Question

Integral help.

Answer 1

\[\int\limits_{}^{}\frac{ dx }{ \sqrt{x^2-4x} }\]

Answer 2

This looks significantly similar to what we want in order to get
\[\sqrt{x^2-a^2}\], x=asec(theta)
right?

Answer 3

Start by completing the square

Answer 4

(x-2)^2-4 ?

Answer 5

As an auxiliary (I was going to prove it but it became quite extensive) it is important to note:
\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{\alpha ^2- (x \pm \beta)^2} }dx = arcsen(\frac{ x \pm \beta }{ \alpha })\]

Answer 6

Also:
\[\int\limits_{}^{}\frac{ dx }{ \sqrt{\alpha^2+(x \pm \beta)^2} }dx=\ln((x \pm \beta) + \sqrt{(x \pm \beta) ^2+\alpha^2})\]

Answer 7

oh woah. I don't know any of that o.0

Answer 8

you have completed the square correctly
so you can write your integral as
\[\int\limits \frac{dx}{\sqrt{(x-2)^2-4}}\]

Answer 9

ok.

Answer 10

so if x-2=2 sec(theta) then what....

Answer 11

so you're putting this as
a^2-x^2
not
x^2-a^2?

Answer 12

there is really only three forms we can compare this to (I consider the other 3 repeats)...
\[\cos^2(\theta)=1-\sin^2(\theta) \\ 1+\tan^2(\theta)=\sec^2(\theta) \\ \sec^2(\theta)-1=\tan^2(\theta)\]

we don't have constant-variable^2 so first one is out
we don't have constant+variable^2 so the second is out
we do have variable^2-constant so the third one is totally in

Answer 13

where the constant is positive

Answer 14

Right, but then if we have
sqrt(x^2-a^2) then we would get x=asec(theta) right?
which would be
x=2sec(theta)
why do you set that equal to x-2?

Answer 15

oh duh. because x = x-2
lol sorry o.0

Answer 16

x=x-2 looks weird to me but I know what you mean

Answer 17

and the 2 in front of the sec came from the a part

Answer 18

x=2sec(theta)+2

Answer 19

if we make that sub I suggested earlier
\[\sqrt{(x-2)^2-4} \\ x-2=2 \sec(\theta) \\ \text{ you will see we get } \\ \sqrt{[2\sec(\theta)]^2-4} \\ \text{ we are going to be able to factor out that 4 inside there } \\ \sqrt{4(\sec^2(\theta)-1)} \\ \sqrt{4} \sqrt{\sec^2(\theta)-1}\]

Answer 20

the whole reason to choose this sub though is to eventually get rid of that radical

Answer 21

which we will be able to if we remember why we chose that sub
we chose that sub because we knew we can write the inside as tan^2(theta)

Answer 22

\[2 \sqrt{\tan^2(\theta)}\]

Answer 23

and that just becomes
2(tan(theta)) yeah?

Answer 24

assuming tan(theta) is positive yes

Answer 25

if we have a definite integral you will have to be careful

Answer 26

but when it comes to indefinite integral for some reason most books and teachers choose to go with the assumption that the trig function is positive
but we do know the trig function can also be negative

Answer 27

right

Answer 28

so since we are dealing with a indefinite integral
we will make the assumption tan(theta) is positive
and write the denominator of our integrand as you have expressed it above

Answer 29

\[\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ \tan(\theta) }d(x)=\frac{ \cos(\theta) }{ \sin(\theta) }d(\theta)\]

Answer 30

\[x-2=2 \sec(\theta) \\ \implies dx=2 \sec(\theta) \tan(\theta) d \theta \]

Answer 31

oh whoops! thank you! o.o

Answer 32

that should have been plugged in way before.

Answer 33

for the numerator

Answer 34

we played with the denominator
we never touched the numerator until now

Answer 35

hm, well when we plugged in x=2sec(theta)+2 we should have immediately replaced the dx as well o.o I think.

anyway so that comes out to
int_ sec(theta)d(theta)?

Answer 36

\[\int\limits \frac{dx}{\sqrt{x^2-4x}}\\ = \int\limits \frac{dx}{\sqrt{(x-2)^2-4}} \\ =\int\limits \frac{ 2 \sec(\theta) \tan(\theta)}{2 \tan(\theta)} d \theta \]

Answer 37

yep everything cancels except the sec(theta) d theta on top

Answer 38

Yay!!
and then that's a whole process too haha
multiplying it by
(sec(theta)tan(theta)/(sec(theta)tan(theta)) right?

Answer 39

you mean sec(theta)+tan(theta) on top and bottom?

Answer 40

or some peeps just let you remember that \[\int\limits \sec(\theta) d \theta=\ln|\sec(\theta)+\tan(\theta)|+C\]

Answer 41

Yeah. So like on an exam, do you think I would have to prove that? like actually show the process? or is it enough of an "identity"?

Answer 42

to some teachers I knew you could just remember that...
I didn't know any teachers but they could exist that you do have to derive that

Answer 43

i do not know your particular teacher :p

Answer 44

kk
the answer i'm given is:
\[\ln|x-2+\sqrt{x^2-4x}|+C\]
but right now I just have
ln|sec(theta)+tan(theta)|+C

Answer 45

you have to write in terms of x

Answer 46

Is it because I have to now switch it to terms of x? so I should draw out a triangle and find the sides?

Answer 47

\[x-2=2 \sec(\theta) \\ \frac{x-2}{2}=\sec(\theta)\]

Answer 48

you can draw a triangle and find tan(theta) with this information

Answer 49

sec= (x-2)/2
tan=sqrt(x^2-4x)/2