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Unformatted text preview: 1 Extra Integrals Show that for the following integrals (all of them exist and are nite): Z 1 1 x β 1 + x 2- 1 x ΒΆ dx =- ln( β 2 + 1) + ln2 , Z Ο/ 2 cos x ln(tan x ) dx =- ln2 , Z Ο/ 2 x sin x cos x tan 2 x + cot 2 x dx = Ο ( Ο- 2) 32 , Z + β sin 3 x x 2 dx = 3 4 ln3 . As an engineer, you probably don't need to care if the integral exists or not, as most of the integrals you would be handling with are well-de ned. But in anycase, for those who are interested, notice that all the 4 above integrands are not "well-de ned" at the end points. For example, sin 3 x x 2 at x = 0 . Note that since the problems occur at the endpoints, the inte- grals could still well exist, since for integration, we do not care about the endpoints itself, but only the limits when the integrands tends to the endpoints. In fact, if we can show that the limit when the integrand tends to the endpoint is nite, then the integral is well de ned in a vicinity of the endpoint. So, consider the rst integral. Z 1 1 x β 1 + x 2- 1 x ΒΆ dx. Then we see that the only problem is at x = 0 , and for all x β (0 , 1] , the function f : x 7β 1 x β 1+ x 2- 1 x is actually continuous. So let's study the behavior when x β . We use a conjugate expression. We have f ( x ) = 1 x β 1 + x 2- 1 x = 1- β 1 + x 2 x β 1 + x 2 =- x β 1 + x 2 (1 + β 1 + x 2 ) , and thus lim x β f ( x ) = 0 , and thus we can conclude that the integral exists. Like I said, the justi cation of the existence of the integral is not too important (for engineers). Anyway to calculate the integral, we will calculate a primitive. The second term gives ln x . The problem is with the rst. We see that the term which irritates the most is actually β 1 + x 2 . So the natural thing to do is to do a substitution z = β 1 + x 2 . So for the rst term we get Z 1 x β 1 + x 2 dx = Z dz z 2- 1 =- 1 2 ln fl fl fl fl 1 + z 1- z fl fl fl fl ....
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- Spring '11
- Limit, Logarithm