Question

@zepdrix @freckles @ganeshie8 @mathmale @mathstudent55 @zepdrix @surjithayer

Answer 1

@Nnesha

Answer 2

Are you just asking what g(x) and g'(x) are?

Answer 3

I think so

Answer 4

I agree on that much.
Might want to experiment a bit. Supposing that g(x)=2x, what would g '(x) be?

I'm not saying that this is the way to go, but you may learn something useful from trying this out.

Answer 5

g'(x) = 2

Answer 6

Actually, g '(x) would be 2dx.

Answer 7

\[Let g(x) = (1/2)\sec(2x)\]
Then \[g'(x) = 2(1/2)\sec2xtan2x = \sec2xtan2x.\]
Is this what you are asking? I hope I didn't just answer the question completely. Just trying to ask what you're looking for.

Answer 8

Maybe u = cos 2x?

Answer 9

I think it best that you actually experiment with these various proposals.

What would happen to the integral on the left were you to replace g(x) with 2x and replace
g '(x) with 2dx?

Answer 10

f(2x)(2)dx

Answer 11

I agree with that. Are you now able to identify what the function f(x) might be?

Answer 12

Hint: sec 2x tan 2x is f(2x). sec u tan u would be f(u). What is f(x)?

Hint #2: The integral of sec u tan u with respect to u is ... what?

Answer 13

Well...wouldn't 2f(2x) = sec 2x tan 2x?

Answer 14

so f(x) = \(\Large\frac{\sec x \tan x}{2}\)

Answer 15

right?

Answer 16

I'll let you answer that. Please find the derivative, with respect to x, of your f(x).

Answer 17

If your f(x) is correct, then this derivative should exactly match the given one on the riight.

Answer 18

ok,. that'll take me a minute. hold on

Answer 19

I don't think it works out because I got \(\Large\frac{1}{2}sec^3 x - \tan^2 x\sec x\)

Answer 20

Sorry, but I think I was off-track there. First of all, we are asked NOT to evaluate the integral. Second of all, we're asked to identify u = g(x) and du = g '(x)dx.

Answer 21

I think u = g(x) = cos x and du = -sin x dx

Answer 22

Again I ask you to consider the possibility that u=g(x) =2x, so that du=2dx.

If you now attempt to write f(g(x))g '(x)dx using u=2dx, do you end up with the integrand on the right side of the given equation?

Answer 23

Of course you can try other g(x) and g '(x) if you like. In the end we have to find g(x) and g '(x) such that we obtain the integral of sec 2x tan 2x dx.

Answer 24

I'm not sure how I'd do that.

Answer 25

Assume that f(u)=sec u tan u, and that g(x)=2x. Then g '(x) =2dx.
Substitute these into the left side of the given (homework) equation. Do your results match the right side of the given equation?

Answer 26

No...?

Answer 27

Also assume that u=g(x)=2x. Sorry for having forgotten the ' u .'

Answer 28

So you get 2 sec 2x tan 2x dx, right?

Answer 29

Let's try this (not really a big jump):

Assume that \[f(u)=\frac{ \sec u \tan u }{ 2 }\]

Answer 30

substitute u=g(x)=2x.

f(2x)=?

Answer 31

\(\Large\frac{\sec 2x \tan 2x}{2}\)

Answer 32

This is pretty much the same as the right side of the given equation. All we need to do now is to figure out why that '2' in the denom. hasn't cancelled out.

Answer 33

hmmm...any ideas, lol?

Answer 34

If g(x)=2x, then g '(x)=2dex, and this 2 will be cancelled out by the 2 in the denom. of our f(x).

Answer 35

Excuse me, that's g '(x)=2dx

Answer 36

Oh, right! I forgot we were multilplying by g'(x) as well

Answer 37

Right.

Answer 38

Awesome, thank you. :)

Answer 39

My great pleasure! Bye for now.