# How do you evaluate the integral #int e^-xcosx#?

# int \ e^(-x) \cosx \ dx = 1/2(e^(-x)sinx - e^(-x)cosx) + C #

Let:

We may apply integration by components:

After that, entering the IBP formula:

gives us

Subsequently, entering the IBP formula provides us with:

When we enter this outcome into [A], we obtain:

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To evaluate the integral ∫e^(-x)cos(x), you can use integration by parts. Let u = e^(-x) and dv = cos(x) dx. Then differentiate u to get du and integrate dv to get v.

du = -e^(-x) dx v = sin(x)

Now, apply the integration by parts formula:

∫u dv = uv - ∫v du

Substitute the values of u, v, du, and dv into the formula:

∫e^(-x)cos(x) dx = e^(-x)sin(x) - ∫sin(x)(-e^(-x)) dx

Simplify the expression:

= e^(-x)sin(x) + ∫e^(-x)sin(x) dx

Now, notice that the integral on the right side is similar to the original integral. So, you can solve for it:

Let I = ∫e^(-x)cos(x) dx

Then, I = e^(-x)sin(x) + ∫e^(-x)sin(x) dx

Rearrange the equation:

I = e^(-x)sin(x) + I

Subtract I from both sides:

0 = e^(-x)sin(x)

Divide both sides by e^(-x)sin(x):

0 = 1

This equation is not true, so there might be a mistake in the calculations. Let's try another approach.

Another method to evaluate this integral is to use the method of integration by parts again on the second integral:

Let u = sin(x) and dv = e^(-x) dx Then, du = cos(x) dx and v = -e^(-x)

Apply integration by parts:

∫e^(-x)sin(x) dx = -e^(-x)sin(x) - ∫(-e^(-x))cos(x) dx

= -e^(-x)sin(x) + ∫e^(-x)cos(x) dx

Now, add the original integral to both sides:

2∫e^(-x)sin(x) dx = -e^(-x)sin(x) + ∫e^(-x)cos(x) dx

Now, solve for the original integral:

∫e^(-x)sin(x) dx = (-1/2)e^(-x)sin(x) + C

Therefore, the integral of e^(-x)cos(x) dx is:

∫e^(-x)cos(x) dx = (-1/2)e^(-x)sin(x) + C

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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