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Zack88
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i know the answer is close to that except for the extra x/2
Evaluating the integral, correct?
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SUMMARY
The discussion focuses on evaluating the integral \(\int x^2 \cos(mx) \, dx\) using integration by parts. The participants detail the steps involved, including the correct assignments for \(u\) and \(dv\), and emphasize the necessity of performing integration by parts multiple times for functions involving polynomials multiplied by trigonometric functions. The final solution is confirmed to be \(\frac{x^2}{m} \sin(mx) - \frac{2}{m^2} \cos(mx) + C\), demonstrating the importance of careful term management throughout the integration process.
PREREQUISITES- Understanding of integration by parts
- Familiarity with trigonometric integrals
- Knowledge of LaTeX for mathematical expressions
- Basic calculus concepts, including differentiation and integration
- Study advanced integration techniques, including repeated integration by parts
- Learn to evaluate integrals involving products of polynomials and trigonometric functions
- Explore the use of substitution methods in integration
- Practice solving integrals with varying degrees of complexity
Students and educators in calculus, mathematicians focusing on integral calculus, and anyone seeking to improve their skills in evaluating complex integrals.
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rocomath
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Your identity is wrong
\cos^2 x=\frac{1}{2}(1+\cos{2x})
\cos^2 x=\frac{1}{2}(1+\cos{2x})
Zack88
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thank you must have wrote it down wrong on paper, but does that matter since i didnt use that identity?
rocomath
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What is this supposed to be ...Zack88 said:thank you must have wrote it down wrong on paper, but does that matter since i didnt use that identity?
cos^2 x = (1 + cos^2 x) / 2
?
That's not an identity.
Zack88
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[integral] x cox^2 x dx
u= x
du = dx
dv= cos^2
v= (sin(2x)) / 4 + x/2
x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du
x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c
i know the answer is
x (sin(2x)) / 4 + cos(2x) / 8 + x^2 / 4 + c
but I got an extra x/2
u= x
du = dx
dv= cos^2
v= (sin(2x)) / 4 + x/2
x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du
x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c
i know the answer is
x (sin(2x)) / 4 + cos(2x) / 8 + x^2 / 4 + c
but I got an extra x/2
rocomath
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No. Use an identity on cos^2 x. You must reduce the power to 1 before you can integrate it.Zack88 said:[integral] x cox^2 x dx
u= x
du = dx
dv= cos^2
v= (sin(2x)) / 4 + x/2
x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du
x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c
i know the answer is
x (sin(2x)) / 4 + cos(2x) / 8 + x^2 / 4 + c
but I got an extra x/2
Zack88
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ok
[integral] x cox^2 x dx
turns into
[integral] x/2(1 + cos2x) dx
[integral] x cox^2 x dx
turns into
[integral] x/2(1 + cos2x) dx
Zack88
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[integral] x/2 (1 + cos2x)
[chain rule]
[integral] x/2 (-2sin2x) + (1 + cos2x)(1/2)
= x^2 /4 (cos2x) + (((sin(2x) / 2) + x)) (x/2)
= (x^2 cos 2x) / 4 + x(sin2x / 2 + x/2)
[chain rule]
[integral] x/2 (-2sin2x) + (1 + cos2x)(1/2)
= x^2 /4 (cos2x) + (((sin(2x) / 2) + x)) (x/2)
= (x^2 cos 2x) / 4 + x(sin2x / 2 + x/2)
rocomath
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Check it by taking the derivative, if I have to tell you again I'm not helping anymore.Zack88 said:[integral] x/2 (1 + cos2x)
[chain rule]
[integral] x/2 (-2sin2x) + (1 + cos2x)(1/2)
= x^2 /4 (cos2x) + (((sin(2x) / 2) + x)) (x/2)
= (x^2 cos 2x) / 4 + x(sin2x / 2 + x/2)
Zack88
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sorry if i was irritating you it has been a year since I've done any math and I am jumping into calc 2, but i finally got it and I have done most of the rest, just three left which I am working on now, so thank you for all your help and putting up with me :)
rocomath
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It wasn't so much irritating, just disappointing. I just want you to get a real good grade :-] I'm mainly pushing you to check your answer to exericse your differentiating skills. I'm more than happy to help!Zack88 said:sorry if i was irritating you it has been a year since I've done any math and I am jumping into calc 2, but i finally got it and I have done most of the rest, just three left which I am working on now, so thank you for all your help and putting up with me :)
Zack88
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hey when completing the square when a square roots is involved, do you square both sides?
rocomath
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Example ...
x^2-\sqrt 2 x+4
This kind of problem?
x^2-\sqrt 2 x+4
This kind of problem?
