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OpenStudy (kayders1997):

Someone please help!

OpenStudy (kayders1997):

Find the integral of sec20tan20do using u substitution

OpenStudy (kayders1997):

@mathmale

OpenStudy (mathmale):

the derivative of what function is sec x tan x ... ?

OpenStudy (mathmale):

Reference: http://www.math.com/tables/derivatives/more/trig.htm

OpenStudy (kayders1997):

Secx

OpenStudy (kayders1997):

@mathmale sorry had to eat

OpenStudy (mathmale):

Good. Hope you're full now. sec 2x*tan 2x *2 is the derivative of what function?

OpenStudy (mathmale):

Be sure to review the Chain Rule if need be.

OpenStudy (kayders1997):

I have no idea :/

OpenStudy (kayders1997):

Tanx? I don't know

OpenStudy (kayders1997):

@zepdrix confused here

OpenStudy (kayders1997):

Sec^2

OpenStudy (mathmale):

but what's your argument? sec x can't stand alone. Must have an argument.

OpenStudy (mathmale):

The derivative of what function is sec 2x tan 2x *2?

OpenStudy (kayders1997):

Ugh

OpenStudy (kayders1997):

2secx I really don't know :(

OpenStudy (mathmale):

the derivative with respect to x of sec 2x) is sec 2x tan 2x * 2. But you were asked to find the integral of only sec 2x tan 2x (no 2!). What function, when differentiated with respect to x, produces the derivative sec 2x tan 2x * 2?

OpenStudy (mathmale):

Hold that for a moment (it's correct, but I wanted to ask you to try something else). Differentiate sec x tan x. Next, differentiate sec 2x tan 2x. Are the results the same or different? You must follow the Chain Rule in the 2nd case.

OpenStudy (mathmale):

and the reason for that is that 2x is itself a separate function.

OpenStudy (kayders1997):

Just give me a second

OpenStudy (kayders1997):

They aren't the same

OpenStudy (mathmale):

Right, they're not. Look: sec x is a function. But so is 2x. Thus, if you want to find the derivative wrt x of (sec 2x)^2, you get sec 2x tan 2x times the derivative of 2x (which happens to be 2). OK with this so far?

OpenStudy (kayders1997):

Yes

OpenStudy (mathmale):

In taking the derivative of 2x, you have successfully applied the Chain Rule. OK: you wanna do substitution? Fine. Let u=2x. Find du/dx. du/dx=?

OpenStudy (kayders1997):

Du/Dx=2

OpenStudy (mathmale):

Right. You want to integrate sec 2x tan 2x dx. Since u=2x, and du/dx=2, solve the 2nd equztion for dx now, please. dx=?

OpenStudy (kayders1997):

1/2du=dx

OpenStudy (mathmale):

Right. So, instead of integrating sec 2x tan 2x dx, you're gonna integrate dy= sec u tan u (du/2).

OpenStudy (mathmale):

Gotta get rid of that /2. So, Move that (1/2) to in front of sec u tan u du.

OpenStudy (kayders1997):

Okay and that makes sense

OpenStudy (mathmale):

so now you are integrating (1/2)[sec u tan u du]. What's the integral? Don't forget that (1/2).

OpenStudy (mathmale):

Remember you have already successfully integrated sec x tan x dx. SAME THING.

OpenStudy (kayders1997):

1/2secu

OpenStudy (mathmale):

Plus C. ;)

OpenStudy (kayders1997):

Oh yeah!

OpenStudy (mathmale):

And put parentheses around that doggone 1/2, please.

OpenStudy (kayders1997):

Okay!

OpenStudy (mathmale):

So your final answer is ... ? Be careful; you did not start with the variable u; you began with 2x. So, your final answer is...?

OpenStudy (kayders1997):

(1/2)sec(2x)

OpenStudy (mathmale):

Plus what?

OpenStudy (kayders1997):

C!

OpenStudy (mathmale):

Kool as all get out.

OpenStudy (mathmale):

Forgive my pestering you, but we want you to get a very high grade in calculus, don't we?

OpenStudy (kayders1997):

Yes

OpenStudy (mathmale):

go for it! Take care, over and out, bye.

OpenStudy (kayders1997):

Thank you! :)

OpenStudy (mathmale):

:)

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