If an anti-derivative of f(x) is `e^x` and that of g(x) is co x, then `intf(x)cosxdx+intg(x)e^xdx` is equal to

Options

A. `f(x).g(x)+C`
B. `f(x)+g(x)+C`
C. `e^xcosx+C`
D. `f(x)-g(x)+C`

Answer

To solve the problem, we need to evaluate the expression:

\[ \int f(x) \cos x \, dx + \int g(x) e^x \, dx \]

Given that the anti-derivative of \( f(x) \) is \( e^x \) and the anti-derivative of \( g(x) \) is \( \cos x \), we can derive \( f(x) \) and \( g(x) \) by differentiating these anti-derivatives.

### Step 1: Find \( f(x) \) and \( g(x) \)

Since the anti-derivative of \( f(x) \) is \( e^x \):

\[ f(x) = \frac{d}{dx}(e^x) = e^x \]

Since the anti-derivative of \( g(x) \) is \( \cos x \):

\[ g(x) = \frac{d}{dx}(\cos x) = -\sin x \]

### Step 2: Substitute \( f(x) \) and \( g(x) \) into the integrals

Now we can substitute \( f(x) \) and \( g(x) \) into the expression:

\[ \int f(x) \cos x \, dx + \int g(x) e^x \, dx = \int e^x \cos x \, dx + \int (-\sin x) e^x \, dx \]

### Step 3: Combine the integrals

This can be rewritten as:

\[ \int e^x \cos x \, dx - \int \sin x \, e^x \, dx \]

### Step 4: Use Integration by Parts

We will use integration by parts to solve both integrals.

For the first integral \( \int e^x \cos x \, dx \):

Let:
- \( u = \cos x \) → \( du = -\sin x \, dx \)
- \( dv = e^x \, dx \) → \( v = e^x \)

Using integration by parts:

\[ \int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx \]

This simplifies to:

\[ \int e^x \cos x \, dx = e^x \cos x + \int e^x \sin x \, dx \]

### Step 5: Substitute back into the original expression

We can now substitute this back into our expression:

\[ \int e^x \cos x \, dx - \int e^x \sin x \, dx = (e^x \cos x + \int e^x \sin x \, dx) - \int e^x \sin x \, dx \]

The integrals of \( e^x \sin x \) cancel out:

\[ \int e^x \cos x \, dx - \int e^x \sin x \, dx = e^x \cos x \]

### Step 6: Add the constant of integration

Finally, we add the constant of integration \( C \):

\[ \int f(x) \cos x \, dx + \int g(x) e^x \, dx = e^x \cos x + C \]

### Final Answer

Thus, the final result is:

\[ \int f(x) \cos x \, dx + \int g(x) e^x \, dx = e^x \cos x + C \]

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