Zack88
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more like 1 / [sqrt]9x^2 + 6x -8[/sqrt]
rocomath
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You don't have to worry about squaring the square root. Just work within your square root and perform your operation as you would.
But don't forget that your coefficient of your highest degree must be 1.
But don't forget that your coefficient of your highest degree must be 1.
Zack88
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rocophysics said:But don't forget that your coefficient of your highest degree must be 1.
I don't know what that means, since i work with the square root do I not pay attention to it until finally getting the final answer. ex.
[sqrt] x^3 + 6x^2 - 4 [/sqrt]
then
x^3 + 6x^2 = 4
x^3 + 6x^2 + 9 = 4 + 9
and so on
rocomath
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You can't complete the square of a polynomial of degree 3.
I meant that the standard quadratic equation needs to have a coefficient of 1 in front of it's highest term.
ax^2+bx+c=0
a=1
I meant that the standard quadratic equation needs to have a coefficient of 1 in front of it's highest term.
ax^2+bx+c=0
a=1
Zack88
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oh ok and i just made up the example so if i have a number higher than 1 I need to divide that number out so ex. 5x^2 + 6x + 5 and then id divide the whole thing by 5.
rocomath
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Yes
x^2+\frac b a x + \frac c a=0
x^2+\frac b a x + \frac c a=0
Zack88
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ok came across a bump.
[integral] dx/ [sqrt]x^2 - 6x + 13[/sqrt]
so i have to complete the square, so
x^2 - 6x + 13 = 0
x^2 - 6x = -13
wait before i continue I always thought anything that was negative and squared it become positive but now my calculator is telling me differently.
the i divide -6 by 2 and then square it (using everything squared is positive) and get 9
x^2 - 6x + 9 = -13 + 9
x^2 + 6x +9 + -4
(x-3)^2 = =4
then square both sides but you can't square -4 w/o getting an imaginary number
[integral] dx/ [sqrt]x^2 - 6x + 13[/sqrt]
so i have to complete the square, so
x^2 - 6x + 13 = 0
x^2 - 6x = -13
wait before i continue I always thought anything that was negative and squared it become positive but now my calculator is telling me differently.
the i divide -6 by 2 and then square it (using everything squared is positive) and get 9
x^2 - 6x + 9 = -13 + 9
x^2 + 6x +9 + -4
(x-3)^2 = =4
then square both sides but you can't square -4 w/o getting an imaginary number
rocomath
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What happen here?
Don't set it equal to 0.
x^2-6x+13
x^2-6x+(\frac{6}{2})^2-(\frac{6}{2})^2+13
x^2 - 6x + 9 = -13 + 9
x^2 + 6x +9 + -4
(x-3)^2 = =4
Don't set it equal to 0.
x^2-6x+13
x^2-6x+(\frac{6}{2})^2-(\frac{6}{2})^2+13
Zack88
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rocophysics said:x^2-6x+(\frac{6}{2})^2-(\frac{6}{2})^2+13
wouldnt the 9 - 9 cancel out? so ur still left with the original problem.
rocomath
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Yep, and from there it simplifies easiy.Zack88 said:would ntthe 9 - 9 cancel out? so ur still left with the original problem.
Zack88
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lol so in other words was the square already completed?
rocomath
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Not completely. I'm going now though, go to this thread ... Find the integral...Please HELP! - https://www.physicsforums.com/showthread.php?t=211212Zack88 said:lol so in other words was the square already completed?
Post 12, pretty much your problem, just different numbers.
Zack88
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k cya thanks again
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