Originally Posted by

**Shyam** you have a sign mistake in your second integral, while integrating.

$\displaystyle \int\limits_{ - \ln 2}^{\ln 2} {\frac{1}

{2}\left[ {e^x - e^{ - x} } \right]} dx \hfill \\$

$\displaystyle = \frac{1}

{2}\left[ {\int\limits_{ - \ln 2}^{\ln 2} {e^x dx + \int\limits_{\ln 2}^{ - \ln 2} {e^{ - x} dx} } } \right] \hfill \\$

$\displaystyle = \frac{1}

{2}\left\{ {\left[ {e^x } \right]_{ - \ln 2}^{\ln 2} + \left[ { - e^{ - x} } \right]_{\ln 2}^{ - \ln 2} } \right\} \hfill \\$

$\displaystyle = \frac{1}

{2}\left[ {e^{\ln 2} - e^{ - \ln 2} - e^{\ln 2} + e^{ - \ln 2} } \right] \hfill \\$

$\displaystyle = \frac{1}

{2}\left[ {2 - 2^{ - 1} - 2 + 2^{ - 1} } \right] \hfill \\$

$\displaystyle = 0 \hfill \\ $

Did you figure out your mistake now? sinewave85?