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# Improper Integrals watch

1. I have been given some questions to do involving improper integrals and I just don't get it

the two questions are " using the substitution x=3tan u find the improper integrals:

∫dx/(9+x^2)
∫dx/(9+x^2)^2

the first is from sqrt 3 to infinity and the other is from 0 to inifinty.

2. What do you get when you make the substitution?
3. for the first one i got ∫3(1+tan^2 u)/9(1+tan^2 u)
4. (Original post by Big_Sam)
for the first one i got ∫3(1+tan^2 u)/9(1+tan^2 u)
cancel the fraction down to lowest terms.
5. Basically when you do a substitution, you have to change the limits too.

So for example, the first one is from x = sqrt3 to x = infinity

Then substitute: 3 tan u = sqrt 3 which implies that your new limit is u = pi/6
And then: 3 tan u = infinity which impliese that tan u = infinity

Since tan u can never actually be equal to infinity, it only tends to infinity, you have to say u = lim arctan (k) as k tends towards infinity.
If you work it out, you'll find that u = pi/2

So your new limits are pi/6 and pi/2

And then you can do the same for the second one
6. thanks guys so does the first one just cancel down to ∫1/3 du which gives 1/3u and so the answer is 1/3[arctan(x/3)] between pi/6 and pi/2?
7. (Original post by Big_Sam)
thanks guys so does the first one just cancel down to ∫1/3 du which gives 1/3u and so the answer is 1/3[arctan(x/3)] between pi/6 and pi/2?
No, leave it as 1/3 u, because you've substituted the limits to be in terms of u, and not x. Just plug your limits into 1/3 u and then you'll get the right answer
8. ah right so it will just end up as pi/6 - pi/18 which gives pi/9 so the answer to the first integral is pi/9?
9. (Original post by Big_Sam)
ah right so it will just end up as pi/6 - pi/18 which gives pi/9 so the answer to the first integral is pi/9?
I believe that is correct
10. thanks! on last question, for the second one i have ended up with the ∫1/3(9sec^2u) which seems a bit of complicated integral to be doing, have I gone wrong somewhere?
11. (Original post by Big_Sam)
thanks! on last question, for the second one i have ended up with the ∫1/3(9sec^2u) which seems a bit of complicated integral to be doing, have I gone wrong somewhere?
What would you differentiate to obtain ?
12. tanu?
13. (Original post by Big_Sam)
tanu?
14. sorry i this seems a stupid question but how does this help?
15. It should make it obvious how to find ∫1/3(9sec^2u) du

Although I'm not certain as to whether the 9sec^2(u) is on the top or the bottm of the fraction. If it's on the top, then it should be easy to integrate, as you know that the integral of sec^2(u) is tan(u)

If it's on the bottom then remember that 1/sec^2(u) = cos^2(u)
16. i think it should be on the bottom so does the integral become 18cos^2(u)?
17. (Original post by Big_Sam)
i think it should be on the bottom so does the integral become 18cos^2(u)?
I think it becomes [cos^2(u)]/27 actually

If you don't know how to integrate cos^2(u) remember:

cos(2u) = 2cos^2(u) - 1
18. how come its over 27?
19. (Original post by Big_Sam)
how come its over 27?
9 x 3 = 27?
20. lol yeh that makes more sense than 3 x 9 = 18.

so after integrated will it become sin(2u)/108 + x/54?

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