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\centerline{MATH 136 -- Calculus 2}
\centerline{Second Practice Day on $u$-substitution}
\centerline{February 5, 2020}
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\noindent
{\it Background}
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Here are a few additional integration formulas that can be used
in conjunction with $u$-substitution:
\begin{itemize}
\item If $b > 0$, then $\displaystyle \int b^x\ dx = \frac{b^x}{\ln(b)} + C$
(This follows by combining the derivative rule for $e^u$ by the chain rule 
with the formula $b^x = (e^{\ln(b)})^x = e^{x\ln(b)}$.)
\item $\displaystyle \int \frac{dx}{|x|\sqrt{x^2 - 1}} = \sec^{-1}(x) + C$
(follows from the derivative rule 
$$\frac{d}{dx}\sec^{-1}(x) = \frac{1}{|x|\sqrt{x^2 - 1}}$$
\end{itemize}
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\noindent
{\it Questions}
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\noindent
Find the following integrals using the formulas above and $u$-substitution as needed: 
\begin{enumerate}
\item $\displaystyle \int 5^{\cos(x)} \sin(x)\ dx$
\item[] {\it Solution:}  Use the rule for $b^x$ with $b = 5$ and $u = \cos(x)$.  Then $du = -\sin(x)\ dx$,
so we have 
$$-\int 5^u\ du = -\frac{5^u}{\ln(5)} + C = -\frac{5^{\cos(x)}}{\ln(5)} + C$$ 
\item $\displaystyle \int \frac{dx}{(x + 8)\ln(2x + 16)}$
\item[] {\it Solution:}  This one is slightly tricky.  We need to notice that if $u = \ln(2x + 16)$, then 
$du = \frac{1}{2x + 16}\cdot 2\ dx = \frac{1}{x + 8}\ dx$.  This means that the integral is
$$\int \frac{1}{u}\ du = \ln|u| + C = \ln|\ln(2x + 16)| + C$$
\item $\displaystyle \int \frac{dx}{x \sqrt{49x^2 - 1}}$ (Take $x > 0$.)
\item[] {\it Solution:}  Let $u = 7x$.  Then $du = 7\ dx$.  The integral becomes
$$\int \frac{du/7}{(u/7) \sqrt{u^2 - 1}} = \int \frac{du}{u \sqrt{u^2 - 1}} = \sec^{-1}(u) + C = \sec^{-1}(7x) + C$$
\item $\displaystyle \int x^2\sqrt{x + 4}\ dx$ (Hint:  Let $u = x + 4$)
\item[] {\it Solution:}  Using the Hint, $du = dx$, $x = u - 4$, and the integrand becomes
\begin{align*}
\int (u - 4)^2\sqrt{u}\ du &= \int u^{5/2} - 8 u^{3/2} + 16u^{1/2}\ du\\
&= \frac{2}{7} u^{7/2} - \frac{16}{5} u^{5/2} + \frac{32}{3} u^{3/2} + C\\
& =\frac{2}{7} (x+4)^{7/2} - \frac{16}{5} (x+4)^{5/2} + \frac{32}{3} (x+4)^{3/2} + C
\end{align*}
\item $\displaystyle \int_0^1 4^{3x}\ dx$ 
\item[] {\it Solution:}  Let $u = 3x$, then converting everything to equivalent expressions
in $u$ (including the limits of integration), the integral is
$$\int_{u=0}^{u=3} \frac{1}{3} 4^u\ du = \left.\frac{1}{3\ln(4)} 4^u \right|_{u=0}^{u=3} = \frac{1}{3\ln(4)}(4^3 - 1)=
\frac{21}{\ln(4)}.$$ 
\end{enumerate}

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