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\centerline{Mathematics 134 -- Intensive Calculus for Science 2}
\centerline{Discussion 1 -- Practice on Integration by Substitution}
\centerline{February 8, 2006}
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{\it Background}
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Last time, we introduced the technique of integration by 
substitution.  Recall that the idea was that many integrals
could be reduced to one of our basic forms by a substitution
$u = g(x)$, $du = g'(x)\ dx$.  If the original integral is
$$\int f'(g(x)) g'(x)\ dx$$
(possibly up to a constant factor), and we do this,
$$\int f'(u)\ du = f(u) + C = f(g(x)) + C$$
(Note:  This will work as long as $\int f'(x)\ dx$ is known
from basic derivative rules.  For instance:  To find
$$\int x \sqrt{1 - 3x^2}\ dx,$$
try $u = 1 - 3x^2$, so $du = -6x$.  Hence the $x\ dx$ is ${-1\over 6}du$,
and 
$$\eqalign{
\int x \sqrt{1 - 3x^2}\ dx &= {-1\over 6}\int \sqrt{u}\ du\cr
                           &= {-1\over 6} {2\over 3} u^{3/2} + C\cr
                           &= {-1\over 9} (1 - 3x^2)^{3/2} + C\cr}$$
Today, we want to practice on a number of additional examples of
this method.  (There's no other way to master this than to 
``practice, practice, practice!'')
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{\it Problems}
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For each integral, identify a likely $u$, compute $du$, and substitute
into the integral to obtain an equivalent integral entirely in terms of $u$.
Integrate, then convert back to $x$.  
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A)  $$\int \sqrt{\sin(4x)}\cos(4x)\ dx$$
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B)  $$\int (3x + 3)^{34}\ dx$$
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C)  $$\int {x + 1\over x^2 + 2x + 29}\ dx$$
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D)  $$\int {e^{\sqrt{x}} \over \sqrt{x}}\ dx$$
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E)  A definite integral -- compute and explain why your answer is 
reasonable:
$$\int_0^2 \sin(5\pi x)\ dx$$
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F)  You'll need substitution, plus some algebra for this one:
$$\int (x + 2)\sqrt{x - 1}\ dx$$   
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{\it Assignment}
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Group writeups due in class on Thursday, February 9. 
\bye