Question

Definite integral:
Bounds are t(top) and 100(bottom)
sec(28x-30)

Answer 1

Secant? Ooo this one is not fun :d

Answer 2

\[\large\rm \int\limits_{100}^t \sec(28x-30)~dx\]

Answer 3

\[\large\rm u=28x-30\qquad\to\qquad du=28~dx\qquad\to\qquad \frac{1}{28}du=dx\]\[\large\rm \int\limits\limits_{100}^t \sec(28x-30)~dx\quad=\quad \frac{1}{28}\int\limits_{u=?}\sec(u)du\]It'll have new bounds, depending on u,
but let's not worry about that.

From here,it requires a clever algebra trick that is fairly unintuitive,
rewrite as a fraction,\[\large\rm \frac{1}{28}\int\limits\limits\frac{\sec(u)}{1}du\]

Answer 4

Multiply top and bottom by sec(u)+tan(u),\[\large\rm \frac{1}{28}\int\limits\frac{\sec u}{1}\cdot\color{royalblue}{\left(\frac{\sec u+\tan u}{\sec u + \tan u}\right)}du\]Giving us,\[\large\rm \frac{1}{28}\int\limits \frac{\sec^2u+\sec u \tan u}{\tan u+\sec u}du\]

Answer 5

This might look like a big mess,
but it turns out that,\[\large\rm \frac{1}{28}\int\limits\limits \frac{\color{orangered}{\sec^2u}+\color{blue}{\sec u \tan u}}{\color{orangered}{\tan u}+\color{blue}{\sec u}}du\]the orange in the top is the derivative of the orange in the bottom,
while the blue in the top is the derivative of the blue in the bottom.
So a substitution is going to work really nicely here.

Answer 6

Let's call it an... m substitution or something, since we already used u, ya?

Answer 7

\[\large\rm m=\tan u+\sec u\]\[\large\rm dm=(\sec^2u+\sec u \tan u)~du\]See how m takes the place of that entire denominator,
while the dm takes the place of the entire numerator??
Works out really nicely!

Answer 8

Any thoughts? :D
You're too quiet broski lol.

I need some foods >.< brb

Answer 9

wow this is a disgusting problem lol

so after all that is done, i integrate 1/m which is ln m
and end up with (tanU + secU)/28 and then sub in u giving me

ln(tan(28x-30)+sec(28x-30)) all over 28

Could you double check my work @Zepdrix

Answer 10

Ooo nice job!\[\large\rm \frac{\ln\left[\tan(28x-30)+\sec(28x-30)\right]}{28}\quad |_{100}^t\]Since you undid your substitution,
you won't need to find new boundary values for u.

But you still gotta plug those boundaries in for x! :) lolol

Answer 11

So all the x's would become t's right?

And would I also have to use the 100 for the subtraction or is that negligible

Answer 12

Are you sure the problem wasn't asking you to `find the derivative` of this integral?
This just looks a whole lot like a Fundamental Theorem of Calculus, Part 1, type problem.

It could be a normal integral too I suppose :)

Answer 13

Ya, x's become t's for the upper bound.
Then x's become 100's and subtract.

Messy messy :d

Answer 14

Oh man, it does. LOL, it says calculate the derivative

Answer 15

Would that change anything in this problem?

Answer 16

Oh :(
We didn't need to go through all of that process then.
Yes it changes it completely lolol

Answer 17

Fundamental Theorem of Calculus, Part 1:\[\large\rm \frac{d}{dx}\int\limits_c^{x}f(t)dt=f(x)\]

Answer 18

Oh oh lemme fix that, so it matches up with the variables we have.

Answer 19

Fundamental Theorem of Calculus, Part 1:\[\large\rm \frac{d}{dt}\int\limits\limits_c^{t}f(x)dt=f(t)\]

Answer 20

Try to remember that integration is `anti-differentiation`.
So you're anti-differetiating,
then you're differetiating,
getting back what you started with.

Answer 21

What's neat about this process though,
is that you're never actually DOING the anti-differentiation.
You just assume that some function, let's say F(x), is the anti-derivative of f(x).

Answer 22

\[\large\rm \frac{d}{dt}\int\limits_c^{t}f(x)dt=\frac{d}{dt}F(x)|_c^t\]So I integrated ^
Now I'll plug in my bounds,\[\large\rm \frac{d}{dt}\left[F(t)-F(c)\right]\]Now differentiate with respect to t,\[\large\rm f(t)-0\]F goes back to f when you differentiate,
now it's in terms of a new variable though.
F(c) is just a constant! So the derivative is 0.

Answer 23

So with our problem, what we would have done is,\[\large\rm \frac{d}{dt}\int\limits\limits_{100}^t \color{orangered}{\sec(28x-30)}~dx\quad=\quad\frac{d}{dt}\int\limits\limits_{100}^t \color{orangered}{f(x)}~dx\]This is just some function of x in here.
Let's not worry about what its anti-derivative is, we'll just call it F(x).
This is like "pseudo integration". Kind of weird, I know. :)

Answer 24

Integrating gives us "some function" F(x),
Plugging in our bounds,\[\large\rm =\frac{d}{dt}\left[F(x)\right]_{100}^t\quad=\quad \frac{d}{dt}\left[F(t)-F(100)\right]\]Differentiating,\[\large\rm =\color{orangered}{f(t)}-0\]\[\large\rm =\color{orangered}{\sec(28t-30)}\]

Answer 25

Mmmmm what do you think?
It's not too bad, ya? :o

Answer 26

Conceptually, it's a little tricky to understand at first.
But mechanically, it's way way easier than what we were doing before lol.

Answer 27

So the F(100) will just become 0 after differentiating and we're left with sec(28t-30) as our final answer?

Answer 28

Yes.
When you plug a number into it, it's no longer varying.
It's held constant.

If we look back at the process we were doing before,
it's actually just this "number",\[\large\rm \frac{\ln\left[\tan(2800-30)+\sec(2800-30)\right]}{28}\]But that's just a constant,
it's a really fancy looking constant,
but it's just a constant.
So derivative is 0.

Answer 29

Ohhhhhhhhhhhhhhhhhhhhhhhh okay I got it. That makes sense
Did all that work for the integral when we just had to find the derivative but I understood both so I guess it wasn't all bad!

Thanks for all the help, it is very much appreciated! :)

Answer 30

np \c:/