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Question: Answered & Verified by Expert
$\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$ is equal to :
MathematicsDefinite IntegrationTS EAMCETTS EAMCET 2006
Options:
  • A 0
  • B 1
  • C $\frac{e^2-1}{2 e}$
  • D $\frac{e^2+2}{2 e}$
Solution:
1489 Upvotes Verified Answer
The correct answer is: $\frac{e^2-1}{2 e}$
$I=\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$
$=\int_{-1}^1 \frac{e^x+e^{-x}}{2\left(1+e^{2 x}\right)} d x\left(\because \cosh x=\frac{e^x+e^{-x}}{2}\right)$
$=\frac{1}{2} \int_{-1}^1 \frac{1+e^{2 x}}{\left(1+e^{2 x}\right) e^x} \cdot d x$
$=\frac{1}{2} \int_{-1}^1 e^{-x} d x=-\frac{1}{2}\left[e^{-x}\right]_{-1}^1$
$=-\frac{1}{2}\left(e^{-1}-e^1\right)=\frac{e^2-1}{2 e}$

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