Solutions - 0080-03

(1) Integral formula for arclength:

$$S={\int }_a^b\sqrt{1+{(f^{\prime })}^2}dx\gg \gg {\int }_0^{1/2}\sqrt{1+e^{2x}}dx$$

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(2) Perform $u$-sub with $u=\sqrt{1+e^{2x}}$ and so $e^{2x}=u^2-1$ and also:

$$\begin{array}{ccc}& u=\sqrt{1+e^{2x}}& \text{(Note A)}\\ & & \\ & \gg \gg e^{2x}=u^2-1,du=\frac{e^{2x}}{\sqrt{1+e^{2x}}}dx& \\ & & \\ & \gg \gg dx=\frac{u}{u^2-1}du& \end{array}$$

Now transform the integral to $u$:

\int_0^{1/2} \sqrt{1+e^{2x}}\,dx \quad \gg\gg \quad \int_\sqrt{2}^\sqrt{1+e} \frac{u^2}{u^2-1}\,du ParseError: Got function '\sqrt' with no arguments as subscript at position 58: …\gg \quad \int_\̲s̲q̲r̲t̲{2}^\sqrt{1+e} …

Note A: Instead of this $u$-sub and partial fractions, one can set $e^x=u$ and obtain \int_1^\sqrt{e}\frac{\sqrt{1+u^2}}{u}\,du ParseError: Got function '\sqrt' with no arguments as superscript at position 8: \int_1^\̲s̲q̲r̲t̲{e}\frac{\sqrt{…. Then trig sub with $u=\tan \theta$ leads to (eventually) the same final answer.

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(3) Integrate: partial fraction decomposition:

Number degree not lower → long division first:

$$\frac{u^2}{u^2-1}\gg \gg 1+\frac{1}{u^2-1}\gg \gg 1+\frac{1}{(u+1)(u-1)}$$

Write general PFD formula:

$$\frac{1}{(u+1)(u-1)}=\frac{A}{u+1}+\frac{B}{u-1}$$

Solve for $A$ and $B$. Cross multiply:

$$1=A(u-1)+B(u+1)$$

• $x=-1\gg \gg A=-1/2$
• $x=1\gg \gg B=1/2$

$${\int }_{\sqrt{2}}^{\sqrt{1+e}}\frac{u^2}{u^2-1}du\gg \gg {\int }_{\sqrt{2}}^{\sqrt{1+e}}1+\frac{-1/2}{u+1}+\frac{1/2}{u-1}du$$

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(4) Evaluate integral:

$$\begin{array}{ccc}& {\int }_{\sqrt{2}}^{\sqrt{1+e}}1-\frac{1}{2(u+1)}+\frac{1}{2(u-1)}du& \\ & & \\ & \gg \gg {u-\frac{1}{2}\ln |u+1|+\frac{1}{2}\ln |u-1||}_{\sqrt{2}}^{\sqrt{1+e}}& \\ & & \\ & \gg \gg [\sqrt{1+e}-\frac{1}{2}\ln (\sqrt{1+e}+1)+\frac{1}{2}\ln (\sqrt{1+e}-1)]& \\ & & \\ & -[\sqrt{2}-\frac{1}{2}\ln (\sqrt{2}+1)+\frac{1}{2}\ln (\sqrt{2}-1)]& \text{(Note B)}\\ & & \\ & \gg \gg \sqrt{1+e}-\sqrt{2}+\frac{1}{2}(\ln \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)+\ln \left(\frac{\sqrt{1+e}-1}{\sqrt{1+e}+1}\right))& \end{array}$$

Note B: This answer is sufficient. It is not necessary to simplify as in the last step.

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Alternative:

(1) Sub $u=e^x$, $du=e^{x}dx$, $dx=\frac{1}{u}du$:

$${\int }_0^{1/2}\sqrt{1+e^{2x}}dx\gg \gg {\int }_1^{\sqrt{e}}\frac{\sqrt{1+u^2}}{u}du$$

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(2) Trig sub $u=\tan \theta$, $du={\sec }^2\theta d\theta$ and simplify the integrand:

$$\begin{array}{c}\frac{\sqrt{1+{\tan }^2\theta }}{\tan \theta }\cdot {\sec }^2\theta d\theta \gg \gg \frac{\sec \theta }{\tan \theta }(1+{\tan }^2\theta )d\theta \\ \\ \gg \gg (\frac{\sec \theta }{\tan \theta }+\sec \theta \tan \theta )d\theta \gg \gg (\csc \theta +\tan \theta \sec \theta )d\theta \end{array}$$

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(3) Integrate:

$$\int (\csc \theta +\tan \theta \sec \theta )d\theta \gg \gg \ln |\csc \theta -\cot \theta |+\sec \theta +C$$

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(4) Back-substitute $\csc \theta =\frac{\sqrt{u^2+1}}{u}$, $\cot \theta =\frac{1}{u}$, $\sec \theta =\sqrt{u^2+1}$:

$$\gg \gg \ln \left|\frac{\sqrt{u^2+1}-1}{u}\right|+\sqrt{u^2+1}+C$$

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(5) Evaluate at the $u$-bounds:

$$\begin{array}{c}\gg \gg \ln \left|\frac{\sqrt{u^2+1}-1}{u}\right|+\sqrt{u^2+1}{|}_1^{\sqrt{e}}\\ \\ \gg \gg \ln \left(\frac{\sqrt{e+1}-1}{\sqrt{e}}\right)+\sqrt{e+1}-\ln (\sqrt{2}-1)-\sqrt{2}\\ \\ \gg \gg \ln (\sqrt{e+1}-1)-\frac{1}{2}+\sqrt{e+1}-\ln (\sqrt{2}-1)-\sqrt{2}\\ \\ \gg \gg \approx 0.8210\end{array}$$