`int t^2 sin(beta t ) dt ` Evaluate the integral

Expert Answers

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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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