Question

can someone please help me with an integral problem.

Answer 1

\[\int\limits_{}^{} \frac{ \sqrt{1+x^2} }{ x} dx\]

Answer 2

how can i start this

Answer 3

\(x= \sqrt{x^2}\), will it work?

Answer 4

i dont think so
I tried this problem by using trignometric substitution

Answer 5

Then let \(x = \tan(t)\) such that \(dx = \sec^2(t) \cdot dt\)

Answer 6

yes. but i will need my t to be a thelta

Answer 7

\[\int\limits \frac{\sqrt{1 + \tan^2(t)}}{\tan(t)} \cdot \sec^2(t) \cdot dt\]

Answer 8

Hey, t and theta are one and same.. replace t by theta..

Answer 9

\[\sqrt{a^2+x^2} ---> X=atan \]

Answer 10

\[\int\limits\limits \frac{\sqrt{1 + \tan^2(\theta)}}{\tan(\theta)} \cdot \sec^2(\theta) \cdot d \theta\]

Answer 11

okay got it

Answer 12

ok

Answer 13

Use :
\(1 + \tan^2(\theta) = \sec^2(\theta)\)

Answer 14

\[\int\limits \frac{\sec^3(\theta)}{\tan(\theta)} d \theta\]

Answer 15

okay

Answer 16

Not working, I think..

Answer 17

\[\int\limits_{}^{} \sec^2(\theta) \csc(\theta) d \theta \]
If I rewrote it correctly I think you can use integration by parts here

Answer 18

make u=csc(theta) and dv=sec^2(theta) d theta

Answer 19

what if we have our sec^3 thelta change its identity to 1/costhelta

Answer 20

from there i thnk i can cancel costhelta with tanthelta=sinthelta/costhelta

Answer 21

freckles : how did you start the integration problem

Answer 22

you will get 1 by cos^2(theta).sin(theta)

Answer 23

\[\sec^3(\theta) \cdot \cot(\theta) \\ \frac{1}{\cos^3(\theta)} \cdot \frac{\cos(\theta)}{\sin(\theta)} \\ \frac{1}{\cos^2(\theta)} \cdot \frac{1}{\sin(\theta)} \\ \sec^2(\theta) \cdot \csc(\theta)\]
is exactly how I rewrote the integrand

Answer 24

just apply integration by parts now and you are home free

Answer 25

\[\int\limits_{}^{} \sec^2(thelta) \csc(thelta) \]

Answer 26

freckles is the pic from above what i use for integration by parts?

Answer 27

Try it, I am going to sleep, will find another way to solve it..
Do check your final answer here:

http://www.wolframalpha.com/input/?i=%5Cint+%5Cfrac%7B+%5Csqrt%7B1%2Bx%5E2%7D+%7D%7B+x%7D+dx&dataset=

Answer 28

Let me know if you need further help but it should be pretty easy ball game now

Answer 29

i got \[(\sec^2\theta)(-\ln \left| cscx+cotx \right| -\int\limits_{}^{}(-\ln \left| cscx+cotx \right| )(\frac{ 1 }{ \cos^2\theta }) d \theta\]

Answer 30

hmmm you did it backwards from what i said

Answer 31

like I suggested u=csc(theta) and dv=sec^2(theta) d theta

Answer 32

\[\tan(\theta)\csc(\theta)-\int\limits\limits_{}^{}\tan(\theta)(-\csc(\theta)\cot(\theta)) d \theta\]

Answer 33

okay. i have a quick question. how would i know which is my u

Answer 34

tan(theta)*cot(theta)=1
so you just have to worry about integrating csc(theta) now

Answer 35

which it looks like you already know how to do

Answer 36

you base your u on whatever makes the integral easiest

Answer 37

not on what makes it more complicated

Answer 38

okay thanks

Answer 39

After you are done with this way
I have another way to show you

Answer 40

it involves no trig sub

Answer 41

\[\int\limits_{}^{}\frac{\sqrt{1+x^2}}{x} dx \\ u=\sqrt{1+x^2} \\ u^2=1+x^2 \\ 2 u du=2 x dx \\ u du =x dx \\ \frac{u}{x} du=dx \\ \text{ now we can rewrite } u \text{ as a function of x } \\ u^2=1+x^2 \\ u^2-1=x^2 \\ \text{ \choose x pos we have } x=\sqrt{u^2-1} \\ \text{ so we are going \to replace } dx=\frac{u}{\sqrt{u^2-1}} du \text{ and } u=\sqrt{1+x^2} \\ \int\limits_{}^{}\frac{\sqrt{1+x^2}}{x} dx=\int\limits_{}^{}\frac{u}{x} dx=\int\limits_{}^{}\frac{u}{\sqrt{u^2-1}} dx=\int\limits_{}^{}\frac{u}{\sqrt{u^2-1}} \frac{u}{\sqrt{u^2-1}} du \\ =\int\limits_{}^{}\frac{u^2}{u^2-1} du=\int\limits_{}^{}\frac{u^2-1+1}{u^2-1} du=\int\limits_{}^{}(1+\frac{1}{u^2-1}) du \\ u+\int\limits_{}^{}\frac{1}{2} \frac{1}{u-1} du +\int\limits_{}^{}\frac{-1}{2} \frac{1}{u+1} du\]
then finish the integration
and replace u back in terms of x
--
this is just another way
your answer may look different from the other way but it is still in the same family of functions

Answer 42

ok on top the anitderivative of sec^2thelta is ??

Answer 43

tan(theta)

Answer 44

tan(thelta) ? i thought it was 1/cos^2 thelta

Answer 45

so you thought the antiderivative of sec^2(theta) is itself?

Answer 46

no...

Answer 47

\[\frac{d}{dx} \tan(x)=\sec^2(x) \\ \int\limits \sec^2(x) dx=\tan(x) +C\]

Answer 48

ok sorry i was looking at the wrong one. i need to remember these identies