`int_(-pi/2)^(pi/2)(e^(|sinx|)cosx)/((1+e^(tanx))`dx is equal to `e+1` (b) `1-e` `e-1` (d) none of these

Options

A. `e+1`
B. `2e`
C. `e-1`
D. `e-2`

Explanation

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^{|\sin x|} \cos x}{1 + e^{\tan x}} \, dx, \]

we will use properties of definite integrals and some substitutions.

### Step 1: Use the property of definite integrals

We can use the property of definite integrals that states:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

In our case, \(a = -\frac{\pi}{2}\) and \(b = \frac{\pi}{2}\). Thus, we can rewrite the integral as:

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

### Step 2: Simplify the expressions

Using the properties of sine and cosine, we have:

- \(\sin(-x) = -\sin x\) so \(|\sin(-x)| = |\sin x|\),
- \(\cos(-x) = \cos x\),
- \(\tan(-x) = -\tan x\).

Thus, we can rewrite the integral:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^{|\sin x|} \cos x}{1 + e^{-\tan x}} \, dx. \]

### Step 3: Combine the two forms of the integral

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

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

Adding these two equations gives:

\[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{e^{|\sin x|} \cos x}{1 + e^{\tan x}} + \frac{e^{|\sin x|} \cos x}{1 + e^{-\tan x}} \right) dx. \]

### Step 4: Simplify the combined expression

The combined expression simplifies to:

\[ \frac{e^{|\sin x|} \cos x \left(1 + e^{-\tan x} + 1 + e^{\tan x}\right)}{(1 + e^{\tan x})(1 + e^{-\tan x})}. \]

This simplifies further to:

\[ \frac{e^{|\sin x|} \cos x (2 + e^{\tan x} + e^{-\tan x})}{(1 + e^{\tan x})(1 + e^{-\tan x})}. \]

### Step 5: Evaluate the integral

Using the property of symmetry, we can evaluate the integral from \(0\) to \(\frac{\pi}{2}\):

\[ I = 2 \int_{0}^{\frac{\pi}{2}} \frac{e^{|\sin x|} \cos x}{1 + e^{\tan x}} \, dx. \]

Now, we can substitute \(t = \sin x\), which gives \(dt = \cos x \, dx\). The limits change from \(0\) to \(1\):

\[ I = 2 \int_{0}^{1} e^{t} \, dt. \]

### Step 6: Compute the integral

Integrating \(e^{t}\) gives:

\[ \int e^{t} \, dt = e^{t} + C. \]

Thus, evaluating from \(0\) to \(1\):

\[ I = 2 \left[ e^{t} \right]_{0}^{1} = 2 \left( e - 1 \right). \]

### Final Answer

Thus, we have:

\[ I = e - 1. \]

### Conclusion

The value of the integral is:

\[ \boxed{e - 1}. \]