sinxcosx]dx=－(1/4)cos2x＋C，其中C是常数。
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