Unformatted text preview: x 2 )(cos x ) and (2 x )(sin x ) and then once straight across, (2)(sin x ), and combine these asx 2 cos x + 2 x sin xZ 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 getx 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” multiplication 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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 Spring '07
 JonathanRogawski
 Math, Calculus, Derivative, Fundamental Theorem Of Calculus, Integration By Parts, Sin, Cos

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