# AntiDeriviative of cos(2x)

I am totally confused about how the anti-derivative of the cos2x ends up being 1/2*sin2x+c. I know the formal definition of the anti-derivative of cosine of X but I am confused as to where the one half comes from. any help?
(Self study-stuck on this example) Edit: the second step is where it loses me (1/2*cos(u)*2

This is a method called integration by u-substitution. Integration by u-substitution is the opposite of the chain rule. This is how it works.
First, you have to look at your function and determine u. In this case, your function is cos(2x) dx, so it is obvious that you should let u=2x.
Then, you must find du in terms of dx. From u=2x, you get du = 2* dx. Once this is done, you need to isolate dx in the equation. So we get dx = du/2.
Finally, we substitute du/2 for dx and u for 2x in the original integrand, getting integrate(cos(u) du/2), which is 1/2* integrate (cos(u) du), which is 1/2*sin(u) +C.
As the last step, you substitute 2x back for u. So you get 1/2*sin (2x) +C.

Thank you! I could not find any way to even come up with that methodology. Now I can progress and prepare myself for Michigan technological university August 24th :)
My eternal gratitude, good sir.
MAy I ask the importance of Simpsons rule?

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Yeah, I think integration by u-substitution is an important part of integral calculus.

And I learned to estimate integrals using the trapezoidal rule, but the wikipedia's article on simpson's rule is very easy to understand. I'm sure you can understand it, although its derivation is not so easy.

And I don't exactly understand what is meant by "importance". Everything in math is important-is the question asking why Simpson's rule is accurate?

Simpson's rule is a way of approximating integrals. Sometimes it is impossible or unnecessary to calculate integrals analytically.

You might not need to know that the integral is 9*pi*sqrt(8). Maybe you just need to know that it's about 55 point something. This is important in engineering and applied physics.

Some integrals it is IMPOSSIBLE to find an antiderivative for, such as e^(x^2).

I am now stuck on a more difficult equation and I'm totally stumped on how to apply what you previously said the anti-chain rule or whatever on dy/dx=tan2xsec^2(2x)
I set u=tan2x
Thus du=sec^2(2x) dx
And dx=du/sec^2(2x)
And I can't figure out where to go
;( preparing for calculus is hard IMO.

My apologies for the double post, but what I meant by importance was the amount of use the formula goes through during actual physical application. Or better said, how frequently is it used in say theoretical or applied physics, and in engineering? A simple example would be much appreciated as I just graduated high school and have ambitions for a physics phd and am trying to get a jump start on the math requirements.

Your du is slightly wrong. Remember chain rule, you're missing a factor in there.
But even then, your sec should cancel out and you should be left with a "u" only.

I can't really think of any way to apply your response although I much appreciate it. What is exactly wrong with my du? I am confused, although it embarrasses me to admit this.
Could it be that du/dx=sec^2(2x)? That's the only thing I can think may be wrong

First, be careful with the differential of u. If u=tan(2x), then du=2*sec^2(2x) because of the chain rule.

Now, you don't ALWAYS solve for dx. You are trying to solve for whatever the second part of the function is-in this case, sec^2(2x). The reason behind this is that when you are substituting, you are trying to eliminate x by using u. When you have u=tan(2x), the tan(2x) part has been taken care of, but not the sec^2(2x) dx. Thus, you are trying to find du in terms of sec^2(2x) dx, so that you can substitute du for it.

That said, du/2=sec^2(2x) dx. So by substitution, integrate tan(2x)*sec^2(2x)=1/2*integrate u du, which is equal to 1/4*u^2+c, which is equal to 1/4*tan^2(2x)+C.

I think my biggest issue is where is the two coming from that you use in du?
That's the thing I don't understand. That 2.
I feel as if it comes out of thin air. Please try to keep in mind I am still a "high school student",and I have a very limited understanding of the chain rule. Just the basics of it.

My apologies for the double post, but what I meant by importance was the amount of use the formula goes through during actual physical application. Or better said, how frequently is it used in say theoretical or applied physics, and in engineering? A simple example would be much appreciated as I just graduated high school and have ambitions for a physics phd and am trying to get a jump start on the math requirements.
"Actual physical application". Well, Simpson's rule is a way of numerically approximating integrals. So what you're asking for is applications of integrals in physics. If you go into physics like you say you want to. In a year or two, you will look back at this question and laugh. :rofl:

I'll give you a very basic example of an application of integration. Newton's Second law states that

ƩF=ma. The sum of the forces on an object is equal to the mass of the object times its acceleration. Now what is acceleration? It's v'. What's v? It's x'. Where v is velocity and x is position.

