Question

Can people here answer my calc 2 questions?

Answer 1

possibly, what is the problem?

Answer 2

calc2? here? people with answers? .... mehbee

Answer 3

Find the Integral of (sqrt(x^2-9)/x^3
You have to use trig substitutions.

Answer 4

what trig subs have you tried?

Answer 5

x = 3sec(t) perhaps?

Answer 6

i know it's 3sec(t), i got pretty far but I got lost at the end

Answer 7

x^3 = sec^3(t)

dx = 3sec(t)tan(t) dt

\[\int \frac{3sec(t)tan(t)}{sec^3(t)}*\sqrt{9sec^2(t)-9}\ dt\]
\[\int \frac{3sec(t)tan(t)}{sec^3(t)}*3tan(t)\ dt\]
\[\int 9\frac{tan^2(t)}{sec^2(t)}\ dt\]
\[\int 9\frac{sin^2(t)}{cos^2(t)}*cos^2(t)\ dt\]
\[9\int sin^2(t)\ dt\]

Answer 8

if i did it right, the rest is simple enough

Answer 9

there are a few ways to tackle a sin^2

reduction formula or indentity

Answer 10

I see what you did and it looks good, but the answer in the back of the book is (1/6)sec^-1(x/3) - sqrt(x^2-9)/(2x^2) + C

Answer 11

It's ridiculous, I'm not gonna waste your entire day trying to figure it out. Thanks for trying though.

Answer 12

the trig sub just makes for an eaiser computation; in the end you still have to turn it back into xs

Answer 13

good luck :)

Answer 14

Thanks