sin平方分之一的积分怎么算啊，∫1/(sin^2 x)dx ∫1/(2+sin^2x)dx
sin\u5e73\u65b9\u5206\u4e4b\u4e00\u7684\u79ef\u5206\u600e\u4e48\u7b97\u554a\uff0c\u222b1/(sin^2 x)dx-cotx+c\uff0cc\u662f\u5e38\u6570\u3002
(tanx)'=(secx)^2
(cotx)'=-(cscx)^2=-1/(sinx)^2
\u222b1/(sin^2 x)dx=-cotx+c
\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
\u222b1/(2+sin^2x)dx\u7684\u89e3\u7b54\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u5176\u4e2dsin²x+cos²x=1\uff0c2=2\u00d71=2\u00d7\uff08sin²x+cos²x\uff09\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u7684\u57fa\u672c\u5173\u7cfb\u5f0f
\u5012\u6570\u5173\u7cfb\uff1atan\u03b1 \u00b7cot\u03b1=1\u3001sin\u03b1 \u00b7csc\u03b1=1\u3001cos\u03b1 \u00b7sec\u03b1=1\uff1b
\u5546\u7684\u5173\u7cfb\uff1a sin\u03b1/cos\u03b1=tan\u03b1=sec\u03b1/csc\u03b1\u3001cos\u03b1/sin\u03b1=cot\u03b1=csc\u03b1/sec\u03b1\uff1b
\u548c\u7684\u5173\u7cfb\uff1asin²\u03b1+cos²\u03b1=1\u30011+tan²\u03b1=sec²\u03b1\u30011+cot²\u03b1=csc²\u03b1\uff1b
\u5e73\u65b9\u5173\u7cfb\uff1asin²\u03b1+cos²\u03b1=1\u3002
\u5e38\u7528\u79ef\u5206\u516c\u5f0f\uff1a
1\uff09\u222b0dx=c
2\uff09\u222bx^udx=(x^(u+1))/(u+1)+c
3\uff09\u222b1/xdx=ln|x|+c
4\uff09\u222ba^xdx=(a^x)/lna+c
5\uff09\u222be^xdx=e^x+c
6\uff09\u222bsinxdx=-cosx+c
7\uff09\u222bcosxdx=sinx+c
8\uff09\u222b1/(cosx)^2dx=tanx+c
9\uff09\u222b1/(sinx)^2dx=-cotx+c
10\uff09\u222b1/\u221a\uff081-x^2) dx=arcsinx+c
11\uff09\u222b1/(1+x^2)dx=arctanx+c
12\uff09\u222b1/(a^2-x^2)dx=(1/2a)ln|(a+x)/(a-x)|+c
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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