20/07/13 17:23:05.81 tR49RdlH.net
>>119
In=∫(sin(Nx)/sin(x))^2dx=∫(1-cos(2Nx))/(sin(x)^2dx
とおいて積和公式から
I(n+1)+I(n-1)
=2∫(1-cos(2Nx)cos2x)/(sin(x)^2dx
=2∫(1-cos(2Nx))/(sin(x)^2dx
. + 2∫cos(2Nx)(1-cos(2x)/(sin(x)^2dx
=2In
∴In=nπ
20/07/13 17:23:05.81 tR49RdlH.net
>>119
In=∫(sin(Nx)/sin(x))^2dx=∫(1-cos(2Nx))/(sin(x)^2dx
とおいて積和公式から
I(n+1)+I(n-1)
=2∫(1-cos(2Nx)cos2x)/(sin(x)^2dx
=2∫(1-cos(2Nx))/(sin(x)^2dx
. + 2∫cos(2Nx)(1-cos(2x)/(sin(x)^2dx
=2In
∴In=nπ