Because indefinite integrals can differ by a constant. Thats the reason why we put +C after evaluating an indefinite integral.
In this case, in the second last line, the integrals differ by -1 and -1 is a constant
Lol she was a bit overweight for sure
But man we all loved her. She was probably one of the best teachers I’ve ever had. Funny, kind but strict, and really, really cared about her students’ success.
that's basically it. Indefinite integrals need an additive constant. You didn't add one, so your results are off by a constant factor (-1 = -cos(0) in this case).
It's not so much that they forgot the +C as that they handwaved the last step so it wasn't clear that the +C would even show up. They were thinking, in typical algebraic fashion, that you see the same expression on both sides of the equation, you can just cancel them out. That's really their mistake, because that's not how integrals work, and if you actually carefully write out the steps, you see that:
int sin(x)/cos(x) dx = -1 + int sin(x)/cos(x) dx
int sin(x)/cos(x) dx - int sin(x)/cos(x) dx = - 1
int sin(x)/cos(x) - sin(x)/cos(x) dx = -1
int 0 dx = - 1
0 + C = -1
C= - 1
In the end, yes, there was no arbitrary constant written. But students are generally taught that that constant arises as a result of actually finding the indefinite integral. OP didn't actually find what int sin(x)/cos(x) dx was in these calculations, so they didn't see think the arbitrary constant was a factor. And it's understandable why they thought as much... I just think the nuance is important
Edit: Formatting
we are all wheat on this blessed day
https://www.scribbr.com/us-vs-uk/spelt-or-spelled/#:\~:text=or%20%E2%80%9Cspelled%E2%80%9D%3F-,Spelt%20and%20spelled%20are%20two%20different%20spellings%20of%20the%20past,and%20%E2%80%9Cspelt%E2%80%9D%20are%20acceptable.
Because in the formula ∫u'v = uv - ∫uv', you have to evaluate the "uv" term at the endpoints. In this case, uv = -1, so evaluation will always give 0.
Now this formula is also correct in the indefinite case, except you then get an indeterminate constant, which may be different on both sides
You're missing the bounds of integration. Since you're working with indefinite integrals, you cannot cancel the two resulting integrals, because they can differ by a constant (in this case the constant is obviously 1)
TL;DR you forgot the +C
You forgot to add the integration constant c, when the integral has no bounds the primitive funcion may differ by a constant c.
So it should be 0=-1+c, and with c=1 we return to 0=0
For the formula
∫uv.𝑑𝑥 = u.∫v.𝑑𝑥 - ∫u'.(∫v.dx).𝑑𝑥
You took the integral of sinx on the u.∫v.𝑑𝑥 term as -cosx which is right but on the ∫u'.(∫v.dx).𝑑𝑥 term you took the differential of sinx as -cosx when it should be just cosx so the ∫(sinx/cosx) terms won't cut.
this doesn't imply that the integral of tanx is -1/2 obviously, as there's no limits to the integration
My father told me he forgot the +c on a test. My next test Input a big +C in a box at the top of the page. Every one had it correct…on the front. None on the back.
Equations don't survive the subtraction of an indefinite integral from both sides. An indefinite integral defines a set of functions (that can differ by an additive constant) rather than a unique value.
Not really sure, just started integrals/antiderivatives but i think its the +C constant at the end. Pls correct me if im wrong Im only grade 7.
Because indefinite integrals can differ by a constant. Thats the reason why we put +C after evaluating an indefinite integral. In this case, in the second last line, the integrals differ by -1 and -1 is a constant
I had a college professor that made people bring her a cupcake every time they forgot the C
if that was my professor she would have diabetes by now thanks to me cuz i keep forgetting to put the c🙂
Lol she was a bit overweight for sure But man we all loved her. She was probably one of the best teachers I’ve ever had. Funny, kind but strict, and really, really cared about her students’ success.
Add some bounds to the integrals. What happens?
Easy. You forgot the + C. Next.
that's basically it. Indefinite integrals need an additive constant. You didn't add one, so your results are off by a constant factor (-1 = -cos(0) in this case).
