Question

I'm having trouble changing limit
\[\int_{0}^{1} x(\sqrt(x^2 +4))\]
my plan is to solve the integral using
\[x=2tan\theta\]
\[dx=2sec^{2}\theta d\theta\]

Answer 1

0=2tan theta so theta=0
1=2sin theta

Answer 2

no it's 2=sin theta

Answer 3

put x^2+4 = t
2*x*dx = dt

Answer 4

what about the limit for some reason I don't seem to know how to change 2=sin theta into theta=? in terms of pi...and yes i tried wolf

Answer 5

\[-1<\sin \theta <1\]
it cann't be 2

Answer 6

well I guess I'm a having a little trouble with the integral
\[16\int tan\theta sec^2 \theta sec^2 theta d\theta\]

Answer 7

16\int tan\theta sec^2 \theta sec^2 \theta d\theta

Answer 8

\[16\int tan\theta sec^2 \theta sec^2 \theta d\theta\]

Answer 9

I think i got it
u=sec^2
du= 2 tan sec^2

Answer 10

\[8\int udu\]

Answer 11

but still changing the limits are an issue...

Answer 12

\[4 sec^4 \theta\]

Answer 13

sometimes putting wrong substitution can make the question worse

Answer 14

what u did last? i think its not correct .check it again

Answer 15

let 2tan \theta = x , x=0, \theta = 0, for x=1 \theta = arctan(1/2)
and i am getting \int 8 tan \theta \sec^3 \theta d\theta as integration

Answer 16

\[2tan \theta = x , x=0, \theta = 0, for x=1 \theta = arctan(1/2)\]
\[\int 8 tan \theta \sec^3 \theta d\theta\]
?

Answer 17

how about dx?

Answer 18

to solve this
let \sec \theta = u, du = \sec \theta \tan \theta, then you have \int u^2 du = 1/3 u^3

Answer 19

so the upper limit of the new limit is?

Answer 20

why would you want to use trig sub for this problem?

Answer 21

it's the section that I'm working on in the chapter....I'm forced to do it

Answer 22

just because the problem is in that section doesn't mean that you are supposed to use that technique

Answer 23

just to double check...what do you get when you do it with a different technique?

Answer 24

sometimes the problems are there to test if you can see that an older technique already learned in a previous section will work better

Answer 25

I would do \[u=x^2+4\]

Answer 26

so yeah that's way easier

Answer 27

the trig subs still bother me though

Answer 28

my answer is way off...so how did you get away with not having dx= 2 sec^2 \theta d\theta?

Answer 29

final answer i have \[\frac 1 3 (5\sqrt 5 +8)\]

Answer 30

\[ \int x \sqrt{x^ 2 + 4} dx = \frac 1 2 \frac{(x^2 + 4)^{3/2}}{3/2}\]
plugin the limits.

Answer 31

\[\frac 1 2 \frac{(x^2 + 4)^{3/2}}{3/2}\]
\[\frac 1 2 \frac{(1 + 4)^{3/2}}{3/2} =5\sqrt5 *\frac 2 6\]
\[\frac 1 2 \frac{( 4)^{3/2}}{3/2} =8*\frac 1 3 \]

Answer 32

lolz...i'm supposed to subtract them...sorry

Answer 33

\[\frac5 3 \sqrt5 -\frac 8 3\]
good enough i guess

Answer 34

well you got
\[ \frac 13 (5 \sqrt 5 - 8) \]

Answer 35

or the other way around

Answer 36

no that's right

Answer 37

http://www.wolframalpha.com/input/?i=integrate+x+sqrt%284%2Bx^2%29+from+0+to+1

Answer 38

yay!

Answer 39

is there something I'm missing?