Question

\[u=\int\frac{\text dx}{x^2\sqrt{1+x^2}}\]

Answer 1

\[\text{let }x=\tan\theta\]\[\text dx=\sec^2\text d\theta\]
\[u=\int\frac{\sec^2\theta\cdot \text d\theta}{\tan^2\theta\sqrt{1+\tan^2\theta}}\]\[u=\int\frac{\sec^2\theta\cdot \text d\theta}{\tan^2\theta\sec\theta}\]\[u=\int\frac{\sec\theta\cdot \text d\theta}{\tan^2\theta}\]

Answer 2

im stuck

Answer 3

perhaps change everything into cos and sin?

Answer 4

\[
\frac{\sec (\theta )}{\tan
^2(\theta )}=\cot (\theta )
\csc (\theta )
\]
The answe is - csx(\theta) + C

Answer 5

\[
-\csc(\theta)
\]

Answer 6

+ Constant

Answer 7

\[u=\int\frac{\sec\theta\cdot \text d\theta}{\tan^2\theta}\]\[=\int\frac 1{\cos \theta}\times\frac{\cos^2\theta}{\sin^2\theta}\text d\theta\]\[=\int\frac{\cos\theta}{\sin^2\theta}\text d\theta\]\[=\int\cot\theta\csc\theta\text d\theta\]

Answer 8

yep

Answer 9

then use the integration table

Answer 10

i hope i never get an integral like this on the test... i've never seen this rule before

Answer 11

how do i integrate \[=\int\cot\theta\csc\theta\text d\theta\]
integration by parts?

Answer 12

not entirely sure

Answer 13

i just see it in my book on the front page with all the basic integrations

Answer 14

how can i show that @eliassaab

Answer 15

Didn't you learn how to find the derivative of csc(x)?
That is easy.

Answer 16

Derivative of csc(x) is - csc(x) cot(x)

Answer 17

i cant remember how

Answer 18

\[\frac{d}{dx}\frac{1}{sin(x)}\]=

\[sin(x)^{-1}\]=-1(sin(x)^{-2}*cos(x)=-1/sin(x)*cos(x)/sin(x)

Answer 19

=-csc(x)cot(x)

Answer 20

so i have to use integration by parts , twice

Answer 21

now let cot(x) = u

u=csc(x)
du=-csc(x)cot(x)

\[\int{}{}-du\]

Answer 22

\[-\int{}{}du=-x+c=-csc(x)+c\]

Answer 23

i find that super strange lol but it works

Answer 24

i've never sen a problem like that amazingly

Answer 25

don't read cot(x)=u .... i meant csc(x)

Answer 26

it just a tiny part in a larger question

Answer 27

are you a teacher or just have homework that you have to do on the computer?!?

Answer 28

this is just question from my DE subject textbook form second year university,
i am opting to work in LaTeX because it is neater

Answer 29

im not a teacher just a physics student

Answer 30

lol you use a lot of function that my teacher doesn't use... such as mew (t) for your integrating factor

Answer 31

that's interesting though... dose your teacher give you not a lot of questions... it seems like that question is a doozie

Answer 32

actually my book uses \(R(x)\) for integrating factors , i have stated using \(\mu(x)\) because it seams to be what most people use , the other common symbol used for the integrating factor i have seen is \(IF(x)\)

Answer 33

most of the questions im by book have ended up taking over a page of latex,

Answer 34

i guess i can use this standard integral

\[\int\sin^m(x)\cos^n(x)\text dx=\frac{\sin^{m+1}(x)\cos^{n-1}}{m+n}+\frac{n-1}{m+n}\int\sin(x)^m\cos^{n-2}(x)\text dx\]

Answer 35

m=-2
n=1

Answer 36

\[=\int\frac{\cos\theta}{\sin^2\theta}\text d\theta=\]\[\int\sin^{-2}(\theta)\cos(\theta)\text d\theta=\frac{\sin^{-1}(\theta)}{-1}=-\csc(\theta)+C\]

Answer 37

how can i show these are equal\[u=-\csc(\theta)+C\]\[u=-\frac{\sqrt{x^2+1}}{x}\]

Answer 38

why not let u= 1+x^2, try that

Answer 39

oh i forgot
\[x=\tan\theta\]
\[\theta=\arctan x\]

Answer 40

figure it out?

Answer 41

\[=-\csc(\arctan x)\]

Answer 42

how can i simplify the trig functions

Answer 43

you let tan(theta)=x correct... well if you construct your triangle for you trig sub