Question

Inverse trig help?

Answer 1

I have to solve the differential equation y' = 1 / (x sqrt(4x^2 - 1))

Answer 2

I am supposed to use u and a substitution so do I use the formula du / u sqrt(u^2 - a^2)

Answer 3

factor out the 4, and that should tell you what a to use

Answer 4

Can you please show me?

Answer 5

show you how to factor out a 4?

Answer 6

lol No walk me through the problem. It's a practice problem for a test.

Answer 7

i can give it a shot ...

Answer 8

The square root of 4 is 2

Answer 9

Obviously. lol

Answer 10

1
--------------
x sqrt(4x^2 - 1)



1
--------------
x sqrt(4(x^2 - 1/4))

....
x sqrt(4) sqrt(x^2 - 1/4)



1
--------------
2x sqrt(x^2 - 1/4)

so a^2 = 1/4, a = 1/2, let u = 1/2 x

Answer 11

Okay but the 1 at the top is supposed to be du right?

Answer 12

eventually yes

Answer 13

once we have u = 1/2 x, we can find du and dx

Answer 14

Okay u = 1/2 x so du = 1/2?

Answer 15

du = 1/2 dx, dx = 2du for the dx sub

Answer 16

Oh sorry.

Answer 17

So what would I do from there?

Answer 18

Do I take the sqrt of the radicand?

Answer 19

well, we might want to solve x in terms of u, since theres gonna be an x laying about:

u = 1/2 x, 2u = x for the x sub

Answer 20

\[\frac{1}{x\sqrt{4x^2-1}}~dx\]
\[\frac{1}{2x\sqrt{\frac14u^2-\frac14}}~dx\]
\[\frac{1}{2x\sqrt{\frac14}\sqrt{u^2-1}}~dx\]
\[\frac{1}{x\sqrt{u^2-1}}~dx\]
\[\frac{2}{x\sqrt{u^2-1}}~du\]
\[\frac{2}{2u\sqrt{u^2-1}}~du\]
\[\frac{1}{u\sqrt{u^2-1}}~du\]

Answer 21

looks very secant to me if memory serves

Answer 22

Thank you!

Answer 23

hope it makes sense

Answer 24

Absolutely thank you so much!

Answer 25

youre welcome :)