Question

How to find this integral?

Answer 1

\[\int\limits_{3}^{8} \sqrt{1+ 9/16 * x^2}\]

Answer 2

HI!!

Answer 3

a little hard to read
is it
\[\int_3^5\sqrt{1+\frac{9}{16}x^2}dx\]

Answer 4

Hi :) Yeah that's right except the upper bound which is 8 :)

Answer 5

Use trigonometric substitution

Answer 6

i think you can get rid of the radical by using
\[x=\frac{3}{4}\tan(\theta)\]

Answer 7

might as well change the limits while you are at it so you don't have to switch back

Answer 8

wait so we just plug in x as that?

Answer 9

Use what misty1212 said and remenber that \(dx = \frac{3}{4}sec^2(\theta )d\theta\)

Answer 10

Ooohohh substitution okay :)

Answer 11

i remember this from a book, not this one but this one
\[\int \sqrt{1+x^2}dx=\frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\log(\sqrt{x^2+1}+x)\] it is a rather involved trig sub ending with integrating \(sec^3(x)\)

Answer 12

Also, \(cos^2\theta + sin^2\theta = 1\), dividing by \(cos^2\theta \) we get \(1+tan^2\theta = sec^2\theta\)

Answer 13

Hey @misty1212 I'm having a bit of a problem deciding what kind of method to use for solving integral questions in general.

Why shouldn't you use (b - a) lim x-> 0 1/n(arithmetic progression) method for this?

Answer 14

Wait so we just substitute x= 3/4 * tan(theta) into the given equation?

Answer 15

yes

Answer 16

to cut to the chase, after a bunch of substitution you are going to have to integrate
\[\int \sec^3(u)du\]
this is a pain, look up the reduction formula in the back of the book

Answer 17

oh maybe not
since this is a definite integral you can change the limits of integration at the start, then you don't have to switch back
do you know what i am talking about?

Answer 18

Yes, I think so

Answer 19

so \[x=\frac{3}{4}\tan(\theta)\] solve
\[3=\frac{3}{4}\tan(\theta)\] for \(\theta\) as your lower limit of integration

Answer 20

pretty much in your head you get \[\theta=\tan^{-1}(4)\]

Answer 21

repeat with
\[8=\frac{3}{4}\tan(\theta)\]

Answer 22

theta= tan^(-1) (32/3) :) Those are the bounds ok

Answer 23

then you are still going to have to integrate \(\sec^3(\theta)\)

Answer 24

look it up in the back of the book
it is a pain to do by hand

Answer 25

which is 1/2 * secutanu + 1/2 * ln|secu + tanu| + C

Answer 26

okay, I'm confused.... what would u be?

Answer 27

if i am not mistaken you can just use your new limits of integration
get a second opinion, i am not 100% sure, but pretty sure

Answer 28

once you make all the changes you do not have to go back or anything

Answer 29

Yes, you don't have to go back to the original variable x since you changed the integration boundaries

Answer 30

of course you do have to figure out what
\[\sec(\tan^{-1}(4))\] is etc
really this is an involved integral, kind of a pain

Answer 31

not really, though since
\[\tan(\tan^{-1}(4))=4\] and
\[\tan(\tan^{-1}(\frac{32}{3}))=\frac{32}{3}\]

Answer 32

\(sec(tan^{-1}(4)) = \sqrt{1+tan^2(tan^{-1}(4))} = \sqrt{1+4^2}\)

Answer 33

that is one method, i favour the triangle method
same answer though

Answer 34

Okay, sounds good :)

Answer 35

I think I got it now, thank you :D