解方程：sin(πarctanx)=cos(πarctanx)

解方程：sin(πarctanx)=cos(πarctanx) ∫arctanx(arctanx)=? 高数 去微积分方程通解求微分方程(1+x²)y'=arctanx的通解解：(1+x²)(dy/dx)=arctanx,分离变量得：dy=[(arctanx)/(1+x²)]dx积分之,即得通解为：y=∫[(arctanx)/(1+x²)]dx=∫(arctanx)d(arctanx)=(1/2)(arc 证明sin(arctanx)=x/根号(1+x^2) y=sin(2x+1)+arctanx的定义域 解方程|sin x|=-sin x tan(arcsinx)=?cos(arcsinx)=?sin(arctanx)=?sin(arccosx)=?tan(arccosx)=? arcsinx+arctanx=π/2 求X 证明恒等式arctanx+arccotx=π/2 y=π+arctanx/2.求出反函数? dy/dx=arctanx的解 非线性方程解析解-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0 -x0*cos(b)-y0*sin(a)*co 求解微分方程通解方程是y''=1/(1+x2),求出来y'=arctanx+c,然后怎么做啊方程是y''=1/(1+x2)，解出y'=arctanx+c，接下来怎么做？ 设y=sin^2•e^arctanx.求dy/dx y=sin(2x+1)+arctanx求下列函数的定义域. tan(arctanX+arctanY)=(X+Y)/(1-XY)想了很久,没转过来......还有 sin(2arctanX)=2X/(1+X^2) 方程sin(x+π/3)=sin(π/3)-sinx的解集 解方程sin(πcosx)=cos(πsinx)