Activity 65.
Evaluate each of the following definite integrals exactly through an appropriate \(u\)-substitution.
(a)
\(\displaystyle \int_1^2 \frac{x}{1 + 4x^2} \, dx\)
Hint.
Let \(u = 1+4x^2\text{.}\)
Answer.
\(\displaystyle \int_{x=1}^{x=2} \frac{x}{1 + 4x^2} \, dx = \frac{1}{8} (\ln(17) - \ln(5))\text{.}\)
Solution.
Let \(u = 1+4x^2\text{,}\) so \(du=8x dx\) and \(xdx = \frac{1}{8} du\text{.}\) Note that \(x=1\) implies \(u=5\) and \(x=2\) implies \(u=17\text{.}\) Thus, \(\int_{x=1}^{x=2} \frac{x}{1 + 4x^2} \, dx = \frac{1}{8} \int_{u=5}^{u=17} \frac{du}{u} = \left. \frac{1}{8} \ln|u| \right|_5^{17} = \frac{1}{8} (\ln(17) - \ln(5))\text{.}\)
(b)
\(\displaystyle \int_0^1 e^{-x} (2e^{-x}+3)^{9} \, dx\)
Hint.
\((2e^{-x}+3)\) and \(e^{-x}\) form a function-derivative pair
Answer.
\(\displaystyle \int_0^1 e^{-x} (2e^{-x}+3)^{9} \, dx = -\frac{1}{20}(2e^{-1}+3)^{10} + \frac{1}{20}(2e^{0}+3)^{10}\text{.}\)
Solution.
First consider the corresponding indefinite integral, \(\int e^{-x} (2e^{-x}+3)^{9} \, dx\text{,}\) and let \(u=2e^{-x}+3\) so that \(du = -2e^{-x}dx\text{.}\) We see \(e^{-x}dx = -\frac{1}{2}du\text{,}\) and thus \(\int e^{-x} (2e^{-x}+3)^{9} \, dx = -\frac{1}{2} \int u^9 \, du = -\frac{1}{2} \cdot \frac{1}{10} u^{10} + C = -\frac{1}{20}(2e^{-x}+3)^{10}\text{.}\) Applying this antiderivative to the definite integral, we see that \(\int_0^1 e^{-x} (2e^{-x}+3)^{9} \, dx = \left. -\frac{1}{20}(2e^{-x}+3)^{10} \right|_0^1 = -\frac{1}{20}(2e^{-1}+3)^{10} + \frac{1}{20}(2e^{0}+3)^{10}\text{.}\)
(c)
\(\displaystyle \int_{2/\pi}^{4/\pi} \frac{\cos\left(\frac{1}{x}\right)}{x^{2}} \,dx\)
Hint.
\(\frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}\)
Answer.
\(\displaystyle \int_{2/\pi}^{4/\pi} \frac{\cos\left(\frac{1}{x}\right)}{x^{2}} \,dx = 1 - \frac{\sqrt{2}}{2}\text{.}\)
Solution.
Using the substitution \(u=\frac{1}{x}\text{,}\) it follows that \(\int \frac{\cos\left(\frac{1}{x}\right)}{x^{2}} \,dx = -\int \cos(u) \, du = -\sin(u) + C\text{.}\) Hence, \(\int_{2/\pi}^{4/\pi} \frac{\cos\left(\frac{1}{x}\right)}{x^{2}} \,dx = \left.-\sin\left(\frac{1}{x}\right) \right|_{2/\pi}^{4/\pi} = -\sin(\frac{\pi}{4} + \sin(\frac{\pi}{2}) = 1 - \frac{\sqrt{2}}{2}\text{.}\)
