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

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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I assume you mean that m and n are integers.  Otherwise this isn't true. 


First we need the following formulas:

`"cos" (a-b) = "cos" a "cos" b + "sin" a "sin" b`

`"cos"(a+b) = "cos"a "cos" b - "sin" a "sin" b`

Subtract the second from the first and you get:

`"cos"(a-b)-"cos"(a+b) = 2 "sin" a "sin" b`


`"sin" a "sin" b = (1/2) ["cos" (a-b) - "cos" (a+b) ]`



`int "sin" mx "sin" nxdx = 1/2 int[ "cos" ((m-n)x) - "cos" ((m+n)x) ]dx`

If m-n = 0, then the first part of the integral becomes:

`int "cos" (0x) dx = int 1 dx = x`

Otherwise, if m-n is not zero:

`int "cos" ((m-n)x)dx = 1/(m-n) "sin" ((m-n)x)`


Similarly for the second integral:

If m+n = 0, then the integral is x.  If not, the integral becomes:

`int "cos" ((m+n)x)dx = 1/(m+n) "sin" ((m+n)x)`


So, if `m != +- n`

`int _(-pi)^pi "sin" mx "sin" nx = `

`1/(m-n) "sin" ((m-n)x) - 1/(m+n) "sin" ((m+n)x)|_(-pi)^pi`

But m,n are integers, so m-n and m+n are integers

`"sin" ("integer"*pi) = 0`

Thus if `m!= +- n` , then the integral is 0


If m=n, then:

`int ... = x - 1/(m+n) "sin" ((m+n)x) |_(-pi)^pi`

As before, the second part is 0, but now the first part is `pi - (-pi) = 2 pi`

Remember though, we want `1/2 int _(-pi)^pi ...`

Thus, for m=n, the integral is `pi`


If m = -n, then:

`int ... = 1/(m-n) "sin"((m-n)x) - x |_(-pi)^pi`

As before, the first part will be 0.  The second part will be:

`-pi - pi = -2pi`

After dividing by 2, we have that, for m=-n, the integral is `-pi/2`