Homework 6

HW 6.pdf

1. Compute the following integrals:

   1. \(\displaystyle\int \left[ \frac{1}{2x - 1}- \frac{1}{2x + 1}\right] dx\)

      \(\displaystyle\int \left[ \frac{1}{2x - 1}- \frac{1}{2x + 1}\right] dx=\frac{1}{2} \int\left[ \frac{1}{x-0.5}-\frac{1}{x+0.5}\right]dx=\frac{1}{2}\int\frac{1}{x-0.5} -\frac{1}{2}\int\frac{1}{x+0.5}dx=\frac{1}{2}\left(\ln|x-0.5|-\ln|x+0.5|\right)+C\) 2. \(\displaystyle\int \frac{\sin(x)}{1 - \cos(x)}dx\)

      Let \(t=1-\cos(x)\), then \(dt=\sin(x)dx\), then \(\displaystyle\int \frac{\sin(x)}{1 - \cos(x)}dx=\displaystyle\int \frac{1}{t}dt= \ln|t|+C=\ln|1-\cos(x)|+C\) 3. \(\displaystyle\int \frac{1}{x^{2} + 9}dx\)

      Since \(\displaystyle\int \frac{1}{x^{2}+ 9}dx=\frac{1}{9}\int \frac{1}{({\frac{x}{3}})^{2}+ 1}dx\), then let \(\frac{x}{3}=t\), then \(\frac{1}{3}dx=dt\)

      Then \(=\displaystyle\frac{1}{3}\int\frac{1}{t^{2}+1}dt=\frac{1}{3}\arctan(t)+C=\frac{1}{3} \arctan(\frac{x}{3})+C\)

2. Compute the following integrals

   1. \(\displaystyle\int \frac{x}{e^{x}}dx\)

      Let \(x=u(x)\), then \(u'(x)=1\). And \(e^{-x}=v^{\prime}(x)\), then \(v(x)=-e^{-x}\)

      Then \({{\displaystyle\int\frac{x}{e^{x}}dx=-xe^{-x}+\int e^{-x}dx}}\) and let \(e^{-x}=t\), then \(-e^{-x}dx=dt\)

      \({{\displaystyle=-xe^{-x}-\int dt=-xe^{-x}-e^{-x}}}+C\) 2. \(\displaystyle\int e^{3x}\sin(2x) dx\)

      Let \(\sin(2x)=u(x)\), then \(u'(x)=2\cos(2x)\). And let \(e^{3x}=v'(x)\), then \(v(x)=\frac{1}{3}e^{3x}\)

      \(\displaystyle\int e^{3x}\sin(2x) dx=\frac{1}{3}\sin(2x)e^{3x}-\frac{2}{3}\int \cos (2x)e^{3x}dx\)

      Again, let \(\cos(2x)=u(x)\), then \(u'(x)=-2\sin(2x)\). And let \(e^{3x}=v'(x)\), then \(v(x)=\frac{1}{3}e^{3x}\)

      \(\displaystyle\int\cos(2x)e^{3x}dx=\frac{1}{3}\cos(2x)e^{3x}+\int\frac{2}{3}\sin( 2x )e^{3x}dx\)

      Thus \(\displaystyle\int e^{3x}\sin(2x) dx=\frac{1}{3}\sin(2x)e^{3x}-\frac{2}{3}\left(\frac{1}{3}\cos(2x)e^{3x}+\int\frac{2}{3}\sin( 2x )e^{3x}dx\right)\)

      \(\frac{13}{9}{\displaystyle\int e^{3x}\sin(2x)dx=\frac{1}{3}\sin(2x)e^{3x}-\frac{2}{9}\cos(2x)e^{3x}}\)

      Then \({\displaystyle\int e^{3x}\sin(2x)dx=\frac{3}{13}\sin(2x)e^{3x}-\frac{2}{13}\cos(2x)e^{3x}}+C\) 3. \(\displaystyle\int (x^{3} - 2x - 5) \sin(2x) dx\)

      Let \(x^3-2x-5=u(x)\), then \(u'(x)=3x^2-2\). And let \(\sin(2x)=v'(x)\), then \(v(x)=-\frac{1}{2}\cos(2x)\)

      \(\displaystyle\int (x^{3}- 2x - 5) \sin(2x) dx=-\frac{1}{2}(x^{3}-2x-5)\cos(2x)+\frac{1}{2} \int(3x^{2}-2)\cos(2x)dx\)

      Again, let \(3x^2-2=u(x)\), then \(u'(x)=6x\). And let \(\cos(2x)=v'(x)\), then \(v(x)=\frac{1}{2}\sin(2x)\)

      \(\displaystyle\int(3x^{2}-2)\cos(2x)dx=\frac{1}{2}(3x^{2}-2)\sin(2x)-3\int x\sin (2x)dx\)

      Again, let \(x=u(x)\) then \(u'(x)=1\). And let \(\sin(2x)=v'(x)\), then \(v(x)=-\frac{1}{2}\cos(2x)\)

