`int t sinh(mt) dt` Evaluate the integral

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marizi eNotes educator| Certified Educator

To help you solve this, we consider the the integration by parts:

`int u * dv = uv - int v* du`

Let `u = t`  and `dv = sinh(mt) dt.`

based from `int t*sinh(mt) dt` for` int u*dv`

In this integral, the "m" will be treated as constant since it is integrated with respect to "t".

 From `u = t` , then `du = dt`

From `dv = sinh(mt) dt` , then  int dv = v

In` int sinh(mt) dt` , let `w = mt`  then `dw= m dt`  or `dt= (dw)/m`

Substitute `w = mt ` and `dt = (dw)/m`

`int sinh(mt) dt ` =` int (sinh(w)dw)/m`

                         =` (1/m) int sinh(w) dw `    

based from  c is constant in`int c f(x) dx=c int f(x) dx +C`

`(1/w) int sinh(w) dw = (1/w) cosh(w) +C`

 Substitute `w = mt` , it becomes `v = 1/(m)cosh(mt)+C`



`u = t `

`du = dt`

`dv = sinh(mt) dt`

`v = 1/(m)cosh(mt)`

Plug into the integration by parts: `int u * dv = uv - int v* du`

`int t* sinh(mt) dt = t*1/(m)cosh(mt) - int 1/mcosh(mt) dt`

                            `= t/mcosh(mt) - 1/mint cosh(mt) dt`

                            `= t/mcosh(mt) - 1/m*1/msinh(mt)+C`

                            = `t/mcosh(mt) - 1/m^2 sinh(mt) +C`

                          =   `(mtcosh(mt) -sinh(mt))/m^2 +C`






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