Since we’ve done quite a few substitution rule integrals to this time we aren’t going to put a lot of effort into explaining the substitution part of things here.

a \( \displaystyle \int_{{\, - 1}}^{{\,5}}{{\left( {1 + w} \right){{\left( {2w + {w^2}} \right)}^5}\,dw}}\)

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The substitution and converted limits are,

\[\begin{align*}u & = 2w + {w^2}\hspace{0.5in}du = \left( {2 + 2w} \right)dw\hspace{0.25in}\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\left( {1 + w} \right)dw = \frac{1}{2}du\\ w & = - 1\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \,\,\,\,\hspace{0.25in}u = - 1\hspace{0.5in}\hspace{0.5in}w = 5\,\,\,\,\,\,\,\,\, \Rightarrow \hspace{0.25in}u = 35\end{align*}\]

Sometimes a limit will remain the same after the substitution. Don’t get excited when it happens and don’t expect it to happen all the time.

Here is the integral,

\[\begin{align*}\int_{{\, - 1}}^{{\,5}}{{\left( {1 + w} \right){{\left( {2w + {w^2}} \right)}^5}\,dw}} & = \frac{1}{2}\int_{{\, - 1}}^{{\,35}}{{{u^5}\,du}}\\ & = \left. {\frac{1}{{12}}{u^6}} \right|_{ - 1}^{35} = 153188802\end{align*}\]

Don’t get excited about large numbers for answers here. Sometimes they are. That’s life.

b \( \displaystyle \int_{{\, - 2}}^{{\, - 6}}{{\frac{4}{{{{\left( {1 + 2x} \right)}^3}}} - \frac{5}{{1 + 2x}}\,dx}}\)

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Here is the substitution and converted limits for this problem,

\[\begin{align*}u & = 1 + 2x\hspace{0.5in}du = 2dx\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,dx = \frac{1}{2}du\\ x & = - 2\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \,\,\hspace{0.25in}\,\,u = - 3\hspace{0.5in}\hspace{0.25in}x = - 6\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\hspace{0.25in}\,\,\,\,u = - 11\end{align*}\]

The integral is then,

\[\begin{align*}\int_{{\, - 2}}^{{\, - 6}}{{\frac{4}{{{{\left( {1 + 2x} \right)}^3}}} - \frac{5}{{1 + 2x}}\,dx}} & = \frac{1}{2}\int_{{\, - 3}}^{{\, - 11}}{{4{u^{ - 3}} - \frac{5}{u}\,du}}\\ & = \left. {\frac{1}{2}\left( { - 2{u^{ - 2}} - 5\ln \left| u \right|} \right)} \right|_{ - 3}^{ - 11}\\ & = \frac{1}{2}\left( { - \frac{2}{{121}} - 5\ln 11} \right) - \frac{1}{2}\left( { - \frac{2}{9} - 5\ln 3} \right)\\ & = \frac{{112}}{{1089}} - \frac{5}{2}\ln 11 + \frac{5}{2}\ln 3\end{align*}\]

c \( \displaystyle \int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y} + 2\cos \left( {\pi y} \right)\,dy}}\)

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This integral needs to be split into two integrals since the first term doesn’t require a substitution and the second does.

\[ \displaystyle \int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y} + 2\cos \left( {\pi y} \right)\,dy}} = \int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y}\,dy}} + \int_{{\,0}}^{{\,\frac{1}{2}}}{{2\cos \left( {\pi y} \right)\,dy}}\]

Here is the substitution and converted limits for the second term.

\[\begin{align*}u & = \pi y\hspace{0.5in}du = \pi dy\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,dy = \frac{1}{\pi }du\\ y & = 0\hspace{0.25in}\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,u = 0\hspace{0.5in}\hspace{0.25in}y = \frac{1}{2}\,\hspace{0.25in}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,u = \frac{\pi }{2}\end{align*}\]

Here is the integral.

\[\begin{align*}\int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y} + 2\cos \left( {\pi y} \right)\,dy}} & = \int_{{\,0}}^{{\,\frac{1}{2}}}{{{{\bf{e}}^y}\,dy}} + \frac{2}{\pi }\int_{{\,0}}^{{\,\frac{\pi }{2}}}{{\cos \left( u \right)\,du}}\\ & = \left. {{{\bf{e}}^y}} \right|_0^{\frac{1}{2}} + \left. {\frac{2}{\pi }\sin u} \right|_0^{\frac{\pi }{2}}\\ & = {{\bf{e}}^{\frac{1}{2}}} - {{\bf{e}}^0} + \frac{2}{\pi }\sin \frac{\pi }{2} - \frac{2}{\pi }\sin 0\\ & = {{\bf{e}}^{\frac{1}{2}}} - 1 + \frac{2}{\pi }\end{align*}\]

d \( \displaystyle \int_{{\,\frac{\pi }{3}}}^{{\,0}}{{3\sin \left( {\frac{z}{2}} \right) - 5\cos \left( {\pi - z} \right)\,dz}}\)

Show Solution
This integral will require two substitutions. So first split up the integral so we can do a substitution on each term.

\[\int_{{\,\frac{\pi }{3}}}^{{\,0}}{{3\sin \left( {\frac{z}{2}} \right) - 5\cos \left( {\pi - z} \right)\,dz}} = \int_{{\,\frac{\pi }{3}}}^{{\,0}}{{3\sin \left( {\frac{z}{2}} \right)\,dz}} - \int_{{\,\frac{\pi }{3}}}^{{\,0}}{{5\cos \left( {\pi - z} \right)\,dz}}\]

There are the two substitutions for these integrals.

\[\begin{align*}u & = \frac{z}{2}\hspace{0.5in}du = \frac{1}{2}dz\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,dz = 2du\\ z & = \frac{\pi }{3}\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,u = \frac{\pi }{6}\hspace{0.5in}z = 0\,\,\,\,\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,u = 0\end{align*}\]
\[\begin{align*}v & = \pi - z\hspace{0.25in}dv = - dz\hspace{0.25in}\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,dz = - dv\\ z & = \frac{\pi }{3}\hspace{0.25in}\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,v = \frac{{2\pi }}{3}\hspace{0.5in}z = 0\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\,\,v = \pi \end{align*}\]

Here is the integral for this problem.

\[\begin{align*}\int_{{\,\frac{\pi }{3}}}^{{\,0}}{{3\sin \left( {\frac{z}{2}} \right) - 5\cos \left( {\pi - z} \right)\,dz}} & = 6\int_{{\,\frac{\pi }{6}}}^{{\,0}}{{\sin \left( u \right)\,du}} + 5\int_{{\,\frac{{2\pi }}{3}}}^{{\,\pi }}{{\cos \left( v \right)\,dv}}\\ & = \left. { - 6\cos \left( u \right)} \right|_{\frac{\pi }{6}}^0 + \left. {5\sin \left( v \right)} \right|_{\frac{{2\pi }}{3}}^\pi \\ & = 3\sqrt 3 - 6 + \left( { - \frac{{5\sqrt 3 }}{2}} \right)\\ & = \frac{{\sqrt 3 }}{2} - 6\end{align*}\]