The value of `int_0^(pi)(e^(cosx)sinx)/((1+cos^2x)(e^(cosx)+e^(-cosx)))dx` is equal to :

Options

A. `pi^2/4`
B. `pi^2/2`
C. `pi/4`
D. `pi/2`

Answer

To solve the integral

\[ I = \int_0^{\pi} \frac{e^{\cos x} \sin x}{(1 + \cos^2 x)(e^{\cos x} + e^{-\cos x})} \, dx, \]

we can use a symmetry property of definite integrals.

### Step 1: Substitute \( x \) with \( \pi - x \)

We start by substituting \( x \) with \( \pi - x \):

\[ I = \int_0^{\pi} \frac{e^{\cos(\pi - x)} \sin(\pi - x)}{(1 + \cos^2(\pi - x))(e^{\cos(\pi - x)} + e^{-\cos(\pi - x)})} \, dx. \]

Using the properties of sine and cosine, we have:

- \( \sin(\pi - x) = \sin x \)
- \( \cos(\pi - x) = -\cos x \)

Thus, the integral becomes:

\[ I = \int_0^{\pi} \frac{e^{-\cos x} \sin x}{(1 + \cos^2 x)(e^{-\cos x} + e^{\cos x})} \, dx. \]

### Step 2: Combine the two expressions for \( I \)

Now we have two expressions for \( I \):

1. \( I = \int_0^{\pi} \frac{e^{\cos x} \sin x}{(1 + \cos^2 x)(e^{\cos x} + e^{-\cos x})} \, dx \)
2. \( I = \int_0^{\pi} \frac{e^{-\cos x} \sin x}{(1 + \cos^2 x)(e^{-\cos x} + e^{\cos x})} \, dx \)

Adding these two equations gives:

\[ 2I = \int_0^{\pi} \frac{e^{\cos x} \sin x + e^{-\cos x} \sin x}{(1 + \cos^2 x)(e^{\cos x} + e^{-\cos x})} \, dx. \]

### Step 3: Simplify the numerator

The numerator simplifies to:

\[ e^{\cos x} + e^{-\cos x} = 2 \cosh(\cos x). \]

Thus, we can write:

\[ 2I = \int_0^{\pi} \frac{2 \sin x \cosh(\cos x)}{(1 + \cos^2 x)(2 \cosh(\cos x))} \, dx. \]

This simplifies to:

\[ 2I = \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx. \]

### Step 4: Solve for \( I \)

Now we can divide both sides by 2:

\[ I = \frac{1}{2} \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx. \]

### Step 5: Change of variable

To evaluate the integral \( \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx \), we can use the substitution \( t = \cos x \), which gives \( dt = -\sin x \, dx \). The limits change from \( x = 0 \) (where \( t = 1 \)) to \( x = \pi \) (where \( t = -1 \)). Thus, we have:

\[ \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx = \int_1^{-1} \frac{-1}{1 + t^2} \, dt = \int_{-1}^{1} \frac{1}{1 + t^2} \, dt. \]

### Step 6: Evaluate the integral

The integral \( \int_{-1}^{1} \frac{1}{1 + t^2} \, dt \) is a standard integral that evaluates to:

\[ \int_{-1}^{1} \frac{1}{1 + t^2} \, dt = \left[ \tan^{-1}(t) \right]_{-1}^{1} = \tan^{-1}(1) - \tan^{-1}(-1) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}. \]

### Step 7: Final result for \( I \)

Thus, we have:

\[ I = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}. \]

### Conclusion

The value of the integral is:

\[ \boxed{\frac{\pi}{4}}. \]