`IfI_1=int_0^(pi/2)(cos^2x)/(1+cos^2x)dx ,I_2=int_0^(pi/2)(sin^2x)/(1+sin^2x)dx` `I_3=int_0^(pi/2)(1+2cos^2xsin^2x)/(4+2cos^2xsin^2x)dx ,t h e n` `I_1=I_2> I_3` (b) `I_3> I_1=I_2` `I_1=I_2=I_3` (d) none of these

Options

A. `I_(1)=I_(2)gtI_(3)`
B. `I_(3)gtI_(1)=I_(2)`
C. `I_(1)=I_(2)=I_(3)`
D. none of these

Explanation

To solve the problem, we need to evaluate the integrals \(I_1\), \(I_2\), and \(I_3\) and compare their values.

### Step 1: Evaluate \(I_1\)

Given:
\[ I_1 = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + \cos^2 x} \, dx \]

Using the property of definite integrals:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
we can rewrite \(I_1\) as:
\[ I_1 = \int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + \sin^2 x} \, dx \]
This integral is exactly \(I_2\).

### Step 2: Show that \(I_1 = I_2\)

Thus, we have:
\[ I_1 = I_2 \]

### Step 3: Evaluate \(I_3\)

Now, we evaluate \(I_3\):
\[ I_3 = \int_0^{\frac{\pi}{2}} \frac{1 + 2 \cos^2 x \sin^2 x}{4 + 2 \cos^2 x \sin^2 x} \, dx \]

### Step 4: Add \(I_1\) and \(I_2\)

Now, we add \(I_1\) and \(I_2\):
\[ I_1 + I_2 = \int_0^{\frac{\pi}{2}} \left( \frac{\cos^2 x}{1 + \cos^2 x} + \frac{\sin^2 x}{1 + \sin^2 x} \right) \, dx \]

Finding a common denominator:
\[ = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x (1 + \sin^2 x) + \sin^2 x (1 + \cos^2 x)}{(1 + \cos^2 x)(1 + \sin^2 x)} \, dx \]

Simplifying the numerator:
\[ = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x + \cos^2 x \sin^2 x + \sin^2 x + \sin^2 x \cos^2 x}{(1 + \cos^2 x)(1 + \sin^2 x)} \, dx \]
\[ = \int_0^{\frac{\pi}{2}} \frac{1 + 2 \cos^2 x \sin^2 x}{(1 + \cos^2 x)(1 + \sin^2 x)} \, dx \]

### Step 5: Relate \(I_1 + I_2\) to \(I_3\)

Notice that:
\[ I_1 + I_2 = \int_0^{\frac{\pi}{2}} \frac{1 + 2 \cos^2 x \sin^2 x}{(1 + \cos^2 x)(1 + \sin^2 x)} \, dx \]

Now, we can express this as:
\[ I_1 + I_2 = 2I_3 \]

### Step 6: Conclusion

Since \(I_1 = I_2\), we have:
\[ 2I_1 = 2I_3 \implies I_1 = I_3 \]

Thus, we conclude:
\[ I_1 = I_2 = I_3 \]

### Final Answer

The correct option is:
(c) \(I_1 = I_2 = I_3\)

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