Q117 / 160 · AP EAMCET 2006 · Mathematics

$\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$ is equal to :

  • A
    0
  • B
    1
  • Correct answer
    C
    $\frac{e^2-1}{2 e}$
  • D
    $\frac{e^2+2}{2 e}$

Solution

$I=\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$ $=\int_{-1}^1 \frac{e^x+e^{-x}}{2\left(1+e^{2 x}\right)} d x\left(\because \cosh x=\frac{e^x+e^{-x}}{2}\right)$ $=\frac{1}{2} \int_{-1}^1 \frac{1+e^{2 x}}{\left(1+e^{2 x}\right) e^x} \cdot d x$ $=\frac{1}{2} \int_{-1}^1 e^{-x} d x=-\frac{1}{2}\left[e^{-x}\right]_{-1}^1$ $=-\frac{1}{2}\left(e^{-1}-e^1\right)=\frac{e^2-1}{2 e}$

Hence, the correct answer is option: 'C'.

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