# Examine what is real number p if `int_0^pi` (x^2-px) sin (2nx)dx=pi^2/n, n natural number?

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You should notice that `sin(2nx) = (-1/(2n)*cos (2nx))'` , hence, you may re-write the integrand, such that:

`int_0^(pi) (x^2-px) sin (2nx)dx = int_0^(pi) (x^2-px) (-1/(2n)*cos (2nx))' dx`

Using integration by parts, yields:

`f(x) = (x^2-px) => f'(x) = 2x - p`

`g'(x) = (-1/(2n)*cos (2nx))' => g(x) = -1/(2n)*cos (2nx)`

`int_0^(pi) (x^2-px) sin (2nx)dx = -(x^2-px)/(2n)*cos (2nx)|_0^(pi) - int_0^pi (2x - p)*(-1/(2n)*cos (2nx))dx`

You need to integrate by parts `int_0^pi (2x - p)*(-1/(2n)*cos (2nx))dx ` such that:

`int_0^pi (2x - p)*(-1/(2n)*cos (2nx))dx = (1/(4n^2))(2x-p)sin 2nx|_0^pi - 1/(2n^2) int_0^(pi) sin (2nx)dx`

`int_0^pi (2x - p)*(-1/(2n)*cos (2nx))dx = (1/(4n^2))(2x-p)sin 2nx|_0^pi - (cos(2nx))/(2n)|_0^pi`

By fundamental theorem of calculus, yields:

`int_0^pi (2x - p)*(-1/(2n)*cos (2nx))dx = (1/(4n^2))(2pi-p)sin 2npi -(1/(4n^2))(2pi-p)sin 0 - (cos(2npi))/(2n) + (cos(0))/(2n)|_0^pi`

`int_0^pi (2x - p)*(-1/(2n)*cos (2nx))dx = 0 - 0 - 1/(2n) + 1/(2n)`

`int_0^pi (2x - p)*(-1/(2n)*cos (2nx))dx = 0`

`int_0^(pi) (x^2-px) sin (2nx)dx = -(x^2-px)/(2n)*cos (2nx)|_0^(pi) + 0`

`int_0^(pi) (x^2-px) sin (2nx)dx = -(pi^2-p*pi)/(2n)*cos (2npi) + ` `0`

`int_0^(pi) (x^2-px) sin (2nx)dx = (p*pi - pi^2)/(2n)`

The problem provides the information that `int_0^(pi) (x^2-px) sin (2nx)dx = (pi^2)/n` , hence, equating the given result and the obtained result, yields:

`(p*pi - pi^2)/(2n) = (pi^2)/n`

Reducing duplicate factors, yields:

`p*pi - pi^2 = 2pi^2 => 3pi^2 = p*pi => p = 3pi`

**Hence, evaluating `p in R` , under the given conditions, yields **`p = 3pi.`

`I=int_0^pi(x^2-2px)sin(2nx)dx`

`=-((x^2-2px)cos(2nx))/(2n)}_0^pi+(1/(2n))int_0^pi(2x-2p)cos(2nx)dx`

`=(2p pi-pi^2)/(2n)+(1/n){(((x-p)sin(2nx))/(2n))_0^pi-int_0^pisin(2nx)/(2n)dx}`

`=(2p pi-pi^2)/(2n)+(1/(2n^2))(cos(2nx))/(2n)}_0^pi`

`=(p pi-pi^2)/(2n)`

But givenĀ

`I=pi^2/n=(p pi-pi^2)/(2n)`

`pi^2/n=(p pi-pi^2)/(2n)`

`p=3pi`

which is required value of p.