      \(\displaystyle\int x\sin(2x)dx=-\frac{x}{2}\cos(2x)+\frac{1}{2}\int\cos(2x)dx=-\frac{x}{2} \cos(2x)+\frac{1}{4}\sin(2x)\)

      Thus \(\displaystyle\int (x^{3}- 2x - 5) \sin(2x) dx=-\frac{1}{2}(x^{3}-2x-5)\cos(2x)+\frac{1}{2} \left(\frac{1}{2}(3x^{2}-2)\sin(2x)-3\left(-\frac{x}{2} \cos(2x)+\frac{1}{4}\sin(2x)\right)\right)\)

      \(\displaystyle\int (x^{3}- 2x - 5) \sin(2x) dx=-\frac{1}{2}(x^{3}-2x-5)\cos(2x)+ \frac{1}{4}(3x^{2}-2)\sin(2x)+\frac{3x}{4}\cos(2x)-\frac{3}{8}\sin(2 x)+C\)

3. Compute the following integrals:

   1. \(\displaystyle\int \frac{5x}{2x^{3} + 6x^{2}}dx\)

      Since \(\frac{5x}{2x^{3}+6x^{2}}=\frac{2.5}{x\left(x+3\right)}=\frac{A}{x}+\frac{B}{x+3} =\frac{Ax+3A+Bx}{x\left(x+3\right)}\), then \(A+B=0,3A=2.5\)

      Then \(A=\frac{5}{6}\) and \(B=-\frac{5}{6}\). Then \(\frac{5x}{2x^{3}+6x^{2}}=\frac{\frac{5}{6}}{x}-\frac{\frac{5}{6}}{x+3}\)

      Thus \(\displaystyle\int \frac{5x}{2x^{3}+ 6x^{2}}dx=\frac{5}{6}\left(\int\frac{1}{x}dx -\int\frac{1}{x+3}dx\right)=\frac{5}{6}\ln|x|-\frac{5}{6}\ln|x+3|+C\) 2. \(\displaystyle\int \frac{x^{4}- x^{3}- x - 1}{x^{3}- x^{2}}dx\)

      Since \(\frac{x^{4}-x^{3}-x-1}{x^{3}-x^{2}}=x-\frac{x+1}{x^{3}-x^{2}}\) and \(\frac{x+1}{x^{3}-x^{2}}=\frac{x+1}{x^{2}\left(x-1\right)}=\frac{A}{x}+\frac{B}{x^{2}} +\frac{C}{x-1}=\frac{\left(A+C\right)x^{2}+\left(B-A\right)x-B}{x^{2}\left(x-1\right)}\)

      Then \(A+C=0,B-A=1,B=-1\), thus \(B=-1,A=-2,C=2\)

      Thus \(\displaystyle\int\left(x+\frac{2}{x}+\frac{1}{x^{2}}-\frac{2}{x-1}\right)dx=\int xdx+\int\frac{2}{x}dx+\int\frac{1}{x^{2}}dx-\int\frac{2}{x-1}dx=\frac{1}{2}x^{2}+ 2\ln|x|-\frac{1}{x}-2\ln|x-1|+C\) 3. \(\displaystyle\int \frac{x^{3} + x^{2} + x + 2}{x^{4} + 3x^{2} + 2}dx\)

      Since \(\frac{x^{3}+x^{2}+x+2}{x^{4}+3x^{2}+2}=\frac{x^{3}+x^{2}+x+2}{\left(x^{2}+2\right)\left(x^{2}+1\right)} =\frac{Ax+B}{x^{2}+2}+\frac{Cx+D}{x^{2}+1}=\frac{Ax^{3}+Ax+Bx^{2}+B+Cx^{3}+2Cx+Dx^{2}+2D}{\left(x^{2}+2\right)\left(x^{2}+1\right)}\)

      Then \(A+C=1,B+D=1,A+2C=1,B+2D=2\)

      Then \(C=0,A=1,D=1,B=0\)

      Then \(\frac{x^{3}+x^{2}+x+2}{x^{4}+3x^{2}+2}=\frac{x}{x^{2}+2}+\frac{1}{x^{2}+1}\), thus

      \(\displaystyle\int \frac{x^{3}+ x^{2}+ x + 2}{x^{4}+ 3x^{2}+ 2}dx=\int\frac{x}{x^{2}+2} dx+\int\frac{1}{x^{2}+1}dx=\int\frac{x}{x^{2}+2}dx+\arctan(x)\)

      For \(\displaystyle\int\frac{x}{x^{2}+2}dx\), then let \(x^2+2=t\), then \(2xdx=dt\)

      Thus \(\displaystyle\int\frac{x}{x^{2}+2}dx=\frac{1}{2}\int\frac{1}{t}dt=\frac{1}{2}\ln (x^{2}+2)\)

      Thus \({\displaystyle\int\frac{x^{3}+x^{2}+x+2}{x^{4}+3x^{2}+2}dx=\frac{1}{2}\ln(x^2+2)+\arctan(x)}+C\)

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