can someone please explain this shit

rC | Mythic Inconceivable!
 
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ayy lmao
http://www.wolframalpha.com/input/?i=is+integral+of+sintcost+%3D%3D+-.5cos%5E2t
http://www.wolframalpha.com/input/?i=is+integral+of+sintcost+%3D%3D+.5sin%5E2t

how the fuck is sin^2(x)/2 NOT the integral of sinxcosx?

here's how i got sin^2(x)/2

integral(sinxcosxdx)
u=sinx
du=cosxdx
integral(udu) = (u^2)/2
sin^2(x)/2 + c


THIS IS FUCKING CORRECT WHY IS IT NOT
Last Edit: November 17, 2015, 02:17:23 PM by rc


 
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i'll do this in like 30 mins.


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ayy lmao
Lol you're retarded
no seriously i have a calc 2 exam in two hours, and this is calc 1 shit, pls help


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Like (sin^2x)/2?

Because sin^2x/2 is just sin^x
Last Edit: November 17, 2015, 02:42:26 PM by Ingloriouswho98


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you can get -0.5*cos^2 and 0.5*sin^2

don't sweat it, you're right.


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how the fuck is sin^2(x)/2 NOT the integral of sinxcosx?

There are two correct answers.

"Not always" doesn't mean "isn't". Wolfram just didn't display both answers in the same question.


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ayy lmao
you can get -0.5*cos^2 and 0.5*sin^2

don't sweat it, you're right.
wolfram alpha says otherwise. also, if you evaluate it from 0 to 2pi (this is just part of a longer problem), you get 0, but it should be 3... it is 3 if you go the cosine route, but yeah.


rC | Mythic Inconceivable!
 
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ayy lmao
how the fuck is sin^2(x)/2 NOT the integral of sinxcosx?

There are two correct answers.

"Not always" doesn't mean "isn't".
also, if you evaluate it from 0 to 2pi (this is just part of a longer problem), you get 0, but it should be 3... it is 3 if you go the cosine route, but yeah.


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ayy lmao


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how the fuck is sin^2(x)/2 NOT the integral of sinxcosx?

There are two correct answers.

"Not always" doesn't mean "isn't".
also, if you evaluate it from 0 to 2pi (this is just part of a longer problem), you get 0, but it should be 3... it is 3 if you go the cosine route, but yeah.


Well your u/du substitution is correct so just use that. Wolfram sucks ass at trig calc. It's different based on which factor you use for u.

.5sin^2(x) + C = -.5cos^2(x) + c, but their constants are not equal.
Last Edit: November 17, 2015, 02:52:34 PM by RadioactiveTurkey


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Because sin^2x/2 is just sin^x
what

Nevermind

I was thinking of something else



Oh | Elite Four Invincible!
 
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you get the same thing with integration by parts, you'll be fine.


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Math sucks.


 
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I learned this in high school and never used it again in five years. Now this thread is literally a foreign language to me.


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Holy shit what the fuck are you guys doing?

Rc asked for help, he gets help, but then these motherfuckers show up and start talking about how they used to do the same thing, and how the world has changed for them.

What is with you people?

What I'm trying to say in my autistic ramblings is that your platitudes aren't doing anything for the thread, so you must be doing it for yourselves, which is fine in some threads, but this thread isn't a thread for discussion about personal experiences or whatever. This is a thread specifically about a certain math problem.

also, 'mirin peoples' math skills.
Last Edit: November 17, 2015, 03:48:35 PM by The Meta Madman Destyyy👌


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Holy shit what the fuck are you guys doing?

Rc asked for help, he gets help, but then these motherfuckers show up and start talking about how they used to do the same thing, and how the world has changed for them.

What is with you people?
le funposting


rC | Mythic Inconceivable!
 
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ayy lmao
how the fuck is sin^2(x)/2 NOT the integral of sinxcosx?

There are two correct answers.

"Not always" doesn't mean "isn't".
also, if you evaluate it from 0 to 2pi (this is just part of a longer problem), you get 0, but it should be 3... it is 3 if you go the cosine route, but yeah.


Well your u/du substitution is correct so just use that. Wolfram sucks ass at trig calc. It's different based on which factor you use for u.

.5sin^2(x) + C = -.5cos^2(x) + c, but their constants are not equal.
Why do I get two different answers when I evaluate it though? I just wanna understand what went wrong.

Because my understanding is that the constant doesn't matter when you're solving a definite integral.
Last Edit: November 17, 2015, 04:00:27 PM by rc


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Holy shit what the fuck are you guys doing?

