Question

help please number 22

Answer 1

evaluate the integral

Answer 2

@agent0smith

Answer 3

@zepdrix

Answer 4

seems like a trig sub

Answer 5

yes , but having trouble

Answer 6

so would it be x= tan x
dx: sec^2 deta

Answer 7

x = tan(t) would be more appropriate

Answer 8

oh hmm would it tan deta ?

Answer 9

it's just a dummy variable. theta is fine if you'd like to be consistent with the book

Answer 10

oh okay so then

would it be like this ?
\[\int\limits_{}^{}\sqrt{(\tan \theta)^2 +1}\]

Answer 11

isn't it sec^2x-1 = tan^2x?

Answer 12

oh yes sorry

Answer 13

what about dx?

Answer 14

\[\int\limits_{}^{}\sqrt{(\sec^2\theta-1)+1} d \theta \]

Answer 15

That was sort of a silly way to apply your Pythagorean Identity :)
Yes, \(\rm \sec^2\theta-1=\tan^2\theta\)
but that means \(\rm \sec^2\theta=\tan^2\theta+1\)

Answer 16

Oh oh, woops you messed something up with your differential I think.

Answer 17

D:

Answer 18

\[\large\rm x=\tan \theta\]\[\large\rm dx=\sec^2\theta~d \theta\]

Answer 19

\[\large\rm \int\limits \sqrt{\color{royalblue}{x}^2+1}~\color{orangered}{dx}\quad=\int\limits \sqrt{\color{royalblue}{\tan}^2\color{royalblue}{\theta}+1}~\color{orangered}{\sec^2\theta~d \theta}\]

Answer 20

And then you apply your Pythagorean Identity,\[\large\rm =\int\limits \sqrt{\sec^2\theta}~\sec^2\theta~d \theta\]

Answer 21

\[\int\limits_{}^{}(\sqrt{1+\tan^2\theta}) ( 1+\tan^2\theta)\]

Answer 22

What? Why turn them back into tangents?
That causes a problem for us.
Now we have the +1 under the root again.

The whole point of using a trig sub...
is to get rid of the addition under the root.

It makes it much easier to take a square root from that point.

Answer 23

oy... no switchy switchy back to tanx

Answer 24

aah lol thats where i get stuck D:

Answer 25

I meant that I was apply Pythagorean Identity from here,\[\large\rm \int\limits \sqrt{\tan^2\theta+1}~\sec^2\theta~d \theta\]to here,\[\large\rm \int\limits \sqrt{\sec^2\theta}~\sec^2\theta~d \theta\]

Answer 26

Now.......... take square root.

Answer 27

hmm wait question what happen to the +1 ?

Answer 28

Our trigonometric Pythagorean Identity gets rid of it.
\(\large\rm tan^2\theta+1=sec^2\theta\)
There is no +1 on the right side of this equation, ya?
We're replacing the tan^2 + 1 with the right side of the equation in our integral.

Answer 29

oh wait nvm i see it .. it was from the identity right ?

Answer 30

oh okay i hope i did it right \[\int\limits_{}^{} \sec^2\theta* \sec ^2 \theta \] d deta

Answer 31

You didn't apply square root.

Answer 32

\(\large\rm \sqrt{x^2}\ne x^2\)