Solutions

Problem 1 (15 points)   Compute the exact value of $\displaystyle \int_0^\pi \sin^2(\theta)\,d\theta $

Solution:

\begin{eqnarray*}\int_0^\pi \sin^2 \theta\,d\theta
&=&\int_0^\pi \frac{1}{2}(1-\...
...{\pi}{2}-0-\frac{1}{2}\sin(2\pi)+\frac{1}{2}\sin 0=\frac{\pi}{2}
\end{eqnarray*}


Problem 2 (10 points)   Compute the exact value of $\displaystyle \int_1^3 \frac{\ln (2x)}{x^2}\,dx $

Solution: Use Integration by Parts: $u=\ln(2x), dv=1/x^2\,dx;\ du=dx/x,v=-1/x$:

\begin{eqnarray*}\int_1^3 \frac{\ln(2x)}{x^2}dx&=&
\left.-\frac{1}{x}\ln(2x)\rig...
...t_1^3-\frac{dx}{x^2}\\
&=&-\frac{\ln 6}{3}+\ln 2 -\frac{1}{3}+1
\end{eqnarray*}


Problem 3 (15 points)   Find $\displaystyle \int e^{-2t}\cos t\,dt $

Solution: We will use Integration by Parts twice. First let $u=e^{-2t}, dv=\cos t\,dt$:

\begin{eqnarray*}\int e^{-2t}\cos t\,dt&=&e^{-2t}\sin t +2\int e^{-2t} \sin t \,dt
\end{eqnarray*}


Now transform the integral on the right via $u=e^{-2t}, dv=\sin t\,dt$:

\begin{eqnarray*}\int e^{-2t}\cos t\,dt&=&e^{-2t}\sin t +2\int e^{-2t} \sin t \,dt\\
&=&e^{-2t}\sin t-2e^{-2t}\cos t-4\int e^{-2t}\cos t\,dt
\end{eqnarray*}


The integrals on the left and the right side are identical. Consequently

\begin{displaymath}5\int e^{-2t}\cos t\,dt=e^{-2t}\sin t-2e^{-2t}\cos t,\end{displaymath}

i.e.

\begin{displaymath}\int e^{-2t}\cos t\,dt=\frac{1}{5}\left(e^{-2t}\sin t-2e^{-2t}\cos t\right)+C.\end{displaymath}

Problem 4 (10 points)   Find $\displaystyle \int \frac{3x^3-17x^2+36x-35}{x^2-4x+4}\,dx $

Solution: First perform polynomial division:

\begin{displaymath}\frac{3x^3-17x^2+36x-35}{x^2-4x+4}=3x-5 +\frac{4x-15}{(x-2)^2}.\end{displaymath}

The denominator is a repeated linear factor, so the setup for the partial fractions method is as follows:

\begin{displaymath}\frac{4x-15}{(x-2)^2}=\frac{A}{(x-2)}+\frac{B}{(x-2)^2}.\end{displaymath}

Solving for the unknowns yields A=4, B=-7. Thus

\begin{eqnarray*}&&\int \frac{3x^3-17x^2+36x-35}{x^2-4x+4}\,dx\\
=&&\int\left( ...
...\
=&&\frac{3}{2} x^2 -5x + 4 \ln \vert x-2\vert+\frac{7}{x-2}+C
\end{eqnarray*}


Problem 5 (10 points)   Find $\displaystyle \int \tan^4(s)\sec^4(s)\,ds $

Solution: We will use the substitution $u=\tan s,\ du=\sec^2 s\,ds$, and the Pythagorean Theorem $1+\tan^2 s=\sec^2 s$.

\begin{eqnarray*}\int \tan^4(s)\sec^4(s)\,ds&=&\int \left(\tan^4(s)\sec^2(s)\rig...
...\,du\\
&=&u^5/5+u^7/7+C=\frac{\tan^5 s}{5}+\frac{\tan^7 s}{7}+C
\end{eqnarray*}


Problem 6 (15 points)   Find $\displaystyle \int\arcsin (x-1)\,dx $

Solution: We will first substitute y=x-1.

\begin{displaymath}\int\arcsin (x-1)\,dx=\int\arcsin y\, dy.\end{displaymath}

Next we use Integration by Parts: $\displaystyle u=\arcsin y, dv=dy;\,du=\frac{dy}{\sqrt{1-y^2}},v=y $. Thus

\begin{displaymath}\int\arcsin y\, dy=y \arcsin y-\int\frac{y}{\sqrt{1-y^2}}dy.\end{displaymath}

Using the substitution w=1-y2, it is easy to see that $\displaystyle \int\frac{y}{\sqrt{1-y^2}}dy=\sqrt{1-y^2} $. Resubstituting we obtain

\begin{displaymath}\int\arcsin (x-1)\,dx=(x-1)\arcsin(x-1)+\sqrt{1-(x-1)^2}+C.\end{displaymath}

Problem 7 (10 points)   Find $\displaystyle \int \sqrt{9-4x^2}\,dx $

Solution: Since $\displaystyle \sqrt{9-4x^2}=3\sqrt{1-\left(\frac{2}{3}x\right)^2} $, we use the trigonometric substitution $\displaystyle \sin t=\frac{2}{3}x,\cos t\,dt=\frac{2}{3}\,dx $.


\begin{eqnarray*}\int \sqrt{9-4x^2}\,dx&=&3\sqrt{1-\left(\frac{2}{3}x\right)^2}\...
...}{4}\arcsin\left(\frac{2}{3}x\right)+\frac{1}{2}x\sqrt{9-4x^2}+C
\end{eqnarray*}


Problem 8 (15 points)   Find $\displaystyle \int\frac{2x-1}{x^2-2x+10}\,dx $

Solution: We split the integral into 2 pieces:

\begin{displaymath}\int\frac{2x-1}{(x-1)^2+9}\,dx=\int\frac{2(x-1)}{(x-1)^2+9}\,dx+\int\frac{1}{(x-1)^2+9}\,dx.\end{displaymath}

For the first part we use the substitution u=(x-1)2+9, for the second part $\displaystyle w=\frac{x-1}{3} $.

\begin{eqnarray*}\int\frac{2x-1}{x^2-2x+10}\,dx&=&\int\frac{2(x-1)}{(x-1)^2+9}\,...
...&=&\ln(x^2-2x+10)+\frac{1}{3}\arctan\left(\frac{x-1}{3}\right)+C
\end{eqnarray*}


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