So his second law states that

ƩF=mx''. The sum of the forces on an object is equal to the mass of the object times the second derivative of its position. Now if I were to give you a "force function", and you wanted to find its position (x) function, you would need to integrate TWICE.

Simpson's Rule is cool because sometimes the functions can become quite complicated. You know the trouble you're having with integrating basic functions? Imagine if they gave you a function that was HUGE or one that WAS IMPOSSIBLE TO FIND AN ANTIDERIVATIVE FOR. You would still need to find the position, but you would have to use it using Simpson's rule.

On another note, not trying to put you down or anything, but one thing at a time. You are just out of high school. A PhD should be the last thing on your mind. Take things one step at a time, just worry about undergrad for the time being. I strongly encourage you to pursue physics and study all you can in it, but don't worry about a PhD. You have no idea what physics is like at that level (or even at an undergrad level, TBH) and it's a little presumptuous to say that you could handle or even WANT one. It's not saying that you're dumb, or anything negative about you personally, but keep your doors open.

It isn't on the forefront of my mind, the PhD, but it is a goal I eventually want to accomplish. at this point I'm just worried about starting in 3 weeks, and finishing the BSC. That's my main objective I'll see where to go from there
EDIT: Thank you so much for the basic, basic application of Simpson's rule. now I think/feel I have a vague but better idea of what it's used for. truly I love this forum :-)

The chain rule says that after you have differentiated tan(2x) into sec^2(2x), you need to multiply the result by the argument (argument means whatever is inside the bracket. In this case, d/dx (2x) =2, so this is the 2.

Similarly, the derivative of (2x)^2 is 4x times the derivative of 2x, which is 4x*2, which is 8x.

Note that (2x)^2 =4x^2, and d/dx 4x^2=8x, which confirms the result.

The chain rule says that after you have differentiated tan(2x) into sec^2(2x), you need to multiply the result by the argument (argument means whatever is inside the bracket. In this case, d/dx (2x) =2, so this is the 2.

Similarly, the derivative of (2x)^2 is 4x times the derivative of 2x, which is 4x*2, which is 8x.

Note that (2x)^2 =4x^2, and d/dx 4x^2=8x, which confirms the result.
Thank you the magic 2 has been found :-) rofl

I think my biggest issue is where is the two coming from that you use in du?
That's the thing I don't understand. That 2.
I feel as if it comes out of thin air. Please try to keep in mind I am still a "high school student",and I have a very limited understanding of the chain rule. Just the basics of it.
Maybe you should stop working on antiderivatives and work on strengthening your differentiation. Antiderivatives are to derivatives as division is to multiplication.

Division is guess and check based on what you know about multiplication. Same thing with computing antiderivatives. Unless your differentiation is strong, you won't be able to antidifferentiate well.

Okay Sorry for the initiative, I was just trying to get a quick run through of Morris Klines "calculus: an intuitive and physical approach" before I started calculus at Michigan Tech in three weeks. I was trying to quickly zoom through the book to get a overview of what calculus will actually entail

It isn't on the forefront of my mind, the PhD, but it is a goal I eventually want to accomplish. at this point I'm just worried about starting in 3 weeks, and finishing the BSC. That's my main objective I'll see where to go from there
EDIT: Thank you so much for the basic, basic application of Simpson's rule. now I think/feel I have a vague but better idea of what it's used for. truly I love this forum :-)
I strongly encourage you to work towards it. If it's what you want, go for it. Just realize things change a lot.

In my area (mathematics), everything changes a lot from one step to the next. I am a 4th year undergrad, and am starting some graduate course work in the fall. Upper division is completely different from lower division. Some people love math until they hit their first upper division class, where everything gets flipped around, then hate it. Also, in studying for my graduate class, it's a lot different as well. Some people can do well in undergrad and hate grad. Or some people can hate undergrad and love grad. Hate lower div, love upper div. Different levels are just so different and require different approaches, levels of thinking.

In lower division, it's a lot about performing computations without error. And some people are really good at that. Some people aren't. Upper division is about proof writing, and some people hate that. Others love it. I really liked both levels, hard to say which I preferred. And my grad class is going to be a lot different as well, I don't know how I'll like that.

I see. Well, thank you for the encouragement. I much appreciate it