(e^2ix + e^-2ix )/4 + e^2ln(sin(x)) + 1/2

asked by guest
on Mar 24, 2025 at 12:56 pm



You asked:

Evaluate the expression: e2ix+e−2ix4+e2ln⁡(sin⁡(x))+12\frac{{e}^{2} i x + {e}^{-2} i x}{4} + {e}^{2} \ln\left( \sin\left( x \right) \right) + \frac{1}{2}

MathBot Answer:

Evaluated



e2ix+e−2ix4+e2ln⁡(sin⁡(x))+12=xi4e2+xie24+e2log⁡(sin⁡(x))+12\displaystyle \frac{{e}^{2} i x + {e}^{-2} i x}{4} + {e}^{2} \ln\left( \sin\left( x \right) \right) + \frac{1}{2} = \frac{x i}{4 e^{2}} + \frac{x i e^{2}}{4} + e^{2} \log{\left(\sin{\left(x \right)} \right)} + \frac{1}{2}

Expanded

e2ix+e−2ix4+e2ln⁡(sin⁡(x))+12=xi4e2+xie24+e2log⁡(sin⁡(x))+12\frac{{e}^{2} i x + {e}^{-2} i x}{4} + {e}^{2} \ln\left( \sin\left( x \right) \right) + \frac{1}{2} = \frac{x i}{4 e^{2}} + \frac{x i e^{2}}{4} + e^{2} \log{\left(\sin{\left(x \right)} \right)} + \frac{1}{2}

Factored

e2ix+e−2ix4+e2ln⁡(sin⁡(x))+12=i(x+xe4−4ie4log⁡(sin⁡(x))−2ie2)4e2\frac{{e}^{2} i x + {e}^{-2} i x}{4} + {e}^{2} \ln\left( \sin\left( x \right) \right) + \frac{1}{2} = \frac{i \left(x + x e^{4} - 4 i e^{4} \log{\left(\sin{\left(x \right)} \right)} - 2 i e^{2}\right)}{4 e^{2}}

ii is the imaginary unit, defined as i2=−1i^2 = -1.

ee is Euler's number, a mathematical constant that is approximately 2.718282.71828.

+ Submit Question