Question

evaluate

Answer 1

\[\int\limits_{}^{}\left(\begin{matrix}1 \\ x \sqrt{x ^{2}-1}\end{matrix}\right)\]

Answer 2

You could do trig substitution. :P

Answer 3

x^2 - 1 = t^2
xdx = tdt

multiply and divide by x

u get
xdx/ (x^2sqrt(x^2-1))

Answer 4

which is
tdt / ((t^2 + 1)t)

Answer 5

which is dt/.(t^2 +1)
which is tan^(-1) (t)

Answer 6

tan^(-1) [sqrt(x^2 -1)]

Answer 7

is it right?

Answer 8

Yeah.

Answer 9

i am working with trig substitutions

Answer 10

itll be messy

Answer 11

i know.... i am half way already.. and stuck

Answer 12

dont do it

Answer 13

use my method..avoid muddles in integration

Answer 14

but is homework...

Answer 15

so?

Answer 16

if ur so guilty about aping it y did u post it

Answer 17

hello,, if i am in the chapter of trig substitution i can go with other method to my professor u dont think so...

Answer 18

then say that

Answer 19

Okay. Doing a trig sub then. You notice you have something in the form of u^2-a^2
So you want to use SECANT.
Set up your reference triangle and your x will be on the hypotenuse, a on the adjacent, and sqrt(u^2-a^2) on the opposite.
So you have:
\[\sec(\theta)=x/1 \rightarrow \sec(\theta)=x\]
Find dx
\[\sec(\theta)\tan(\theta)d \theta=dx\]
Then:
\[\tan(\theta)=\sqrt{x^2-1}\]
Replace all that in the integral and you get:
\[\int\limits \frac{\sec(\theta)\tan(\theta) d \theta}{\sec(\theta)\tan(\theta)}\]
Which just gives you:
\[\int\limits d \theta=\theta+C\]
But you know that:
\[\tan(\theta)=\sqrt{x^2-1} \rightarrow \theta=\arctan(\sqrt{x^2-1})\]
So the integral is:
\[\arctan(\sqrt{x^2-1})+C\]

Answer 20

yeah hes right then

Answer 21

u shldve mentioned man..my bad

Answer 22

Tell me if my trig sub doesn't make sense^^^

Answer 23

wait... i think when u go back to find the value of theta u got that sec @=x

Answer 24

\[\theta=\sec ^{-1}x\]