Prove that the indefinite integral of I = int(exp(x).cos(x))dx is (1/2)exp(x).sin(x) + (1/2)exp(x).cos(x) + C

Starting with the initial integral of int(exp(x).cos(x))dx we can see that this is going to have to be integrated by parts. This states that the integral of (u . dv/dx)dx is equal to u.v - int(v . du/dx)dx

Therefore, by applying this equation we can determine that u=exp(x), dv=sin(x), du=exp(x), v=-cox(x), as integrating sin(x) will give us -cos(x)

This gives us int(I) = exp(x).sin(x) - int(exp(x).sin(x))dx

As can be seen, this changes the form of the equation but it hasnt become any simpler. At this point we integrate once more by parts.

By looking at the 'int(exp(x).sin(x))dx' which we obtained, this can be integrated again.

int(exp(x).sin(x))dx = -exp(x).cos(x) + int(exp(x).cos(x))dx

Substituting this into the first integral we worked out will give us:

I = exp(x).sin(x) + exp(x).cos(x) - int(exp(x).cos(x))

It may seem that we have once again achieved nothing, but by inspecting the equation closely, we can see that we have ended up with the initial integral we were presented with on the RHS of the equation. By moving this negative integral to the other side we can see that we are going to have 2I. Dividing the whole equation by 2 will give us I = (exp(x).sin(x) + exp(x).cos(x))/2 + C (dont forget the constant!).

Hence we have obtained an answer to this cyclic integral.

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