121616949-math.183

# 121616949-math.183 - x 2-cos x and-2 x-sin x and then once...

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8.4 Integration by Parts 169 sign u dv x 2 sin x - 2 x - cos x 2 - sin x - 0 cos x or u dv x 2 sin x - 2 x - cos x 2 - sin x 0 cos x To form the first table, we start with u at the top of the second column and repeatedly compute the derivative; starting with dv at the top of the third column, we repeatedly compute the antiderivative. In the first column, we place a “ - ” in every second row. To form the second table we combine the first and second columns by ignoring the boundary; if you do this by hand, you may simply start with two columns and add a “ - ” to every second row. To compute with this second table we begin at the top. Multiply the first entry in column u by the second entry in column dv to get - x 2 cos x , and add this to the integral of the product of the second entry in column u and second entry in column dv . This gives: - x 2 cos x + Z 2 x cos x dx, or exactly the result of the first application of integration by parts. Since this integral is not yet easy, we return to the table. Now we multiply twice on the diagonal, (
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Unformatted text preview: x 2 )(-cos x ) and (-2 x )(-sin x ) and then once straight across, (2)(-sin x ), and combine these as-x 2 cos x + 2 x sin x-Z 2 sin x dx, giving the same result as the second application of integration by parts. While this integral is easy, we may return yet once more to the table. Now multiply three times on the diagonal to get ( x 2 )(-cos x ), (-2 x )(-sin x ), and (2)(cos x ), and once straight across, (0)(cos x ). We combine these as before to get-x 2 cos x + 2 x sin x + 2 cos x + Z dx =-x 2 cos x + 2 x sin x + 2 cos x + C. Typically we would ﬁll in the table one line at a time, until the “straight across” multipli-cation gives an easy integral. If we can see that the u column will eventually become zero, we can instead ﬁll in the whole table; computing the products as indicated will then give the entire integral, including the “+ C ”, as above....
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