Question

integral ((x^2 + 1)^1/2)/x from 1 to 2

Answer 1

u=x^2+1
du =2x dx
du/2x=dx

Answer 2

this is really crazy looking but can't you u just be be the top

Answer 3

I would use u substitution here. Try making:
u=(x^2+1)

Answer 4

\[\int\limits_{1}^{2} \sqrt{x ^{2} - 1} / x\]

Answer 5

I'm supposed to do it by setting up a right triangle of some sort

Answer 6

x=(u-1)^(1/2)

Answer 7

ohhhhh

Answer 8

trignometric substitution

Answer 9

yes!

Answer 10

u= asec(theta)

Answer 11

let x = 1sec(theta)

Answer 12

dx=sec(x)tan(x)

Answer 13

how did you come up with =1sec(theta?

Answer 14

there is special instances ... like how i remember is that when you were taught inverse trig integrals

Answer 15

i'm not sure how they work again but it is triangles that describe the relationship

Answer 16

for inverse
A^2+u^2 =was how you noticed arctan correct well the same for trig substitution only with a squareroot

Answer 17

when you did inverse trig when you saw sqrt of u^2-a^2.... secant was the first you thought of

Answer 18

http://www.youtube.com/watch?v=3lC5AuCFK4c - here is a video on it

Answer 19

oh okay, so in this problem, x would be hypotenuse, adjacent would be 1, and opposite would be root x^2 -1?

Answer 20

yes once you get all the way at the end you have to remember you used substitution so you have to put it back into x form

Answer 21

okay so sec (theta) = x, so dx = sec(theta)tan(theta)

Answer 22

tan (theta) = root x^2 - 1

Answer 23

then substituting: tan(theta)sec(theta)tan(theta)/sec(theta)

Answer 24

so far looks good

Answer 25

so cancel sec(theta), and then its just the integral of tan^2 (theta)? is that right?

Answer 26

i believe so

Answer 27

how do i do integral of tan^2 theta?

Answer 28

however you cannot take the integral of tan^2 so you are going to have to do

Answer 29

trignometric integrals which is what you were doing before

Answer 30

okay how would i start that?

Answer 31

with the identity
\[\cos^2(x)+\sin^2(x)=1\]
\[1+\tan^2(x)=\sec^2(x)\]

Answer 32

okay so tan^2 x = sec^2 x -1

Answer 33

exactly

Answer 34

so the integral of that is tan x - x?

Answer 35

well now you integrated it... but however you used substituion much like you did with u..... so when you evaluate you have to do something

Answer 36

and your x should be thetas

Answer 37

oh yeah thanks

Answer 38

\[\tan \theta - \theta from 1 \to 2\]

Answer 39

so would i substitute something in for theta?

Answer 40

those are the integration limits for x

Answer 41

you can either change your integration limits to theta or x

Answer 42

ill change theta to x, its easier for me

Answer 43

what would theta change to though?

Answer 44

arcsec

Answer 45

use your substitution that u started with to solve for theta

Answer 46

1.22

Answer 47

okay so: \[(\sqrt{x^{2}} - 1) - arcsec x from 1 t\]

Answer 48

that looks correct

Answer 49

okay how do i solve arcsec x now?

Answer 50

integral sqrt(1+x^2)/x dx
For the integrand, sqrt(x^2+1)/x substitute x = tan(u) and dx = sec^2(u) du. Then sqrt(x^2+1) = sqrt(tan^2(u)+1) = sec(u) and u = tan^(-1)(x):
= integral csc(u) sec^2(u) du
For the integrand csc(u) sec^2(u), use the trigonometric identity sec^2(u) = tan^2(u)+1:
= integral (tan^2(u)+1) csc(u) du
Expanding the integrand (tan^2(u)+1) csc(u) gives csc(u)+tan(u) sec(u):
= integral (csc(u)+tan(u) sec(u)) du
Integrate the sum term by term:
= integral csc(u) du+ integral tan(u) sec(u) du
For the integrand tan(u) sec(u), substitute s = sec(u) and ds = tan(u) sec(u) du:
= integral 1 ds+ integral csc(u) du
The integral of csc(u) is -log(cot(u)+csc(u)):
= integral 1 ds-log(cot(u)+csc(u))
The integral of 1 is s:
= s-log(cot(u)+csc(u))+constant
Substitute back for s = sec(u):
= sec(u)-log(cot(u)+csc(u))+constant
Substitute back for u = tan^(-1)(x):
= sqrt(x^2+1)-log((sqrt(x^2+1)+1)/x)+constant
An alternative form of the integral is:
= sqrt(x^2+1)-csch^(-1)(x)+constant
Which is equivalent for restricted x values to:
= sqrt(x^2+1)-log(sqrt(x^2+1)+1)+log(x)+constant
now you aply the limits u'll get the answer as 1.22

Answer 51

the book said by using trig substitution...

Answer 52

s this ur ans!!!!1

Answer 53

okay im kinda stuck, I'm at :\[\sqrt{3 - arcsec 2} - \sqrt{arcsec 1}\]

Answer 54

http://www.wolframalpha.com/input/?i=integral+%28%28x^2+%2B+1%29^1%2F2%29%2Fx&x=5&y=7

Answer 55

shouldn't arc bet outside the sqr rt

Answer 56

wait a second...

Answer 57

okay so i tan theta - theta from 1 to 2 is what we are trying to solve

Answer 58

arccos(1/x)

Answer 59

yess from there you are correct

Answer 60

and tan theta = root x^2 - 1 and theta is arc sec x

Answer 61

so yes i messed up there

Answer 62

\[\sqrt{3} - arcsec 2 - acrsec 1\] is the answer?

Answer 63

i believeso let met check with my imaginary calculator... let me go get it

Answer 64

it doesn't come out right when i do it

Answer 65

i know tan theta - theta is correct

Answer 66

yeah same, I'm just not sure after that

Answer 67

you could always go into theta

Answer 68

what do you mean

Answer 69

x=sec(x) correct?

Answer 70

so 2 = sec(theta) and 1=sec(theta)

Answer 71

sec(theta) = x

Answer 72

will get you the limits in theta

Answer 73

find the angles in which sec(theta) =1 and 2

Answer 74

one will be when cos=1/2

Answer 75

other will be when cos =1

Answer 76

so 0 and pi/3

Answer 77

are your upper and lower

Answer 78

okay so then what is the equation?

Answer 79

if you switch the limits you don't have to take it out of theta

Answer 80

so tan(theta)-theta)]from 0 to pi/3

Answer 81

did your teacher ever teach you that? you can keep your u substitutions in u form if you plug x into your substitution

Answer 82

so root 3 - pi/3 is the answer?

Answer 83

i think that is right actually!

Answer 84

yes that looks right idk how she got 1.22

Answer 85

yeS!!!! you are the best

Answer 86

glad i was help

Answer 87

btw i believe the way i solved your other questions is the way your teacher wants you to solve it as in my book, trig integrals is right before trig substitution

Answer 88

and parts you probably don't even have a clue yet

Answer 89

okay thank you!