# What is limit of F(x)= integral (0 to x) e^(-t)(cos t+sin t)dt, x>0?

*print*Print*list*Cite

### 1 Answer

You need to evaluate the primitive `F(x)` using the property of linearity of integral, such that:

`F(x) = int_0^x e^(-t)(cos t) dt + int_0^x e^(-t)(sin t) dt`

You should come up with the following notations, such that:

`I_1 = int_0^x e^(-t)(cos t) dt`

`I_2 = int_0^x e^(-t)(sin t) dt`

You need to use parts to evaluate the definite integrals `I_1` and `I_2` , such that:

`u = e^(-t) => du = -e^(-t)dt`

`dv = cos t => v = sin t`

Using the formula of integration by parts yields:

`int udv = uv - int vdu`

Reasoning by analogy yields:

`I_1 = e^(-t)sin t|_0^x + int_0^x e^(-t)(sin t) dt`

You need to use again parts to evaluate `int_0^x e^(-t)(sin t) dt` , such that:

`u = e^(-t) => du = -e^(-t)dt`

`dv = sin t => v = -cos t`

`int_0^x e^(-t)(sin t) dt = -e^(-t)cos t|_0^x - I_1`

`I_1 = e^(-t)sin t|_0^x + (-e^(-t)cos t|_0^x - I_1)`

`I_1 = e^(-t)sin t|_0^x - e^(-t)cos t|_0^x - I_1`

`2I_1 = e^(-x) sin x - 0 - e^(-x)cos x - 1`

`2I_1 = e^(-x) sin x - e^(-x)cos x + 1`

`I_1 = (e^(-x) sin x - e^(-x)cos x + 1)/2`

Evaluating `I_2` yields:

`I_2 = -e^(-t)cos t|_0^x - (e^(-t)sin t|_0^x + I_2)`

`2I_2 = -e^(-t)cos t|_0^x - e^(-t)sin t|_0^x`

`2I_2 = -e^(-x)cos x + 1 - e^(-x)sin x`

`I_2 = (-e^(-x)cos x + 1 - e^(-x)sin x)/2`

`F(x) = I_1 + I_2`

`F(x) = (e^(-x) sin x - e^(-x)cos x + 1 - e^(-x)cos x + 1 - e^(-x)sin x)/2`

Reducing duplicate terms yields:

`F(x) = (2 - 2e^(-x)cos x)/2`

`F(x) = 1 - e^(-x)cos x`

You need to evaluate the limit of the function, such that:

`lim_(x->oo) (1 - e^(-x)cos x) = lim_(x->oo) 1 - lim_(x->oo) e^(-x)cos x`

`lim_(x->oo) (1 - e^(-x)cos x) = 1 - lim_(x->oo) (cos x)/(e^x)`

Since the limit is indeterminate, `lim_(x->oo) (cos x)/(e^x) = oo/oo` , and `|cos x| <= 1` , you may use squeeze principle to evaluate `lim_(x->oo) (cos x)/(e^x), ` such that:

`lim_(x->oo) -1/e^x <= lim_(x->oo) (cos x)/(e^x) <= lim_(x->oo) 1/e^x`

`0 <= lim_(x->oo) (cos x)/(e^x) <= 0 => lim_(x->oo) (cos x)/(e^x) = 0`

`lim_(x->oo) (1 - e^(-x)cos x) = 1 - 0`

`lim_(x->oo) (1 - e^(-x)cos x) = 1`

**Hence, evaluating the limit of `F(x)` , under the given conditions, yields **`lim_(x->oo) (int_0^x e^(-t)(cos t) dt + int_0^x e^(-t)(sin t) dt) = 1.`