Unformatted text preview: I: % dx = ln(——(—2)) ——1n(—(—4)) : 1n(2) —1n(4). The other way to approach this problem is to simply use the indefinite integral
J‘%dx =1n(x)+ C.
Using this formula, we obtain
72
L idx = ln(—2) — 11’1(—4i) = 111(2) — 111(4). x
Some Basic Integrals The following indefinite integrals (which come from basic differentiation formulas from
differential calculus) should be memorized. You will need to know them as we study
integration techniques that will allow us to compute “less basic” integrals such as [Webs
1. lfK is any constant, then Ide = Kx + C.
2. If n is any constant except n = —1, then ix" dx = ﬁx’m + C.
3. Babe =1n(x)+ C.
4. If a is any positive constant except a = 1, then [ax dx = 1 at + C.
5 1n(a)
. A special case of formula 4 (the nicest case) is when we use base 6. In this case
the formula is x = 1 x
Jedx 1n(e)t2+C or simply
Iexdx = ex + C. 6. _i cos(x)dx = sin(x) + C. 7. .iSin(x) dx = —cos(x) + C. 8. l sec2(x)dx = tan(x) + C. 9. jesc2(x)dx = cot(x) + C. 10. l sec(x)tan(x)dx = sec(x) + C.
11. Icsc(x) c0t(x) dx = —csc(x) + C. 12. I hi? dx = arcsin(x) + C. 13. I 1 dx = arctan(x) + C. 1+x2
The last two integrals in this basic list involve inverse trigonometric functions. Since
these two are usually the least familiar to students who have completed a differential
calculus course, we will give a derivation of formula number 13. (The derivation of
formula 12 is left as a homework exercise.) First, we recall the deﬁnition of the inverse tangent function, which we denote by
“arctan”. (Another notation for the inverse tangent function is “tan—1”.) The statement 16 ...
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 Fall '08
 HOOVER
 Calculus

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