Show that `int_(-L)^L ` cos((m`pi` *x)/L)*cos((n`pi` *x)/L)dx =L if m=n?

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You need to perform the following conversion to integrand, using the trigonometric identity, such that:

`cos alpha cos theta = (1/2)(cos(alpha - theta) + cos(alpha + theta))`

Reasoning by analogy , yields:

`cos((m*pi*x)/L)*cos((n*pi*x)/L) = (1/2)(cos(((m - n)*pi*x)/L) + cos(((m + n)*pi*x)/L))`

The problem provides the information that `m = n` , such that:

`cos((m*pi*x)/L)*cos((n*pi*x)/L) = (1/2)(cos(((m - m)*pi*x)/L) + cos(((m + m)*pi*x)/L))`

`cos((m*pi*x)/L)*cos((n*pi*x)/L) = (1/2)(cos 0 + cos((2m*pi*x)/L)))`

`cos((m*pi*x)/L)*cos((n*pi*x)/L) = (1/2)(1 + cos((2m*pi*x)/L)))`

Integrating both sides yields:

`int_(-L)^L cos((m*pi*x)/L)*cos((n*pi*x)/L) dx = (1/2)int_(-L)^L (1 + cos((2m*pi*x)/L))) dx`

Using the property of linearity of integral, yields:

`int_(-L)^L cos((m*pi*x)/L)*cos((n*pi*x)/L) dx = (1/2)int_(-L)^Ldx + (1/2)int_(-L)^L (cos((2m*pi*x)/L))) dx`

`int_(-L)^L cos((m*pi*x)/L)*cos((n*pi*x)/L) dx = (1/2)( x|_(-L)^L + (L/(2m*pi))(sin((2m*pi*x)/L))|_(-L)^L )`

`int_(-L)^L cos((m*pi*x)/L)*cos((n*pi*x)/L) dx = (1/2)(L - (-L) + (L/(2m*pi))(sin((2m*pi*L)/L - (L/(2m*pi))(sin((2m*pi*(-L))/L))`

`int_(-L)^L cos((m*pi*x)/L)*cos((n*pi*x)/L) dx = (1/2)(2L + (L/(2m*pi))(sin(2m*pi) + sin(2m*pi)))`

`int_(-L)^L cos((m*pi*x)/L)*cos((n*pi*x)/L) dx = (1/2)(2L + (L/(2m*pi))(0 + 0))`

`int_(-L)^L cos((m*pi*x)/L)*cos((n*pi*x)/L) dx = (1/2)(2L)`

`int_(-L)^L cos((m*pi*x)/L)*cos((n*pi*x)/L) dx = L `

Hence, evaluating the given definite integral, using the condition m = n, yields that the statement `int_(-L)^L cos((m*pi*x)/L)*cos((n*pi*x)/L) dx = L` is valid.

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