Edexel_2017_8

Edexel_2017_8

(a)

We have by simple manipulation of logs:

y = \ln(e^x + 1) - \ln (e^x - 1)

So:

\begin{align*}\frac {\mathrm dy} {\mathrm dx} &= \frac {e^x} {e^x + 1} - \frac {e^x} {e^x - 1} \\ & = \frac {e^x(e^x - 1) - e^x(e^x + 1)} {(e^x + 1) (e^x - 1)} \\ & = \frac {e^{2x} - e^x - e^{2x} - e^x} {e^{2x} - 1} \\ & = -\frac {2e^x} {e^{2x} - 1}\end{align*}

(b)

The length of the curve is given by:

\begin{align*}l &= \int_{\ln 2}^{\ln 3} \sqrt {1 + \left(\frac {\mathrm dy} {\mathrm dx}\right)^2} \mathrm dx \\ & = \int_{\ln 2}^{\ln 3} \sqrt {1 + \left(-\frac {2e^x} {e^{2x} - 1}\right)^2} \mathrm dx \\ & = \int_{\ln 2}^{\ln 3} \sqrt {1 + \frac {4e^{2x}} {(e^{2x} - 1)^2}} \mathrm dx \\ &= \int_{\ln 2}^{\ln 3} \sqrt {\frac {(e^{2x} - 1)^2 + 4e^{2x}} {(e^{2x} - 1)^2}} \mathrm dx \\ &= \int_{\ln 2}^{\ln 3} \sqrt {\frac {e^{4x} - 2e^{2x} + 1 + 4e^{2x}} {(e^{2x} - 1)^2}} \mathrm dx \\ & = \int_{\ln 2}^{\ln 3} \sqrt {\frac {(e^{2x} + 1)^2} {(e^{2x} - 1)^2}} \mathrm dx \\ &= \int_{\ln 2}^{\ln 3} \frac {e^{2x} + 1} {e^{2x} - 1} \mathrm dx\end{align*}

(note here that e^{2x} > 1 for x > 0, so there is no worry of confusion with signs and the square root [the numerator is positive since e^x > 0 for all real x])

There are a few ways to tackle this, including substitution of u = e^{2x} - 1 to get a rational integrand which can then be sorted with partial fractions. However, it is simple enough to write (multiplying numerator and denominator by e^{-x}):

\displaystyle \int_{\ln 2}^{\ln 3} \frac {e^{2x} + 1} {e^{2x} - 1} \mathrm dx = \int_{\ln 2}^{\ln 3} \frac {e^x + e^{-x}} {e^x - e^{-x}} \mathrm dx = \int_{\ln 2}^{\ln 3} \frac {(e^x - e^{-x})'} {e^x - e^{-x}} \mathrm dx

So that:

\begin{align*}\int_{\ln 2}^{\ln 3} \frac {(e^x - e^{-x})'} {e^x - e^{-x}} \mathrm dx & = [\ln(e^x - e^{-x})]_{\ln 2}^{\ln 3} \\ & = \ln\left(3 - \frac 1 3\right) - \ln \left(2 - \frac 1 2\right) \\ & = \ln \left(\frac 8 3 \times \frac 2 3\right) \\ & = \ln \left(\frac {16} 9\right)\end{align*}

remembering that \int \frac {f'} f \mathrm dx = \ln f.

Alternatively you can observe that the integrand is \coth x, which integrates to \ln(\sinh x). (which differs from \ln(e^x - e^{-x}) by only a constant of course)

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