Monday, April 10, 2017

`int_0^t e^s sin(t-s) ds` Evaluate the integral



You need to use integration by parts, such that:


`int udv = uv - int vdu`


`u = e^s => du = e^s ds`


`dv = sin(t-s) => v = (-cos(t-s))/(-1)`


`int e^s sin (t-s) ds = e^s*cos(t-s) - int e^s*cos(t-s)ds`


You need to use parts again, for `int e^s*cos(t-s)ds` , such that:


`u = e^s => du = e^s ds`


`dv = cos(t-s) => v = (sin(t-s))/(-1)`


`int e^s*cos(t-s)ds = -e^s*sin (t-s)...



You need to use integration by parts, such that:


`int udv = uv - int vdu`


`u = e^s => du = e^s ds`


`dv = sin(t-s) => v = (-cos(t-s))/(-1)`


`int e^s sin (t-s) ds = e^s*cos(t-s) - int e^s*cos(t-s)ds`


You need to use parts again, for `int e^s*cos(t-s)ds` , such that:


`u = e^s => du = e^s ds`


`dv = cos(t-s) => v = (sin(t-s))/(-1)`


`int e^s*cos(t-s)ds = -e^s*sin (t-s) + int e^s*sin (t-s) ds`


Replacing back, yields:


`int e^s sin (t-s) ds = e^s*cos(t-s) - (-e^s*sin (t-s) + int e^s*sin (t-s) ds)`


You need to use the substitution `int e^s sin (t-s) ds= I:`


`I = e^s*cos(t-s) + e^s*sin (t-s)- I`


`2I = e^s*cos(t-s) + e^s*sin (t-s) => I = (e^s*(cos(t-s)+sin(t-s)))/2`


You need to evaluate the definite integral, using the fundamental theorem of calculus, such that:



`int_0^t e^s sin (t-s) ds = (e^s*(cos(t-s)+sin(t-s)))/2|_0^t`


`int_0^t e^s sin (t-s) ds = (e^t*(cos(t-t)+sin(t-t)) - e^0*(cos(t-0)+sin(t-0)))/2`



`int_0^t e^s sin (t-s) ds = (e^t - cos t - sin t)/2`


Hence, evaluating the definite integral, using integration by parts, yields  `int_0^t e^s sin (t-s) ds = (e^t - cos t - sin t)/2.`

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