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\begin{document}
	
	\mbox{}
	\title{Solutions for Quiz 1}
	
	\maketitle

	%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%\section{Introduction} \label{sec:intro}
	\noindent {\footnotesize Note: Most of the problems were taken from the textbook} \cite{jS15}.
		
	\begin{problem}
		Solve the integral
		\[
			\int \cos^5 t \sin^6 t \, dt.
		\]
	\end{problem}
	
	\noindent {\it Solution:} Taking $u = \sin t$ one has that $du = \cos t \,dt$ and, therefore,
		\begin{align*}
			\int \cos^5 t \sin^6 t \, dt &= \int (1 - \sin^2 t)^2 \sin^6 t \, \cos t \, dt \\
													   &= \int (1 - u^2)^2 u^6 \, du \\
													   &= \int u^6 - 2u^8 + u^{10} \, du \\
													   &= \frac{u^7}{7} - 2\frac{u^9}{9} + \frac{u^{11}}{11} + C \\
													   &= \frac{\sin^7 t}{7} - 2\frac{\sin^9 t}{9} + \frac{\sin^{11}t}{11} + C.										   
		\end{align*}
	\hfill \qed
	
	\begin{problem}
		Solve the integral
		\[
			\int x^25^x \, dx.
		\]
	\end{problem}
	
	\noindent {\it Solution:}
	Let us use integration by parts taking $u = x^2$ and $dv = 5^x \, dx$. In this case, $du = 2xdx$ and $v = 5^x/\ln 5$, which implies that
	\begin{equation} \label{eq:problem 2}
		\int x^25^x \, dx = \frac{x^2 5^x}{\ln 5} - \frac{2}{\ln 5} \int x5^x \, dx.
	\end{equation}
	To compute the integral in the right-hand side of \eqref{eq:problem 2}, we applying again integration by parts setting this time $u = x$ and $dv = 5^x \, dx$. Then $du = dx$ and $v = 5^x/\ln 5$, and we obtain
	\[
		\int x^25^x \, dx = \frac{x^2 5^x}{\ln 5} - \frac{2}{\ln 5} \bigg(\frac{x5^x}{\ln 5} - \frac{1}{\ln x} \int 5^x \, dx \bigg) = \frac{x^2 5^x}{\ln 5} - \frac{2x5^x}{(\ln 5)^2} + \frac{2\cdot5^x}{(\ln 5)^3} + C.
	\]
	\hfill \qed
	
	\newpage
	
	\begin{problem}
		Evaluate the integral
		\[
			\int_0^2 \frac{dt}{\sqrt{4 + t^2}}.
		\]
	\end{problem}
	
	\noindent {\it Solution:} Taking $t = 2 \tan u$, we get $\sqrt{4 + t^2} = 2 \sec u$ and $dt = 2 \sec^2 u \,du$. As result,
	\[
		\int_0^2 \frac{dt}{\sqrt{4 + t^2}} = \int_0^{\pi/4} \sec u \, du = \ln |\sec u + \tan u|\bigg{|}_0^{\pi/4} = \ln(\sqrt{2} + 1).
	\]
	\hfill \qed
	
	\begin{problem}
		Solve the integral
		\[
			\int \frac{x^3 - 3x^2 + 9x + 14}{x^2 - 4x + 13} \, dx.
		\]
	\end{problem}
	
	\noindent {\it Solution:} \hfill
	\begin{align*}
	\int \frac{x^3 - 3x^2 + 9x + 14}{x^2 - 4x + 13} \, dx
		&= \int \bigg(x+1 + \frac{1}{x^2 - 4x + 13} \bigg)\, dx \\
		&= \frac{x^2}2 + x + \int \frac{dx}{(x-2)^2 + 3^2} \\
		&= \frac{x^2}2 + x + \frac 1{3} \tan^{-1}\bigg( \frac{x-2}{3}\bigg) + C.									   
	\end{align*}
	\hfill \qed
	
	\begin{problem}
		Solve the integral
		\[
			\int \sec^4 x \tan^3 x \, dx.
		\]
	\end{problem}
	
	\noindent {\it Solution:}  Taking $u = \tan x$ we obtain $du = \sec^2 x \,dx$. Thus,
	\begin{align*}
		\int \sec^4 x \tan^3 x \, dx 
			&= \int (1 + \tan^2 x) \tan^3 x \, \sec^2 x \, dx \\
			&= \int (1 + u^2) u^3 \, du \\
			&= \frac{u^4}{4} + \frac{u^6}{6} + C \\
			&= \frac{\tan^4 x}{4} + \frac{\tan^6 x}{6} + C.
	\end{align*}
	\hfill \qed
	
	\newpage
	
	\begin{problem}
		Solve the integral
		\[
			\int \frac{dx}{\sqrt{x^2 + 6x + 10}}.
		\]
	\end{problem}
	
	\noindent {\it Solution:} Taking $x+3 = \tan u$, one finds that $dx = \sec^2 u \, du$. Therefore
	\begin{align*}
		\int \frac{dx}{\sqrt{x^2 + 6x + 10}} 
			&= \int \frac{dx}{\sqrt{(x+3)^2 + 1}} \\
			&= \int \sec u \, du \\
			&= \ln |\sec u + \tan u| + C \\
			&= \ln |\sqrt{x^2 + 6x + 10} + x + 3| + C.
	\end{align*}
	\hfill \qed
	
	\begin{problem}
		Solve the integral
		\[
			\int \frac{x^2 - 7}{x^2 + 4x + 3} \, dx.
		\]
	\end{problem}
	
	\noindent {\it Solution:} \hfill
	\begin{align*}
		\int \frac{x^2 - 7}{x^2 + 4x + 3} \, dx
			&= \int \bigg(1 - \frac{4x + 10}{x^2 + 4x + 3} \bigg) \, dx \\
			&= x  - \int \frac{3}{x+1}\, dx - \int \frac{1}{x+3} \, dx \\
			&= x - 3\ln|x+1| - \ln|x+3| + C.							   
	\end{align*}
	\hfill \qed
	
	\begin{thebibliography}{20}
		
		\bibitem{jS15} J.~Stewart: \emph{Single Variable Calculus} 8th Edition, Cengage Learning, Boston 2015.
		
	\end{thebibliography}
	
\end{document}