Homework Help

Prove the following formula: integrate of sin(mx)sin(nx)dx = 0 if n does not equal m...

user profile pic

rugby4life | Student | (Level 1) Honors

Posted October 10, 2012 at 11:16 AM via web

dislike 1 like

Prove the following formula:

integrate of sin(mx)sin(nx)dx = 0 if n does not equal m

integrate of sin(mx)sin(nx)dx = pi if n=m

2 Answers | Add Yours

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted October 10, 2012 at 4:09 PM (Answer #1)

dislike 1 like

The problem is inconsistent since it does not provide the limits of integration because only evaluating a definite integral yields a value such 0 or `pi` .

You only may evaluate the indefinite integral converting the product `sin(mx)sin(nx)`  into a sum such that:

`sin(mx)sin(nx) = (1/2)(cos (mx - nx) - cos(mx + nx))`

`int sin(mx)sin(nx)dx = (1/2)int cos(m-n)x dx - (1/2)int cos(m+n)x dx`

`int sin(mx)sin(nx)dx = (1/2)((sin(m-n)x)/(m-n) - (sin(m+n)x)/(m+n)) + c`

Considering `m!=n`  yields:

`int sin(mx)sin(nx)dx = (1/2)((sin(m-n)x)/(m-n) - (sin(m+n)x)/(m+n)) + c`

Considering `m=n`  yields:

`int sin(mx)sin(nx)dx = 0/0 ` invalid

Hence, evaluating the given indefinite integral if `m!=n`  yields `int sin(mx)sin(nx)dx = (1/2)((sin(m-n)x)/(m-n) - (sin(m+n)x)/(m+n)) + c` , but if m=n, the result is invalid.

user profile pic

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted October 10, 2012 at 12:03 PM (Answer #2)

dislike 0 like

I suppose the problem is to integrate from 

-Pi to Pi 

or equivalently from

0 to 2Pi.

and also m and n are positive integers.


I will use the formula:

Sin(A)Sin(B) = (1/2)[Cos(A-B) - Cos(A+B)]

 

Therefore,

Sin(mx)Sin(nx) =(1/2)[Cos((m-n)x) - Cos((m+n)x)] 

So gievn,

I = Integral[ Sin(mx) Sin(nx) dx]  0 to 2Pi

  = Integral[(1/2){Cos((m-n)x) - Cos((m+n)x)} dx] 0 to 2Pi

  = Integral[(1/2){Cos((m-n)x)] 0 to 2Pi

     - (1/2) Sin((m+n)x)/(m+n)] 0 to 2Pi

as m+n =/= 0 always. So the second term in the result will be zero both at 0 and 2Pi. 


therefore,

I = Integral[(1/2){Cos((m-n)x)] 0 to 2Pi

  = Integral[(1/4) { Exp[(m-n) x] + Exp[ - (m-n) x]}]

  =(1/4)  (2Pi) Integral[(1/2Pi){ Exp[(m-n) x] + Exp[ - (m-n) x]}]

  = (Pi/2) Integral[(1/2Pi){ Exp[(m-n) x] + Exp[(n-m) x]}]

  = (Pi/2) * [CroneckerDelta(m,n) + CroneckerDelta(n,m)]

  = (Pi/2) * 2 * CroneckerDelta(m,n)

  = Pi * CroneckerDelta(m,n)

Where I  have used the definition of Cronecker Delta function:

CroneckerDelta(m,n)=Integral[(1/2Pi){ Exp[(m-n) x]


and

CroneckerDelta(m,n)

= CroneckerDelta(n,m) = 1 {if m=n}

                                      = 0 {if m =/= n} 


Hence The Result:

I = Integral[ Sin(mx) Sin(nx) dx]  0 to 2Pi

  = Pi * CroneckerDelta(m,n)

  = Pi   {if m=n}

  = Pi *0 = 0 for m =/= n

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes