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\problemgrouptitle{The Substitution Rule}

\begin{problemblock}[2]{Evaluate the following indefinite integrals using the Substitution Rule.}

\item $\displaystyle\int\frac{x^2+2x}{\left(x^3+3x^2+4\right)^5}\d{x}$
\answer{$\displaystyle-\frac{1}{12}\left(x^3+3x^2+4\right)^{-4}+C$}

\item $\displaystyle\int\left(5x+1\right)\sqrt{5x^2+2x}\,dx$
\answer{$\displaystyle\frac{1}{3}\left(5x^2+2x\right)^{\frac{3}{2}}+C$}

\item $\displaystyle\int\left(\frac{\cos\sqrt{t}}{\sqrt{t}}+t^3\right)\d{t}$
\answer{$2\sin(\sqrt{t})+\frac{1}{4}t^4+C$}

\item $\displaystyle\int\cos(\theta)\sqrt{3-\sin\theta}\d{\theta}$
\answer{$\displaystyle-\frac{2}{3}(3-\sin\theta)^\frac{3}{2}+C$}

\item $\displaystyle\int x\sqrt{4x+1}\d{x}$
\answer{$\displaystyle\frac{1}{40}(4x+1)^\frac{5}{2}-\frac{1}{24}(4x+1)^\frac{3}{2}+C$}

\item $\displaystyle \int\sec^2\!\left(2x-\frac{\pi}{3}\right)dx$
\answer{$\displaystyle\frac{1}{2}\tan\left(2x-\frac{\pi}{3}\right)+C$}

\end{problemblock}

\begin{problemblock}[2]{Evaluate the following definite integrals using the Substitution Rule.}

\item $\displaystyle\int_0^2 x^3\sqrt{x^4+9}\d{x}$
\answer{Using $u=x^4+9$, integral is $\displaystyle=\int_9^{25}u^{\frac{1}{2}}\frac{du}{4}=\frac{49}{3}$}

\item $\displaystyle\int_0^\frac{\pi}{4} \tan^4\!\theta\sec^2\!\theta\d{\theta}$
\answer{Using $u=\tan\theta$, integral is $\displaystyle=\int_0^1u^4\d{u}=\frac{1}{5}$}

\item $\displaystyle\int_0^1\left[1+x+(1-x)^5\right]\,dx$
\answer{Breaking the integral into three separate integrals (one per term) and using $u=1-x$ on the last one that integral equals $=\int_1^0u^5(-du)=\frac{1}{6}$\,.  Combining this with the definite integral of the first two terms gives the final answer $\frac{5}{3}$\,.  Alternatively, one can find the indefinite integral of the third term (i.e. substitute back to $x$) to get the antiderivative of the entire integrand as $x+\frac{1}{2}x^2-\frac{1}{6}(1-x)^6+C$ and evaluate that at the original limits of $x$.}

\item $\displaystyle\int_1^3\frac{x}{\left(2x^2+1\right)^2}\d{x}$
\answer{Using $u=2x^2+1$, integral is $\displaystyle=\int_3^{19} u^{-2}\frac{du}{4}=\frac{4}{57}$}

\item $\displaystyle\int_{\frac{\pi}{4}}^\frac{\pi}{3}\frac{\sin t}{\cos^\frac{2}{3}\!t}\d{t}$
\answer{Using $u=\cos t$, integral is $\displaystyle=\int_{\frac{1}{\sqrt{2}}}^{\frac{1}{2}}u^{-\frac{2}{3}}(-du)=3\left(\frac{1}{\sqrt[6]{2}}-\frac{1}{\sqrt[3]{2}}\right)$}

\item $\displaystyle\int_0^{\frac{T}{3}}\cos\left(\frac{2\pi t}{T}\right)dt$\ \  ($T>0$ is constant)
\answer{Using $\displaystyle u=\frac{2\pi t}{T}$, integral is $\displaystyle=\int_0^{\frac{\pi}{3}}\cos(u)\frac{T\,du}{2\pi}=\frac{\sqrt{3}\,T}{4\pi}$}


\end{problemblock}
