谁知道不定积分∫1/(sin^2xcos^2x)dx是多少 求不定积分∫(cos2x)/(sin^2x)(cos^2x)...
\u6c42\u4e0d\u5b9a\u79ef\u5206(1/sin^2xcos^2x)dx\u539f\u5f0f=\u222b4dx/(2sinxcosx)²
=4\u222bdx/sin²2x
=2\u222bcsc²2xd2x
=-2cot2x+C
1\u3002\u5c06\u5206\u6bcd\u53d8\u4e3asin2x\u5373\u539f\u5f0f\u4e3a\u222b[\uff084cos2x/sin^2\uff082x\uff09\uff09]dx
2.\u8fdb\u884c\u6362\u5143\u53732x\u53d8\u4e3at\uff0c\u539f\u5f0f\u53d8\u4e3a\u222b[\uff082cos2x/sin^2t\uff09]dt.
3\u7ee7\u7eed\u6362\u5143\uff0c\u53ef\u89c2\u5bdf\u5230\uff08sin
t\uff09'=cost\u3002\u6240\u4ee5\u539f\u5f0f\u7b49\u4e8e2\u222b[\uff081/sin^2t]d\uff08sint\uff09.
4.\u5f97\u51fa\u7b54\u6848\u4e3a\uff1a\uff08-2/sint\uff09+c
5.\u5c06t\u6362\u56de\u4e3a2x\u6709\uff08-2/sin2x\uff09+c\u3002
\u624b\u6253\u5f88\u7d2f\uff0c\u671b\u91c7\u7eb3\u3002
∫1/(sin^2xcos^2x)dx=-2cot2x+C。
解答过程如下：
∫1/(sin^2xcos^2x)dx
=∫dx/(sinxcosx)^2
=∫4dx/(sin2x)^2
=2∫d2x/(sin2x)^2
=2∫(csc2x)^2 d2x
= -2cot2x+C
扩展资料：
分部积分：
(uv)'=u'v+uv'
得：u'v=(uv)'-uv'
两边积分得：∫ u'v dx=∫ (uv)' dx - ∫ uv' dx
即：∫ u'v dx = uv - ∫ uv' d,这就是分部积分公式
也可简写为：∫ v du = uv - ∫ u dv
常用积分公式：
1）∫0dx=c
2）∫x^udx=(x^(u+1))/(u+1)+c
3）∫1/xdx=ln|x|+c
4）∫a^xdx=(a^x)/lna+c
5）∫e^xdx=e^x+c
6）∫sinxdx=-cosx+c
7）∫cosxdx=sinx+c
8）∫1/(cosx)^2dx=tanx+c
9）∫1/(sinx)^2dx=-cotx+c
10）∫1/√（1-x^2) dx=arcsinx+c
(sinx*cosx)^2=0.25*sin(2x)^2
积分=-2/sin(2*x)*cos(2*x)+C
原式=∫1/(1+(cosx)^2) dx 分子分母同除以(cosx)^2
=∫(secx)^2/((secx)^2+1) dx
=∫1/((secx)^2+1) d (tanx)
=∫1/((tanx)^2+2) d (tanx)
套公式
=1/√2*arctan((tanx)/√2)+C
∫1/sin²xcos²x dx
=∫1/sin²x dx+∫1/cos²x dx
=－cotx + tanx + c
=tanx－cotx + c
是平方吗
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