Integral with sine(x) and x multiplied

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Homework Statement


I'm doing a physics problem, but I'm stuck on the math part. My integral is

$$ \frac {2 \sqrt {15}}{a^{3}} \int_{0}^{a} \hspace{0.01 in} sin(\frac {n \pi}{a} x)x(a-x) dx $$
Where ##a## is a constant, and ##n = 1,2,3,...##

Homework Equations

The Attempt at a Solution


I will separate these into two integrals

$$ \frac {2 \sqrt {15}}{a^{3}} \Big [ \int_{0}^{a} ax \hspace{0.01 in} sin(\frac {n \pi}{a}x) dx - \int_{0}^{a} x^{2} \hspace {0.01 in} sin( \frac {n \pi}{a} x) \Big ] dx $$

But from here, I know I could look up on the internet some sort of integral table, but the one I found

upload_2015-3-1_0-13-19.png

upload_2015-3-1_0-14-0.png

Is this integral reasonable to solve by hand, or is it just better to use the table?
 
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Maylis said:

Homework Statement


I'm doing a physics problem, but I'm stuck on the math part. My integral is

$$ \frac {2 \sqrt {15}}{a^{3}} \int_{0}^{a} \hspace{0.01 in} sin(\frac {n \pi}{a} x)x(a-x) dx $$
Where ##a## is a constant, and ##n = 1,2,3,...##

Homework Equations

The Attempt at a Solution


I will separate these into two integrals

$$ \frac {2 \sqrt {15}}{a^{3}} \Big [ \int_{0}^{a} ax \hspace{0.01 in} sin(\frac {n \pi}{a}x) dx - \int_{0}^{a} x^{2} \hspace {0.01 in} sin( \frac {n \pi}{a} x) \Big ] dx $$

But from here, I know I could look up on the internet some sort of integral table, but the one I found

View attachment 79722
View attachment 79723
Is this integral reasonable to solve by hand, or is it just better to use the table?

It's just integration by parts. Try doing x*sin(x) using the method. It would be good practice.
 
Okay, that was a fun little exercise to do the integration by parts (it's been a couple years)

So for the first term
$$ \frac {2 \sqrt {15}}{a^{3}} \int_{0}^{a} ax \hspace{0.01 in} sin(\frac {n \pi}{a}x) \hspace {0.01 in} dx $$
I will set ##u = x##, ##du = dx##, ##dv = sin(\frac {n \pi}{a} x) dx##, ##v = - \frac {a}{n \pi} \hspace {0.01 in} cos (\frac {n \pi}{a} x)##
So using integration by parts (and cancelling the ##a## term in the integral with the one outside)
$$ \int u \hspace {0.01 in} dv = uv - \int v \hspace {0.01 in} du $$

$$ \frac {2 \sqrt {15}}{a^{2}} \Big [ x \frac {-a}{n \pi} \hspace {0.01 in} cos (\frac {n \pi}{a} x) - \int - \frac {a}{n \pi} \hspace {0.01 in} cos (\frac {n \pi}{a} x) \hspace {0.01 in} dx \Big ] $$
$$ \frac {2 \sqrt {15}}{a^{2}} \Big [ x \frac {-a}{n \pi} \hspace {0.01 in} cos (\frac {n \pi}{a} x) + \frac {a^{2}}{n^{2} \pi^{2}} \hspace {0.01 in} sin(\frac {n \pi}{a} x) \Big ] \bigg |_{0:a} $$
$$ \frac {2 \sqrt {15}}{a^{2}} \Big [ \frac {-a^{2}}{n \pi} \hspace {0.01 in} cos (n \pi) + \frac {a^{2}}{n^{2} \pi^{2}} \hspace {0.01 in} sin(n \pi) \Big ] $$
The first term reduces to
$$ \frac {-2 \sqrt {15}}{n \pi} \hspace {0.01 in} cos (n \pi) + \frac {2 \sqrt {15}}{n^{2} \pi^{2}} \hspace {0.01 in} sin(n \pi) $$

