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\centerline{Mathematics 134 -- Intensive Calculus for Science 2}
\centerline{Discussion 2 -- Practice on Integrals by Parts {\it Answers}}
\centerline{February 14, 2006}
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{\it Integrals}
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\leftline{Group 1:}  $$\int x \cos(4x)\ dx$$
{\it Answer:}  Let $u = x$, $dv = \cos(4x)\ dx$
$${x\over 4}\sin(4x) + {1\over 16}\cos(4x) + C$$
\leftline{Group 2:}  $$\int x(\ln x)^2 \ dx$$
{\it Answer:} Let $u = (\ln x)^2$, $dv = x\ dx$ to start.  After two 
applications of parts you should get:
$${x^2\over 2}(\ln x)^2 - {x^2\over 2}\ln x + {x^2\over 4} + C$$
\leftline{Group 3:}  $$\int x\sqrt{5x + 3}\ dy$$
{\it Answer:} Let $u = x$, $dv = \sqrt{5x + 3}\ dx$
$${2x\over 15} (5x+3)^{3/2} - {4\over 375}(5x + 3)^{5/2} + C$$
\leftline{Group 4:}  $$\int \theta^2\sin(2\theta)\ d\theta$$
{\it Answer:} Let $u = \theta^2$, $dv = \sin(2\theta)\ d\theta$
to start.  After two applications of parts, you should get:
$${-\theta^2\over 2}\cos(2\theta) + {\theta\over 2}\sin(2\theta) + {1\over 4}\cos(2\theta) + C$$
\leftline{Group 5:}  $$\int t^2 e^{-3t}\ dt$$
{\it Answer:} Let $u = t^2$, $dv = e^{-3t}$
to start.  After two applications of parts, you should get:
$${-t^2\over 3}e^{-3t} - {2t\over 9}e^{-3t} - {2\over 27} e^{-3t} + C$$
\leftline{Group 6:}  $$\int {\ln(x)\over x^2}\ dx$$
{\it Answer:} Let $u = \ln x$, $dv = x^{-2}\ dx$
$${-1\over x}\ln x - {1\over x} + C$$
\leftline{Group 7:}  $$\int x^5\cos(x^3)\ dx$$
\leftline{Hint: The $\cos(x^3)$ should go in $dv$, 
but what ``else'' do you need to be able to find $v$?}
{\it Answer:} Let $u = x^3$ and $dv = x^2\cos(x^3)\ dx$.  The idea is to include
the derivative of the $x^3$ inside the cosine (up to a constant),
so that you can find $v = \int dv$ by substitution.  
  $${x^3\over 3} \sin(x^3) + {1\over 3}\cos(x^3) + C$$
\leftline{Group 8:}  $$\int \arcsin(x) dx$$
{\it Answer:} Let $u = \arcsin(x)$ and $dv = dx$.  The idea is to include
the derivative of the $x^3$ inside the cosine (up to a constant),
so that you can find $v = \int dv$ by substitution.  
$$x\arcsin(x) + \sqrt{1-x^2} + C$$
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