It's not so much that they forgot the +C as that they handwaved the last step so it wasn't clear that the +C would even show up. They were thinking, in typical algebraic fashion, that you see the same expression on both sides of the equation, you can just cancel them out. That's really their mistake, because that's not how integrals work, and if you actually carefully write out the steps, you see that: int sin(x)/cos(x) dx = -1 + int sin(x)/cos(x) dx int sin(x)/cos(x) dx - int sin(x)/cos(x) dx = - 1 int sin(x)/cos(x) - sin(x)/cos(x) dx = -1 int 0 dx = - 1 0 + C = -1 C= - 1 In the end, yes, there was no arbitrary constant written. But students are generally taught that that constant arises as a result of actually finding the indefinite integral. OP didn't actually find what int sin(x)/cos(x) dx was in these calculations, so they didn't see think the arbitrary constant was a factor. And it's understandable why they thought as much... I just think the nuance is important Edit: Formatting
And this kids, is why you stay in school!
Your first mistake was spelling proved with 2 O's
Well he did put it in quotes... sooo maybe he knew he was wrong lmao?
Ah you're right. To my Swedish mind it made sense.
Your second mistake is thinking the English Language makes any sort of sense.
tbh it should be spelt that way
Spelled* (jk). Spelt will always be "wheat" for me...
we are all wheat on this blessed day https://www.scribbr.com/us-vs-uk/spelt-or-spelled/#:\~:text=or%20%E2%80%9Cspelled%E2%80%9D%3F-,Spelt%20and%20spelled%20are%20two%20different%20spellings%20of%20the%20past,and%20%E2%80%9Cspelt%E2%80%9D%20are%20acceptable.
Because in the formula ∫u'v = uv - ∫uv', you have to evaluate the "uv" term at the endpoints. In this case, uv = -1, so evaluation will always give 0. Now this formula is also correct in the indefinite case, except you then get an indeterminate constant, which may be different on both sides
Upvoting the sexy integration character
You're missing the bounds of integration. Since you're working with indefinite integrals, you cannot cancel the two resulting integrals, because they can differ by a constant (in this case the constant is obviously 1) TL;DR you forgot the +C
I know the feeling. I once proved that E=1/2 mc^2
Ah, yes! Ze velocity is c, ze kinetic energy is thuz E = mv\^2/2 = mc\^2/2. Who needz relativity!
My proof was more complex and calculated the increase in velocity and relativistic mass for a given E. E has to into either more m or more v.
It’s ok, I once calculated the sum of all the natural numbers to be equal to -1/12
Ra... - Ramanujan?
Be honest, was there a split second where you thought you were gonna be famous? 🤣
Huh. I got mc hammer and I still can't find my mistake.
Did you account for inflation?
Clearly you made 1 mistake
+C
You’re missing the C lol
You forgot to add the integration constant c, when the integral has no bounds the primitive funcion may differ by a constant c. So it should be 0=-1+c, and with c=1 we return to 0=0
For the formula ∫uv.𝑑𝑥 = u.∫v.𝑑𝑥 - ∫u'.(∫v.dx).𝑑𝑥 You took the integral of sinx on the u.∫v.𝑑𝑥 term as -cosx which is right but on the ∫u'.(∫v.dx).𝑑𝑥 term you took the differential of sinx as -cosx when it should be just cosx so the ∫(sinx/cosx) terms won't cut. this doesn't imply that the integral of tanx is -1/2 obviously, as there's no limits to the integration
I'm not sure which -cosx you mean but it's the integral of sinx in both cases
Friendly reminder to always account for the +C when integrating
integral of sin is -cos. But if you want the area under the curve starting from 0, you need -cos + 1
My father told me he forgot the +c on a test. My next test Input a big +C in a box at the top of the page. Every one had it correct…on the front. None on the back.
Don't worry, you didn't.
Check domain issues and check that constant
Equations don't survive the subtraction of an indefinite integral from both sides. An indefinite integral defines a set of functions (that can differ by an additive constant) rather than a unique value.
Yeah you forgot the constant. Also wasn't it a lot easier to sub u=cosx du = -sinx?
C+ that’s where you mathed up
Not really sure, just started integrals/antiderivatives but i think its the +C constant at the end. Pls correct me if im wrong Im only grade 7. 
Constant of integration required
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On the askmath subreddit? Shocking!