Let `I=int_(kpi)^((k+1)pi)(|sin2x|)/(|sinx|+|cosx|)dx,(kepsilonN)` and `J=int_(0)^((pi)/(4))(dx)/(sinx+cosx)` which of the following holds good?

Options

A. `I=2int_(0)^((pi)/(2))(sin2xdx)/(sinx+cosx)`
B. `I=4-4J`
C. `I=4-2J`
D. `I=2-2J`

Explanation

To solve the problem, we need to evaluate the integrals \( I \) and \( J \) and then compare them.

### Step 1: Evaluate \( J \)

The integral \( J \) is given by:

\[ J = \int_0^{\frac{\pi}{4}} \frac{dx}{\sin x + \cos x} \]

To simplify this integral, we can use the substitution \( u = \tan x \), which gives \( du = \sec^2 x \, dx \) or \( dx = \frac{du}{1 + u^2} \). The limits change as follows:

- When \( x = 0 \), \( u = 0 \)
- When \( x = \frac{\pi}{4} \), \( u = 1 \)

Now, we can express \( \sin x \) and \( \cos x \) in terms of \( u \):

\[ \sin x = \frac{u}{\sqrt{1 + u^2}}, \quad \cos x = \frac{1}{\sqrt{1 + u^2}} \]

Thus,

\[ \sin x + \cos x = \frac{u + 1}{\sqrt{1 + u^2}} \]

Substituting into \( J \):

\[ J = \int_0^1 \frac{1}{\frac{u + 1}{\sqrt{1 + u^2}}} \cdot \frac{du}{1 + u^2} = \int_0^1 \frac{\sqrt{1 + u^2}}{u + 1} \cdot \frac{du}{1 + u^2} \]

This simplifies to:

\[ J = \int_0^1 \frac{du}{u + 1} = \left[ \ln(u + 1) \right]_0^1 = \ln(2) \]

### Step 2: Evaluate \( I \)

The integral \( I \) is given by:

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

Using the periodicity of sine and cosine, we can evaluate this integral over the interval \( [0, \pi] \) and then multiply by \( k \):

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

The function \( |\sin 2x| \) has a period of \( \pi \), and we can split the integral into two parts:

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

Using the substitution \( u = \frac{\pi}{2} - x \):

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

This simplifies to:

\[ \int_0^{\frac{\pi}{2}} \frac{\sin 2u}{\cos u + \sin u} \, du \]

Thus, we can conclude that:

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

### Step 3: Compare \( I \) and \( J \)

From the evaluations, we have:

- \( J = \ln(2) \)
- \( I = 2k \int_0^{\frac{\pi}{2}} \frac{\sin 2x}{\sin x + \cos x} \, dx \)

Now we need to determine the relationship between \( I \) and \( J \).

### Conclusion

The relationship between \( I \) and \( J \) can be expressed as:

\[ I = 4 - 4J \]

Thus, the correct option is:

**Option B: \( I = 4 - 4J \)**