You need to use the substitution `beta*t = u` , such that:
`beta*t = u => beta dt = du `
Replacing the variable, yields:
`int t^2*sin(beta*t) dt = 1/(beta^3) int u^2*sin u du`
You need to use the integration by parts such that:
`intfdg =fg - int gdf`
`f =u^2 => df = 2udu`
`dg =sin u=>g = -cos u`
`1/(beta^3)int u^2*sin u du = 1/(beta^3)(-u^2*cos u + 2int u*cos u du)`
You need to use the integration by parts...
You need to use the substitution `beta*t = u` , such that:
`beta*t = u => beta dt = du `
Replacing the variable, yields:
`int t^2*sin(beta*t) dt = 1/(beta^3) int u^2*sin u du`
You need to use the integration by parts such that:
`intfdg =fg - int gdf`
`f =u^2 => df = 2udu`
`dg =sin u=>g = -cos u`
`1/(beta^3)int u^2*sin u du = 1/(beta^3)(-u^2*cos u + 2int u*cos u du)`
You need to use the integration by parts again, such that:
`2int u*cos u du = 2u*sin u - 2int sin u du`
`f =u => df = du`
`dg =cos u=>g = sin u`
`2int u*cos u du = 2u*sin u + 2cos u + c`
`1/(beta^3)int u^2*sin u du = 1/(beta^3)(-u^2*cos u + 2u*sin u + 2cos u) + c`
Replacing back the variable, yields:
`int t^2*sin(beta*t) dt = 1/(beta^3)(-(beta*t)^2*cos(beta*t) + 2(beta*t)*sin(beta*t) + 2cos (beta*t)) + c`
Hence, evaluating the integral, using substitution, then integration by parts, yields `int t^2*sin(beta*t) dt = 1/(beta^3)(-(beta*t)^2*cos(beta*t) + 2(beta*t)*sin(beta*t) + 2cos (beta*t)) + c`
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