# How do you integrate #int e^x cos x^2 dx #?

# int \ e^x (cosx)^2 \ dx = 1/2 e^x + 1/5 e^xsin2x +1/10 e^xcos2x + C#

Assuming that we seek:

The first integral is trivial, for the second we apply integration by parts:

Let # { (u,=e^x, => (du)/dx,=e^x), ((dv)/dx,=cos2x, => v,=1/2 sin2x ) :}#

Then plugging into the IBP formula:

We have:

Now if we perform a second application of Integration By Parts:

Let # { (u,=e^x, => (du)/dx,=e^x), ((dv)/dx,=sin2x, => v,=-1/2 cos2x ) :}#

Then plugging into the IBP formula, we have:

Substituting result [C] into [B] we have:

Now, substituting this result into [A] and integrating we get:

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See below.

Assuming the question reads

Now

so finally

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To integrate ( \int e^x \cos(x^2) , dx ), we can use the method of integration by parts. Let ( u = e^x ) and ( dv = \cos(x^2) , dx ). Then, ( du = e^x , dx ) and ( v = \int \cos(x^2) , dx ).

The integral ( \int \cos(x^2) , dx ) does not have a elementary antiderivative. Therefore, we need to leave it as is.

Applying the integration by parts formula ( \int u , dv = uv - \int v , du ), we get:

[ \int e^x \cos(x^2) , dx = e^x \int \cos(x^2) , dx - \int \left( \int \cos(x^2) , dx \right) e^x , dx ]

So, the integral becomes:

[ \int e^x \cos(x^2) , dx = e^x \int \cos(x^2) , dx - \int e^x \left( \int \cos(x^2) , dx \right) , dx ]

This is as simplified as we can get. The integral cannot be expressed in terms of elementary functions.

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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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