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Feb 10 2014 05:37pm
Integral of (2sinxcosx)

Solve it two ways

first way
let u = sinx
du = cosx



second way
let u = cosx
du = -sinx


I keep getting two different answers when I solve it.
Show step by step what you get with the same answers.

If the answer/steps are right/make sense, you get 100fg
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Feb 10 2014 05:50pm
I'm guessing you keep getting (sin x)^2 + c and -(1/2)cos(2x) + c
This in fact can be worked different ways and the result is 2 different answers.
However, these 2 answers are equivalent by the double angle identity
(sin x)^2 = (1/2)(1 - cos(2x))
after fully expanding the right side of the equation, you have:
(sin x)^2 = ((1/2) - (1/2)cos(2x))

As you can see, the only difference in this equation and the 2 bolded ones is the (1/2), which is included in the "+c" portion of the answer

Hope this helps

This post was edited by TritonV8 on Feb 10 2014 05:50pm
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Posts: 101
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Feb 10 2014 05:58pm
I am getting sin(x)^2 for when I do u=sinx

However, i've come to the conclusion that this answer is wrong.
Here is why.
When I put in the integral to solve on wolframalpha, I get that the answer is -cos(x)^2
I am getting -cos(x)^2 for when I do u=cosx



If sin(x)^2 = -cos(x)^2 then that should mean thatsin(x)^2 + cos(x)^2 = 0 correct?
The issue is that sin(x)^2+cos(x)^2 = 1, it doesn't = 0 so they can't be equal.

Am I misunderstanding something or am I going crazy or what?

This post was edited by Pindrought on Feb 10 2014 06:11pm
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Feb 10 2014 06:10pm
Quote (Pindrought @ Feb 10 2014 07:58pm)
I am getting sin(x)^2 for when I do u=sinx

However, i've come to the conclusion that this answer is wrong.
Here is why.
When I put in the integral to solve on wolframalpha, I get that the answer is -cos(x)^2
I am getting -cos(x)^2 for when I do u=cosx

( Note: -cos(x)^2 = -(1/2)cos(2x) )


If sin(x)^2 = -cos(x)^2 then that should mean thatsin(x)^2 + cos(x)^2 = 0 correct?
The issue is that sin(x)^2+cos(x)^2 = 1, it doesn't = 0 so they can't be equal.

Am I misunderstanding something or am I going crazy or what?


actually

-cos^2(x) = (-1/2)(1 + cos(2x))

This post was edited by rx7drifter on Feb 10 2014 06:10pm
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Posts: 101
Joined: Jan 19 2014
Gold: 255.00
Feb 10 2014 06:12pm
My bad on that note, you're right about that. Any idea about my issue though?
Member
Posts: 15,275
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Feb 10 2014 06:13pm
Quote (Pindrought @ Feb 10 2014 06:58pm)
I am getting sin(x)^2 for when I do u=sinx

However, i've come to the conclusion that this answer is wrong.
Here is why.
When I put in the integral to solve on wolframalpha, I get that the answer is -cos(x)^2
I am getting -cos(x)^2 for when I do u=cosx

( Note: -cos(x)^2 = -(1/2)cos(2x) )


If sin(x)^2 = -cos(x)^2 then that should mean thatsin(x)^2 + cos(x)^2 = 0 correct?
The issue is that sin(x)^2+cos(x)^2 = 1, it doesn't = 0 so they can't be equal.

Am I misunderstanding something or am I going crazy or what?


the way you're trying to compare them can't be done.
You can't leave off the "+c" terms when setting them equal to each other.
They are equal to each other through trig identities, so when you get answers like this, always tweak with identities to try and get the answers the same
But yes, doing it that way you tried it works as well, AS LONG as you keep the "+c" in both of your answers.
So, (sin x)^2 + c = -(cos x)^2 + c (note that both c's are NOT necessarily equal)
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Feb 10 2014 06:17pm
Quote (TritonV8 @ Feb 10 2014 08:13pm)
the way you're trying to compare them can't be done.
You can't leave off the "+c" terms when setting them equal to each other.
They are equal to each other through trig identities, so when you get answers like this, always tweak with identities to try and get the answers the same
But yes, doing it that way you tried it works as well, AS LONG as you keep the "+c" in both of your answers.
So, (sin x)^2 + c = -(cos x)^2 + c (note that both c's are NOT necessarily equal)


this :)
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Feb 10 2014 06:31pm
100fg to both, my mind got fucking blown
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