Studying Mathematics Thread

[polymath257]

General relativity, eh? I'll eventually catch up to what you're saying, but for now, I'm personally going to address a confusion some have with the concept of the constant of integration C. In the derivation of the formula for integration by parts, C just isn't there anywhere in the final formula. Where did it magically disappear?

Lo and behold, here's my derivation of the formula that puts this in perspective.

https://pasteboard.co/HiCvx8S.png

Fucking math nerds, the bunch here.

It 'disappears' into the new integral.

So,

int u dv

is only determined up to a constant.

Same for uv-int v du.

[GrandizerII]

This took quite a bit of effort, but here's how you derive the integral of secx "from scratch". Whoever came up with such solution was pretty damn smart.

Here's the work (thanks to YouTube, of course!):
https://pasteboard.co/HiEZbdk.png

That's the cursor at the end of the last line (not an absolute value bar).

[GrandizerII]

Now that I've done the work for the integral of secx, time to integrate (secx)^3, which requires knowing the integral of secx itself.

https://pasteboard.co/HiI4fYE.png

Next is the integral of sqrt(1+x^2), which is made easy by knowing the integral of (secx)^3.

[polymath257]

This took quite a bit of effort, but here's how you derive the integral of secx "from scratch". Whoever came up with such solution was pretty damn smart.

Here's the work (thanks to YouTube, of course!):
https://pasteboard.co/HiEZbdk.png

That's the cursor at the end of the last line (not an absolute value bar).

Much easier is to realize that the derivative of

sec x + tan x

is

sec x tan x + sec^2 x = sec x(sec x + tan x).

So, write

int sec x dx = int [ sec x (sec x + tan x) ]/(sec x + tan x) dx

and do a substitution u= sec x + tan x. The result falls out.

[GrandizerII]

This took quite a bit of effort, but here's how you derive the integral of secx "from scratch". Whoever came up with such solution was pretty damn smart.

Here's the work (thanks to YouTube, of course!):
https://pasteboard.co/HiEZbdk.png

That's the cursor at the end of the last line (not an absolute value bar).

Much easier is to realize that the derivative of

sec x + tan x

is

sec x tan x + sec^2 x = sec x(sec x + tan x).

So, write

int sec x dx = int [ sec x (sec x + tan x) ]/(sec x + tan x) dx

and do a substitution u= sec x + tan x. The result falls out.

Yeah, that seems to be the better way of arriving at the expected answer, but requires thinking about differentiating secx+tanx in the first place. There's a reason I call solving these problems a form of art. Lots of creative thinking needed. No way I can do all this on my own without any assistance (at this stage, at least).

Here's my solution to the integral of sqrt(1+x^2). The triangle method comes in handy here.

https://pasteboard.co/HiNzLeB.png

[polymath257]

Much easier is to realize that the derivative of

sec x + tan x

is

sec x tan x + sec^2 x = sec x(sec x + tan x).

So, write

int sec x dx = int [ sec x (sec x + tan x) ]/(sec x + tan x) dx

and do a substitution u= sec x + tan x. The result falls out.

Yeah, that seems to be the better way of arriving at the expected answer, but requires thinking about differentiating secx+tanx in the first place. There's a reason I call solving these problems a form of art. Lots of creative thinking needed. No way I can do all this on my own without any assistance (at this stage, at least).

Here's my solution to the integral of sqrt(1+x^2). The triangle method comes in handy here.

https://pasteboard.co/HiNzLeB.png

I think a lot of these are discovered by simply playing around enough. So, for example, take a derivative of

ln( tan (x/2) )

to get

1/tan(x/2) * sec^2 (x/2) * 1/2

which is

cos(x/2)/sin(x/2) * 1/cos^2 (x/2) * 1/2

which simplifies to

1/[ 2 sin(x/2) cos(x/2) ]

Which, by the double angle formula, is

1/sin(x) = csc(x).

So, the integral of csc(x) is ln ( tan(x/2) ) +C.

But now,

sec(x) = csc (pi/2 - x), so we can do a substitution to find that the integral of sec x is

- ln( tan ( pi/4 - x/2 ) ) +C

A challenge: show this is the same as the 'usual' anti-derivative for sec x.

[GrandizerII]

But now,

sec(x) = csc (pi/2 - x), so we can do a substitution to find that the integral of sec x is

- ln( tan ( pi/4 - x/2 ) ) +C

A challenge: show this is the same as the 'usual' anti-derivative for sec x.

I think I just did it. I did have to consult the more complex trig identities, though (I don't have time for now to derive them for myself).

Feel free to check my work (note it's rather confusing in terms of the flow, so do rely on your intuition skills to figure out the flow):

https://pasteboard.co/Hj12NGv.png

[polymath257]

Much easier is to realize that the derivative of

sec x + tan x

is

sec x tan x + sec^2 x = sec x(sec x + tan x).

So, write

int sec x dx = int [ sec x (sec x + tan x) ]/(sec x + tan x) dx

and do a substitution u= sec x + tan x. The result falls out.

Yeah, that seems to be the better way of arriving at the expected answer, but requires thinking about differentiating secx+tanx in the first place. There's a reason I call solving these problems a form of art. Lots of creative thinking needed. No way I can do all this on my own without any assistance (at this stage, at least).

Here's my solution to the integral of sqrt(1+x^2). The triangle method comes in handy here.

https://pasteboard.co/HiNzLeB.png

A related integral is that of sqrt(1-x^2). But this one can be done by 'elementary' methods. No substitution.

Hint: interpret as an area.

[GrandizerII]

Yeah, that seems to be the better way of arriving at the expected answer, but requires thinking about differentiating secx+tanx in the first place. There's a reason I call solving these problems a form of art. Lots of creative thinking needed. No way I can do all this on my own without any assistance (at this stage, at least).

Here's my solution to the integral of sqrt(1+x^2). The triangle method comes in handy here.

https://pasteboard.co/HiNzLeB.png

A related integral is that of sqrt(1-x^2). But this one can be done by 'elementary' methods. No substitution.

Hint: interpret as an area.

The area of the upper semicircle? I can do the definite integral from -1 to 1 intuitively, but indefinite integral without substitution I will have to think about for a while. But first, need to go to work now. So later.

[polymath257]

A related integral is that of sqrt(1-x^2). But this one can be done by 'elementary' methods. No substitution.

Hint: interpret as an area.

The area of the upper semicircle? I can do the definite integral from -1 to 1 intuitively, but indefinite integral without substitution I will have to think about for a while. But first, need to go to work now. So later.

Yes, the indefinite integral. If you can do it from -1 to 1, you have part of the basic idea.