7.5: Integration Tables and Calculators

Solution

If we look at integration tables, we see that several formulas contain expressions of the form \(\displaystyle \sqrt{a^2−u^2}.\) This expression is actually similar to \(\displaystyle \sqrt{16−e^{2x}},\) where \(\displaystyle a=4\) and \(\displaystyle u=e^x\). Keep in mind that we must also have \(\displaystyle du=e^x\). Multiplying the numerator and the denominator of the given integral by \(\displaystyle e^x\) should help to put this integral in a useful form. Thus, we now have

\(\displaystyle ∫\frac{\sqrt{16−e^{2x}}}{e^x}\,dx=∫\frac{\sqrt{16−e^{2x}}}{e^{2x}}e^x\,dx\)

Substituting \(\displaystyle u=e^x\) and \(\displaystyle du=e^x\,dx\) produces \(\displaystyle ∫\frac{\sqrt{a^2−u^2}}{u^2}du\)

From the integration table (#88 in Appendix A),

\(\displaystyle ∫\frac{\sqrt{a^2−u^2}}{u^2}\,du=−\frac{\sqrt{a^2−u^2}}{u}−\sin^{−1}\left(\frac{u}{a}\right)+C\)

Thus,

\[\begin{align*} ∫\frac{\sqrt{16−e^{2x}}}{e^x}\,dx&=∫\frac{\sqrt{16−e^{2x}}}{e^{2x}}e^x\, dx\\[4pt]
&=∫\frac{\sqrt{4^2−u^2}}{u^2}\,du \\[4pt] 
&=−\frac{\sqrt{4^2−u^2}}{u}−\sin^{−1}\left(\frac{u}{4}\right)+C \\[4pt]
&=\boxed{−\frac{\sqrt{16−e^{2x}}}{e^x}−\sin^{−1}\left(\frac{e^x}{4}\right)+C} \end{align*} \]

Solution

Using Wolfram Alpha, we obtain

\(\displaystyle ∫\frac{dx}{\sqrt{x^2−4}}=\boxed{\ln \left|\sqrt{x^2−4}+x\right| +C}\)

Notice that

\(\displaystyle \ln \left|\frac{\sqrt{x^2−4}}{2}+\frac{x}{2}\right| +C=\ln \left|\frac{\sqrt{x^2−4}+x}{2}\right| +C=\ln \left|\sqrt{x^2−4}+x\right| −\ln (2)+C\)

Since these two antiderivatives differ by only a constant, the solutions are equivalent. We could have also demonstrated that each of these antiderivatives is correct by differentiating them.