求解∫(x+2sinxcosx)/(1+cos2x) dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 18:23:44
x){ŏ:VkThgU$Whӯa
dUh*T$RΆSj4*5|Te
hR@6HN#u!@#p07Rӝ2 .фl n<(pi B#lmgLpND3#vX
<;PZ
求解∫(x+2sinxcosx)/(1+cos2x) dx
求解∫(x+2sinxcosx)/(1+cos2x) dx
求解∫(x+2sinxcosx)/(1+cos2x) dx
∫ (x + 2sinxcosx)/(1 + cos2x) dx
= ∫ x/(1 + cos2x) dx + ∫ 2sinxcosx/(1 + cos2x) dx
= ∫ x/(1 + 2cos²x - 1) dx + ∫ sin2x/(1 + cos2x) dx
= (1/2)∫ xsec²x dx - (1/2)∫ d(cos2x)/(1 + cos2x)
= (1/2)∫ x d(tanx) - (1/2)∫ d(1 + cos2x)/(1 + cos2x)
= (1/2)xtanx - (1/2)∫ tanx dx - (1/2)ln(1 + cos2x) + C
= (1/2)xtanx + (1/2)ln(cosx) - (1/2)ln(1 + cos2x) + C
= (1/2)xtanx + (1/2)ln(cosx) - (1/2)ln(1 + 2cos²x - 1) + C
= (1/2)[xtanx - ln(cosx)] + C
求解∫(x+2sinxcosx)/(1+cos2x) dx
∫cosx^2/1+sinxcosx dx如何求解
求解不定积分41∫x²ln[x-3]*dx2∫sinxcosx分之cos2x*dx
求解高中数学f(x)=2sinxcosx+2cos平方x为啥化简后是这样
∫dx/(1+sinxcosx)^2
∫sinxcosx/(1+sin^4x)dx
化简2*(sinxcosx-cos平方x)+1
Mathematica求解方程sinxcosx-x^2=0的所有根,麻烦给个完整代码,
当x属于(0,π)时,化简√ 1-2sinxcosx+√ 1+2sinxcosx
sin^2xtanx+cos^2x/tanx+2sinxcosx-(1+cosx/sinxcosx)化简
sin^2xtanx+(cos^2x/tanx)+2sinxcosx-1+cos/sinxcosx
三角函数计算题求解,3sinX-cosX=0.求sin^2X+2sinxcosx+2cos^2x=?
f(x)=sin^2x+sinxcosx-1/2,化简
求证 1+2sinxcosx/sin^2x-cos^2x=sin^2x-cos^2x/1-2sinxcosx
证明1-2sinxcosx/cos^2x-sin^2x=cos^2x-sin^2/1+2sinxcosx
∫dx/sinxcosx求解并要求有过程
2sin^2x-sinxcosx-cos^2=1
化简y=2sinxcosx+2cos^2x-1