Rc asked for help, he gets help, but then these motherfuckers show up and start talking about how they used to do the same thing, and how the world has changed for them.

What is with you people?

What I'm trying to say in my autistic ramblings is that your platitudes aren't doing anything for the thread, so you must be doing it for yourselves, which is fine in some threads, but this thread isn't a thread for discussion about personal experiences or whatever. This is a thread specifically about a certain math problem.

also, 'mirin peoples' math skills.
Read the thread, people helped him

Good God man read before you type a wall


Desty | Mythic Inconceivable!
 
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Holy shit what the fuck are you guys doing?

Rc asked for help, he gets help, but then these motherfuckers show up and start talking about how they used to do the same thing, and how the world has changed for them.

What is with you people?

What I'm trying to say in my autistic ramblings is that your platitudes aren't doing anything for the thread, so you must be doing it for yourselves, which is fine in some threads, but this thread isn't a thread for discussion about personal experiences or whatever. This is a thread specifically about a certain math problem.

also, 'mirin peoples' math skills.
Read the thread, people helped him

Good God man read before you type a wall
Read the reply, I wrote that people helped him.


 
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Holy shit what the fuck are you guys doing?

Rc asked for help, he gets help, but then these motherfuckers show up and start talking about how they used to do the same thing, and how the world has changed for them.

What is with you people?

What I'm trying to say in my autistic ramblings is that your platitudes aren't doing anything for the thread, so you must be doing it for yourselves, which is fine in some threads, but this thread isn't a thread for discussion about personal experiences or whatever. This is a thread specifically about a certain math problem.

also, 'mirin peoples' math skills.
Read the thread, people helped him

Good God man read before you type a wall
Read the reply, I wrote that people helped him.

who cares if we mention we used to do calculus?

just you


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how the fuck is sin^2(x)/2 NOT the integral of sinxcosx?

There are two correct answers.

"Not always" doesn't mean "isn't".
also, if you evaluate it from 0 to 2pi (this is just part of a longer problem), you get 0, but it should be 3... it is 3 if you go the cosine route, but yeah.


Well your u/du substitution is correct so just use that. Wolfram sucks ass at trig calc. It's different based on which factor you use for u.

.5sin^2(x) + C = -.5cos^2(x) + c, but their constants are not equal.
Why do I get two different answers when I evaluate it though? I just wanna understand what went wrong.

Because my understanding is that the constant doesn't matter when you're solving a definite integral.
Probably has to do with the integral of Sinx cancelling out, if you remember the unit circle you can go to something like pi/2 and multiply the integral by 4, because you are integrating 4 values of pi/2  = to one value of 2pi.


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http://www.wolframalpha.com/input/?i=is+integral+of+sintcost+%3D%3D+-.5cos%5E2t
http://www.wolframalpha.com/input/?i=is+integral+of+sintcost+%3D%3D+.5sin%5E2t

how the fuck is sin^2(x)/2 NOT the integral of sinxcosx?

here's how i got sin^2(x)/2

integral(sinxcosxdx)
u=sinx
du=cosxdx
integral(udu) = (u^2)/2
sin^2(x)/2 + c


THIS IS FUCKING CORRECT WHY IS IT NOT
the integral of sinxcosxdx isn't sin^2(x)/2
it's sin^2(x)/2 + C

sin^2(x) + cos^(x) = 1
sin^2(x) = 1 - cos^2(x)
sin^2(x)/2 = .5 - cos^2(x)/2

sin^2(x)/2 + C is equivalent to  -cos^2(x)/2 + C
Last Edit: November 17, 2015, 09:53:19 PM by Baha


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ayy lmao
http://www.wolframalpha.com/input/?i=is+integral+of+sintcost+%3D%3D+-.5cos%5E2t
http://www.wolframalpha.com/input/?i=is+integral+of+sintcost+%3D%3D+.5sin%5E2t

how the fuck is sin^2(x)/2 NOT the integral of sinxcosx?

here's how i got sin^2(x)/2

integral(sinxcosxdx)
u=sinx
du=cosxdx
integral(udu) = (u^2)/2
sin^2(x)/2 + c


THIS IS FUCKING CORRECT WHY IS IT NOT
the integral of sinxcosxdx isn't sin^2(x)/2
it's sin^2(x)/2 + C

sin^2(x) + cos^(x) = 1
sin^2(x) = 1 - cos^2(x)
sin^2(x)/2 = .5 - cos^2(x)/2

sin^2(x)/2 + C is equivalent to  -cos^2(x)/2 + C
but the constant doesn't matter in a definite integral, right?