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

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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.

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