A.f=cos(2x)/(cos(x)+sin(x)); int((f,x)
B.f=cos(2*x)/(cos(x)+sin(x)); int(f,x)
C.syms x; f=cos(2x)/(cos(x)+sin(x)); int(f,x)
D.syms x; f=cos(2*x)/(cos(x)+sin(x)); int(f,x)
A.f=cos(2x)/(cos(x)+sin(x)); int((f,x)
B.f=cos(2*x)/(cos(x)+sin(x)); int(f,x)
C.syms x; f=cos(2x)/(cos(x)+sin(x)); int(f,x)
D.syms x; f=cos(2*x)/(cos(x)+sin(x)); int(f,x)
A.syms x;f=1/(x^2*cos(2x));int(f,x, -inf ,+inf)
B.syms x;f= 1/x^2*cos(2*x);int(f,x, -inf ,+inf)
C.syms x;f=1/x^2cos(2*x);int(f,x, -inf ,+inf)
D.syms x;f= 1/(x^2*cos(2x));int(f,x, -inf ,+inf)
A.Cancel[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
B.TrigFactor[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]
C.Simplify[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
D.TrigReduce[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]
化简三角函数表达式
A.Cancel[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
B.TrigFactor[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]
C.Simplify[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
D.TrigReduce[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]
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