## anonymous one year ago how do you integrate sqrt(8x^2)

1. anonymous

|dw:1442604743169:dw|

2. IrishBoy123

that's$\int \sqrt{8} \, \sqrt{x^2} \, dx$

3. anonymous

how would do this using u subs

4. IrishBoy123

$\sqrt{8} = ??$ $\sqrt{x^2} = ??$

5. anonymous

|dw:1442605850479:dw|

6. anonymous

|dw:1442605975556:dw|

7. IrishBoy123

yes, i would see it that way initially but the problem is i have a feeling that $$\sqrt{x^2}$$ is a trap let me ask some others. sorry about this, but i think it might be worthwhile @ganeshie88

8. IrishBoy123

@ganeshie8

9. anonymous

ok

10. IrishBoy123

@Empty

11. Empty

Yeah I definitely think it's a trap :P

12. anonymous

i try solving this using the u substitution where u=8x^2, du=16x dx and now the problem is i have an x hanging with 16

13. Empty

I don't think a substitution is the right choice here. I think @IrishBoy123 has the right strategy here, separate the constant out front like this: $\sqrt{8} \int \sqrt{x^2}dx$ The problem is specifically this subtle fact: $\sqrt{x^2} \ne x$ Here's the proof, let x be some negative number such as -2 why not that's a possible value x can take: $\sqrt{(-2)^2} = \sqrt{2^2} \ne -2$ So really what we have is the absolute value function! This is kind of tricky to deal with, but easier if you have bounds on your integral: $\int \sqrt{x^2} dx = \int |x| dx = \int_{-a}^0 -x dx + \int_0^b x dx$ A picture will hopefully clarify this a bit! |dw:1442607260478:dw|

14. anonymous

Thank you very much

15. Empty

Cool so was that all no questions?! :O

16. anonymous

there are lol ill try to do them myself and if i cant ill come back for help

17. IrishBoy123

ah! thanks a million, @Empty

18. anonymous

how do i integrate this now |dw:1442608670564:dw|

19. anonymous

@ganeshie8

20. anonymous

@Empty

21. IrishBoy123

@magepker728 make a new thread for this new question this one is a tan sub [or maybe, just for fun, a hyperbolic: $$cosh^2(x) − sinh^2(x) = 1$$].

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