Solving a Physics Homework Problem with Sin^n Integration

[vorcil]

Homework Statement

Please I am just going to list the question then show my solving, It is a physics question but I need help on integration, on parts of the question, thank you

quesiton:

A particle in the infinite square well, (i.e V(x) = \left{ \begin{array}{ccc} 0 & \textrm({ if 0 <= x <= a} \\ \infty & otherwise \end{array}\right
has the initial wave function

\Psi(x,0) = Asin^3(\frac{\pi x}{a} \textrm{ for (0<=x<=a) }

Determine A, find \Psi(x,t) \textrm{ and calculate &lt;x&gt; as a function of time, what is the expectation value of the energy? (hint: sin^n \theta and cos^n \theta can be reducted by repeated application of trigonometric sum formulas, to linear combinatiosn of sin(m\theta) and cos(m\theta), with m = 0,1,2...n }<br />

Homework Equations

1 = \int_{-\infty}^{\infty} |\Psi(x,0)|^2 dx

The Attempt at a Solution

My attempt, well first I normalize it,

so the solution to the question is,\int_0^a |A sin^3(\frac{\pi x}{a}|^2 = 1

now this brings up something I do not know how to solve,
since the sin would go to the ^6 power

I was told by my course instructor that there are many ways to solve integrals of this type,
and the question states Sin^n theta can be reduced by repeated application of the trigonometric sum formulas, to linear combinations of sin

so I have to use some trigonometric identities to solve/simplify the integral basically

 

[darkchild]

Ok, so what is the problem you are having? Look up the trigonometric sum formulas and find one that matches sine squared.

 

[vorcil]

I need to know the integral of sin^3 x

to reduce it to sin^2

\int sin^3x dx = \int [(sin^2 x dx)(sin x)

umm

using the trigonometric identity (the only one I know, sin^2 x + cos^2 x = 1

I'm going to substitute in 1-cos^2 x = sin^2x

so uh, sin^3 x = (1-cos^2 x) sin x

can I just move the sin to the other side? so it cancels out and becomes

\frac{sin^3 x}{sin x} = (1-cos^2 x)
sin^2 x = 1-cos^2x

oh wow, it works and just becomes the trig identity lol
jeese well that's helpful
yesssssss

now i make it simplier?

 

[darkchild]

The problem in your original post had a sine squared, now you're working on sine cubed...which is it?

The hint you posted says use the trig sum identities. If you don't know them, look them up.

 

[vorcil]

so my equation

\int_0^a sin^3\frac{\pi x}{a}

\int_0^a sin^2(\frac{\pi x}{a} = \int_0^a (1-cos^2\frac{\pi x}{a} )

now I need to find a trig identity for sin^2 x to make it simplier

 

[vorcil]

The problem in your original post had a sine squared, now you're working on sine cubed...which is it?

The hint you posted says use the trig sum identities. If you don't know them, look them up.

Sorry it was a typo

it was A sin^3 x

 

[darkchild]

so my equation

\int_0^a sin^3\frac{\pi x}{a}

\int_0^a sin^2(\frac{\pi x}{a} = \int_0^a (1-cos^2\frac{\pi x}{a} )

now I need to find a trig identity for sin^2 x to make it simplier

You're confusing yourself. LOOK UP THE TRIGONOMETRIC SUM IDENTITIES.

 

[vorcil]

<br /> \int_0^a sin^2(\frac{\pi x}{a}) = \int_0^a (1-cos^2(\frac{\pi x}{a} )) <br />

Normalizing the wave function,
\int_0^a \left| (1-cos^2(\frac{\pi x}{a})) \right|^2 =1

i'm lost I've done something wrong

i'm going to restart and do it properly

 

[vorcil]

Found a trigonometric identity that I like

sin 3\theta =3sin\theta -4sin^3 \theta

letting theta be my \frac{\pi x}{a}

sin 3(\frac{\pi x}{a}) = 3sin(\frac{pi x}{a})

re arranging it so sin^3 theta is the subject

adding 4 sin^3 theta to each side

sin 3\theta + 4sin^3 \theta = 3sin\theta

subtracting sin 3 theta from each side

4sin^3 \theta = 3sin\theta - sin 3\theta

diving the both sides by 4

sin^3 \theta = \frac{3}{4} sin\theta - \frac{1/4}sin 3\theta

plugging back in my theta value I get

sin^3 (\frac{\pi x}{a}) = \frac{3}{4}sin(\frac{\pi x}{a}) - \frac{1}{4} sin 3 (\frac{\pi x}{a})

ok that seems better

 

[vorcil]

now to normalize this wave function, I square it, and the absolute value =1

\int_0^a \left| sin^3 (\frac{\pi x}{a}) \right|^2 dx= \left| \int_0^a \frac{3}{4}sin(\frac{\pi x}{a}) - \frac{1}{\4} sin 3 (\frac{\pi x}{a}) \right|^2 dx = 1

squaring each of the components
Can someone check this part please, not sure how squaring in this context works

\left( \frac{3}{4}sin(\frac{\pi x}{a}) - \frac{1}{\4} sin 3 (\frac{\pi x}{a}) \right)^2
=
\left( \frac{3}{4}sin^2(\frac{\pi x}{a})\right) - \left( \frac{1}{4} sin^2 (\frac{3\pi x}{a}) \right)
(can I just square the sin part? and leave the coefficients i.e 3/4 and 1/4 constant? without squaring them? )

 

[Cyosis]

Hint:

(a+b)^2=a^2+b^2+2ab

 

[vorcil]

\left( \frac{3}{4}sin{\frac{\pi x}{a}} - \frac{1}{4}sin{3\frac{\pi x}{a} \right)^2

=

\left( \frac{3}{4}sin{\frac{\pi x}{a}} - \frac{1}{4}sin{3\frac{\pi x}{a} \right)\left( \frac{3}{4}sin{\frac{\pi x}{a}} - \frac{1}{4}sin{3\frac{\pi x}{a} \right)

=

\left( \frac{3}{4}\frac{3}{4} sin^2{\frac{\pi x}{a} \right) + \left( \frac{1}{4}\frac{1}{4}sin^2 (3*3(\frac{\pi x}{a} ) ) \right) -2\left(\frac{1}{4}\frac{3}{4}sin^2 (3\frac{\pi x}{a} \right)

=

\left( \frac{9}{14}sin^2 {\frac{\pi x}{a} \right) + \left( \frac{1}{16}sin^2 9\frac{\pi x}{a} \right) -2\left(\frac{3}{16} sin^2 3\frac{\pi x}{a} \right)

so my equation is

1 = A^2 \int_0^a \left[ \left( \frac{9}{14}sin^2 {\frac{\pi x}{a} \right) + \left( \frac{1}{16}sin^2 9\frac{\pi x}{a} \right) -2\left(\frac{3}{16} sin^2 3\frac{\pi x}{a} \right) \right]dx

what do I do from here?