求 ∫ sin²(x)cosxdx ∫ 下面是0上面是π/2 多谢 步骤请写详细一点 不胜感激! ∫sin2x·cosdx=? ∫1.0 x^2/(1+x^2...

\u6c42 \u222b cos(x)sin²(x)dx \u222b \u4e0b\u9762\u662f0\u4e0a\u9762\u662f\u03c0(pai) \u591a\u8c22 \u6b65\u9aa4\u8bf7\u5199\u8be6\u7ec6\u4e00\u70b9 \u4e0d\u80dc\u611f\u6fc0\uff01

\u7528\u7b2c\u4e00\u7c7b\u6362\u5143\u79ef\u5206\u6cd5\u5148\u6c42\u51facos(x)sin²(x)\u7684\u4e00\u4e2a\u539f\u51fd\u6570
\u4ee4sinx=t,\u5219 dt=cosxdx
\u222b cos(x)sin²(x)dx =\u222bt²dt=1/3t^3+c=1/3(sinx)^3+c
\u518d\u7b97\u5b9a\u79ef\u5206\u5f97\u7ed3\u679c\u4e3a0
\u6216\u76f4\u63a5\u7528\u5b9a\u79ef\u5206\u7684\u6362\u5143\u79ef\u5206\u6cd5
\u4ee4sinx=t\uff0cdt=cosxdx
\u5f53x=0,t=0, \u5f53x=\u03c0,t=0
\u222b cos(x)sin²(x)dx \u222b \u4e0b\u9762\u662f0\u4e0a\u9762\u662f\u03c0(pai) \u7ecf\u6362\u5143\u540e\u79ef\u5206\u4e0a\u4e0b\u9650\u5168\u662f0
\u7ed3\u679c\u5c31\u662f0


\u222bsin(2x)cosxdx
=\u222b2sinxcosxcosxdx
=2\u222bsinxcos²xdx
=2\u222bcos²xd(-cosx)
=-2\u222bcos²xd(cosx)
=(-⅔)cos³x+C

\u222b[0\uff1a1][x²/(1+x²)]dx
=\u222b[0\uff1a1][(x²+1-1)/(1+x²)]dx
=\u222b[0\uff1a1][1- 1/(1+x²)]dx
=\u222b[0\uff1a1]dx -\u222b[0\uff1a1][1/(1+x²)]dx
=x|[0\uff1a1]-arctanx|[0\uff1a1]
=1-0-(arctan1-arctan0)
=1-(\u03c0/4 -0)
=1- \u03c0/4

d(sinx)=cosx*dx
替换后得到结果1/3
详细见图,满意请采纳

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