Question

I'm stuck on this trig substitution problem. integral of sqrt(x^2-64)/x dx
I got the theta down to the integral of 8tan^3(x). So I did that and ended up with:
8((cos^2(x)/2) + log(sin(x))
Am I on the right track? Where do I go from here?

Answer 1

Hmm, I don't think you did something correctly. How about using \(x=8\sec u\) and \(dx = 8\tan(u) \sec(u)\,du\) as your substitution?

Answer 2

@ganeshie8 can help you with this, I have to call it a night...

Answer 3

I'll be here :) but I see he already got what he needs exactly :)

Answer 4

\(\large \int \frac{\sqrt{x^2-64}}{x} dx\)

\(\large \int \frac{\sqrt{(8\sec u)^2-64}}{8\sec u } (8 \sec u \tan u du)\)

\(\large \int \sqrt{(8\sec u)^2-64}~~(\tan u du)\)

\(\large \int 8\sqrt{(\sec u)^2-1}~~(\tan u du)\)

\(\large \int 8\sqrt{(\tan u)^2}~~(\tan u du)\)

\(\large \int 8(\tan u)~~(\tan u du)\)

\(\large \int 8 (\tan u)^2 du\)

Answer 5

it simplifies to that right ?

Answer 6

Yep, I see what I did now. I forgot to take the root of tan^2

Answer 7

:) see if u can manage the rest
u will need to rewrite tan in terms of sec and take the integral

Answer 8

also dont forget to convert u's to x's... good luck !

Answer 9

Thanks! Hopefully I won't mess it up somewhere down the line. I'm pretty new at trig substitution. Let me see what I come up with.

Answer 10

So you go back to the x= 8sec(u) formula and say u= sec^-1(x/8) right?
I ended up with tan(sec^-1(x/8))-8sec^-1(x/8)

Answer 11

technically, thats correct... but that doesnt look neat right ?
so to make it look neat, u need to draw a triangle.

Answer 12

\(\large \int 8 (\tan u)^2 du\)

\(\large 8\int (\sec u)^2 - 1 du\)

\(\large 8\tan u - 8u + c \)

Answer 13

you got that after taking integral right ?

Answer 14

to convert u's to x's,
you need to draw a triangle

Answer 15

Yes I did.

Answer 16

Would you use x as the hypotenuse and 8 as the adjacent side? That's what it looks like according to my other answer.

Answer 17

right !

\(x = 8 \sec u\)