Solutions

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

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 $$\int_1^3 \frac{\ln (2x)}{x^2}\,dx$$

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 $$\int e^{-2t}\cos t\,dt$$

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

$$5\int e^{-2t}\cos t\,dt=e^{-2t}\sin t-2e^{-2t}\cos t,$$

i.e.

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

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

Solution: First perform polynomial division:

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

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

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

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 $$\int \tan^4(s)\sec^4(s)\,ds$$

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 $$\int\arcsin (x-1)\,dx$$

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

$$\int\arcsin (x-1)\,dx=\int\arcsin y\, dy.$$

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

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

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

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

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

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

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

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

Solution: We split the integral into 2 pieces:

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

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

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

[Calculus] [CyberExam]

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