Now for the second term
##u = x^{2}##, ##du = 2x \hspace {0.01 in} dx##, ##dv = sin(\frac {n \pi}{a} x) dx##, ##v = - \frac {a}{n \pi} \hspace {0.01 in} cos (\frac {n \pi}{a} x)##
$$ \frac {2 \sqrt {15}}{a^{3}} \Big [ x^{2} \frac {-a}{n \pi} \hspace {0.01 in} cos (\frac {n \pi}{a} x) + \frac {1}{2} \int \frac {a}{n \pi} \hspace {0.01 in} cos (\frac {n \pi}{a} x) \hspace {0.01 in} dx \Big ]$$
$$ \frac {2 \sqrt {15}}{a^{3}} \Big [ x^{2} \frac {-a}{n \pi} \hspace {0.01 in} cos (\frac {n \pi}{a} x) + \frac {a^{2}}{2n^{2} \pi^{2}} \hspace {0.01 in} sin (\frac {n \pi}{a} x) \Big ] \bigg |_{0:a} $$
$$ \frac {2 \sqrt {15}}{a^{3}} \Big [ \frac {-a^{3}}{n \pi} \hspace {0.01 in} cos (n \pi) + \frac {a^{2}}{2n^{2} \pi^{2}} \hspace {0.01 in} sin (n \pi) \Big ]$$
The second term reduces to
$$ \frac {-2 \sqrt{15}}{n \pi} \hspace {0.01 in} cos (n \pi) + \frac {\sqrt {15}}{an^{2} \pi^{2}} \hspace {0.01 in} sin (n \pi) $$

Now going all the way back, I have to subtract these terms,
$$\Big [ \frac {-2 \sqrt {15}}{n \pi} \hspace {0.01 in} cos (n \pi) + \frac {2 \sqrt {15}}{n^{2} \pi^{2}} \hspace {0.01 in} sin(n \pi) \Big ] - \Big [\frac {-2 \sqrt{15}}{n \pi} \hspace {0.01 in} cos (n \pi) + \frac {\sqrt {15}}{an^{2} \pi^{2}} \hspace {0.01 in} sin (n \pi) \Big ] $$

Simplifying to
$$ c_{n} = \frac {\sqrt {15}}{n^{2} \pi^{2}} \hspace {0.01 in} sin(n \pi) \big (2 - \frac {1}{a} \big ) $$
 
Last edited:
I think the author just used a table, not sure if his solution is equivalent to mine
upload_2015-3-1_1-21-22.png
 
Maylis said:
I think the author just used a table, not sure if his solution is equivalent to mine
Are you aware that ##\sin(n\pi)=0## so you got ##0##?
 
Maylis said:

Homework Statement


I'm doing a physics problem, but I'm stuck on the math part. My integral is

$$ \frac {2 \sqrt {15}}{a^{3}} \int_{0}^{a} \hspace{0.01 in} sin(\frac {n \pi}{a} x)x(a-x) dx $$
Where ##a## is a constant, and ##n = 1,2,3,...##

Homework Equations

The Attempt at a Solution


I will separate these into two integrals

$$ \frac {2 \sqrt {15}}{a^{3}} \Big [ \int_{0}^{a} ax \hspace{0.01 in} sin(\frac {n \pi}{a}x) dx - \int_{0}^{a} x^{2} \hspace {0.01 in} sin( \frac {n \pi}{a} x) \Big ] dx $$

But from here, I know I could look up on the internet some sort of integral table, but the one I found

View attachment 79722
View attachment 79723
Is this integral reasonable to solve by hand, or is it just better to use the table?

A useful alternative method is to use "differentiation under the integral sign". Here is how it works.

We have ##\sin(\theta) = \text{Im}\: e^{i \theta}##, where ##\text{Im}## denotes the imaginary part. Also: the integral of the imaginary part equals the imaginary part of the integral (for integration over real numbers). So, if we write
F(k) = \int_0^a e^{i x k/a} \, dx
we can use the facts that
\frac{d}{dk} e^{i k x /a} = \frac{i x}{a} e^{i k x/a}
and
\frac{d^2}{dk^2} e^{i k x/a} = -\frac{x^2}{a^2} e^{i k x/a}
to get
\int_0^a x(a-x) e^{i k x/a} \, dx = \left( 1 + \frac{a^2}{i} \frac{d}{dk} + a^2 \frac{d^2}{dk^2} \right) F(k)
You can easily find ##F(k)##, take the derivatives above, and simplify. Then you can take the imaginary part, to get the integral with ##\sin(k x /a)## in the integrand. Finally, you can set ##k = n \pi## and simplify further.
 
Last edited:
Where did I go wrong in my integration? Everything seems to check out...I overlooked that I got zero, but at least it is connect if ##n## is even...but my solution is for all ##n##
 
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