If `int e ^(x) ((2tan x)/(1+tan x)+ cosec ^(2)(x+(pi)/(4)))dx =e ^(x). g(x)+k,` then `g ((5pi)/(4))=`

Options

A. 0
B. 1
C. `-1`
D. `2`

Answer

To solve the problem, we need to evaluate the integral given in the question and find the function \( g(x) \) at \( x = \frac{5\pi}{4} \).

### Step-by-Step Solution:

1. **Set Up the Integral**:
We start with the integral:
\[ I = \int e^x \left( \frac{2 \tan x}{1 + \tan x} + \csc^2\left(x + \frac{\pi}{4}\right) \right) dx \]

2. **Simplify the Cosecant Term**:
Recall that \( \csc^2(x + \frac{\pi}{4}) = \frac{1}{\sin^2(x + \frac{\pi}{4})} \). Using the sine addition formula:
\[ \sin\left(x + \frac{\pi}{4}\right) = \sin x \cos\frac{\pi}{4} + \cos x \sin\frac{\pi}{4} = \frac{\sin x + \cos x}{\sqrt{2}} \]
Thus,
\[ \csc^2\left(x + \frac{\pi}{4}\right) = \frac{2}{\sin^2 x + \cos^2 x + 2\sin x \cos x} = \frac{2}{1 + \sqrt{2} \sin(2x)} \]

3. **Combine Terms**:
Now, we can rewrite the integral:
\[ I = \int e^x \left( \frac{2 \tan x}{1 + \tan x} + \frac{2}{\sin^2 x + \cos^2 x + 2\sin x \cos x} \right) dx \]

4. **Let \( f(x) = \frac{2 \sin x}{\sin x + \cos x} \)**:
We define:
\[ f(x) = \frac{2 \sin x}{\sin x + \cos x} \]
We need to find \( f'(x) \).

5. **Differentiate \( f(x) \)**:
Using the quotient rule:
\[ f'(x) = \frac{(2 \cos x)(\sin x + \cos x) - (2 \sin x)(\cos x - \sin x)}{(\sin x + \cos x)^2} \]
Simplifying gives:
\[ f'(x) = \frac{2}{\sin x + \cos x} \]

6. **Integrate**:
The integral can be expressed using integration by parts:
\[ I = e^x f(x) + C \]

7. **Identify \( g(x) \)**:
From the equation \( I = e^x g(x) + k \), we find that \( g(x) = f(x) \).

8. **Evaluate \( g\left(\frac{5\pi}{4}\right) \)**:
Now we need to evaluate:
\[ g\left(\frac{5\pi}{4}\right) = \frac{2 \sin\left(\frac{5\pi}{4}\right)}{\sin\left(\frac{5\pi}{4}\right) + \cos\left(\frac{5\pi}{4}\right)} \]
Since \( \sin\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} \) and \( \cos\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} \):
\[ g\left(\frac{5\pi}{4}\right) = \frac{2 \left(-\frac{1}{\sqrt{2}}\right)}{-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}} = \frac{-\frac{2}{\sqrt{2}}}{-\frac{2}{\sqrt{2}}} = 1 \]

### Final Answer:
Thus, the value of \( g\left(\frac{5\pi}{4}\right) \) is:
\[ \boxed{1} \]