verify if y=e^x sin x is primitive of y=e^x (sin x +cos x )?

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You may test if a function `F(x)` is primitive of other function `f(x)` , if differentiating `F(x)` with respect to x,yields the equation of `f(x)` .

Hence, you need to differentiate `F(x) = e^x*sin x` , with respect to x, using the product rule, such that:

`F'(x) = (e^x)'*sin x + e^x*(sin x)'`

`F'(x) = e^x*sin x + e^x*cos x`

Factoring out `e^x` yields:

`F'(x) = e^x(sin x + cos x) = f(x)`

You may also solve the problem using integration method, hence, if `int f(x) dx = F(x)` yields that `F(x)` is primitive of `f(x)` .

Hence, evaluating the indefinite integral of the function `f(x) = e^x(sin x + cos x)` yields:

`int e^x(sin x + cos x) dx = int e^x*sin x dx + int e^x*cos x dx`

You need to use integration by parts, such that:

`int udv = uv - int vdu`

`u = e^x => du = e^x dx`

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

`int e^x*sin x dx = -e^x*cos x + int e^x*cos x`

You need to integrate by parts `int e^x*cos x` , such that:

`u = e^x => du = e^x dx`

`dv = cos x => v = sin x`

`int e^x*sin x dx = -e^x*cos x + (e^x*sin x - int e^x*sin xdx)`

You need to use the following notation, such that:

`int e^x*sin x dx =I`

`I = -e^x*cos x + e^x*sin x - I => 2I = e^x*sin x - e^x*cos x`

`I = (e^x*sin x - e^x*cos x)/2`

`int e^x*cos x = e^x*sin x - int e^x*sin xdx`

`int e^x*cos x = e^x*sin x - (-e^x*cos x + int e^x*cos x)`

You need to use the following notation, such that:

`int e^x*cos x = E`

`E = e^x*sin x + e^x*cos x - E => 2E = e^x*sin x + e^x*cos x => E = (e^x*sin x + e^x*cos x)/2`

`int e^x(sin x + cos x) dx = I + E`

`int e^x(sin x + cos x) dx = (e^x*sin x - e^x*cos x + e^x*sin x + e^x*cos x)/2`

`int e^x(sin x + cos x) dx = (2e^x*sin )/2`

`int e^x(sin x + cos x) dx = e^x*sin x = F(x)`

Hence, either differentiating F(x) with respect to x, or integrating f(x) yields, in both cases, that the statement that F(x) represents the primitive function of f(x) is valid.

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