Question

how do i integrate sqrt(x^2-49)/x^4

Answer 1

@modphysnoob quotient rule right?

Answer 2

I know I have to use trig substitution, but not sure how

Answer 3

this is integration , quotient rule is for differentiation

Answer 4

oops, sorry I haven't done calc in over 2 months :/
But yea now I remember

Answer 5

this probably involves something way more than trig sub

Answer 6

i get stuck after i make the substitution using secant

Answer 7

i know i have to be familiar with sohcahtoa

Answer 8

Can't you rewrite and work backwards from there?
\[\int\limits (x^2-49)^{1/2}x^{-4}dx\]

Answer 9

you have to use trig? if so, then id go with either a sin or cos setup

Answer 10

or tangent/cotangnet stuff

Answer 11

sec^2 - 1 = tan^2

Answer 12

for this particular problem I have make a substitution using secant

Answer 13

let x = 7sec(t)

Answer 14

I have that but after i plug everything in I get a little lost

Answer 15

i am stuck at sqrt (7sec(t))^2/(7 sec(t))^4 * 7sec(t)tan(t)

Answer 16

x = 7sec(t)

dx = 7sec(t)tan(t) dt
\[\frac{\sqrt{(x)^2-49}}{(x)^4}~(dx)\]
replace parts
\[\frac{\sqrt{(7sec(t))^2-49}}{(7sec(t))^4}~(7sec(t)tan(t)~dt)\]
\[\frac{\cancel{7sec(t)}tan(t)~\sqrt{49}\sqrt{sec^2(t)-1}}{7^3 sec^3(t)}~~dt\]
\[\frac{7sec(t)~tan(t)~tan(t)}{7^3 sec^3(t)}~~dt\]
\[\frac{tan^2(t)}{7^2 sec^2(t)}~~dt\]

Answer 17

\[tan=\frac{sin}{cos}\]
\[\frac1{sec}=cos\]
\[\frac{tan}{sec}=\frac{sin}{cos}cos=sin\]

Answer 18

\[\frac1{49}\int sin^2(t)~dt\]

Answer 19

theres a pretty simple trig identity for sin^2:

1/2 - cos(2t)/2

Answer 20

I see. So i wouldn't need to draw a triangle because my professor showed us examples using sohcahtoa

Answer 21

the triangle would be useful for after the integration to compare it back to a function without trigs

Answer 22

I understand. So the triangle isn't absolutely necessary ?

Answer 23

no, its not absolutely necessary; just convenient is all

Answer 24

\[x = 7~sec(t)\]
\[sec(t) = x/7\]