Question

\[evaluate \int\limits\limits \frac{dx}{4x \sqrt{x ^{2}-16}}\]

Answer 1

didnt you just post that question?

Answer 2

I have a question for that.

Answer 3

what should I let???

Answer 4

let u = 4x???

Answer 5

and du = 4dx???

Answer 6

isn't this some inverse trig function?

Answer 7

It's a trig sub.
\(x=4sec\theta\)

Answer 8

what @Mimi_x3 ???

Answer 9

\(dx=4sec\theta tan\theta\)

Answer 10

What's wrong?

Answer 11

I can't get it. :(

Answer 12

oh so it is

Answer 13

what should i let???

Answer 14

\[\frac{1}{4} \int\limits\frac{4\tan\theta \sec\theta}{4\sec\theta\sqrt{16\sec^{2}\theta-16}} => \frac{1}{4} \int\limits\frac{4\tan\theta \sec\theta}{4\sec\theta\sqrt{16(\sec^{2}\theta-1)}} \]
=> \[\frac{1}{4} \int\limits\frac{4\tan\theta \sec\theta}{4\sec\theta\sqrt{16\tan^{2}\theta}} \]

Answer 15

\[\frac{1}{4} \int\limits\frac{4\tan\theta \sec\theta}{4\sec\theta4\tan\theta} \]
maybe..if i didntt make a mistake

Answer 16

wow what an amazing amount of cancellation! just get the integral of 1

Answer 17

lol yeah; a very easy integral

Answer 18

which leads me to believe we could do this without a trig sub, if you recall that
\[\frac{d}{dx}\sec^{-1}(x)=\frac{1}{x\sqrt{x^2-1}}\]
but maybe not

Answer 19

table of standard integrals?

Answer 20

@satellite73 was right.

Answer 21

lol sorry i went the long way; i didnt know that sec was in the table of standard integrals

Answer 22

you don't need to do it in standard integral. but it's okay @Mimi_x3

Answer 23

Thanks @Mimi_x3 and @satellite73 :)

Answer 24

no it is not the long way
if you happen do remember the derivative of inverse secant then i think you can use \(u=\frac{x}{4}\) and get
\[\frac{1}{u\sqrt{16u^2-16}}\] or something like that
probably constants are wrong