# 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

sciencesolve | Certified Educator

You should use the formula that converts the product into a difference or a sum such that:

`sin alpha sin beta= (1/2)(cos(alpha - beta) - cos(alpha + beta))`

Reasoning by analogy yields:

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

Factoring out x yields:

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

Considering `m!=n`  and integrating both sides yields:

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

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

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

Converting the sum or difference of sines into products yields:

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

`int sin(mx)sin(nx) dx = (-mcos mx*sin nx + nsin mx*cos nx)/(m^2-n^2)`

Notice that if `m!=n` , the integral does not yield 0.

Considering `m=n`  yields:

`int sin(mx)sin(nx) dx = (-mcos mx*sin mx + msin mx*cos mx)/(m^2-m^2)= 0/0`  invalid

Hence, evaluating the given integral yields that`int sin(mx)sin(nx) dx = (-mcos mx*sin nx + nsin mx*cos nx)/(m^2-n^2) ` if `m!=n`  and an invalid result if `m=n` .