`int_0^(pi//2)(tan^2x)/(1+tan^2x)dx` is equal to

Options

A. `oo`
B. 0
C. `pi/4`
D. `pi/2`

Answer

To solve the definite integral \( I = \int_0^{\frac{\pi}{2}} \frac{\tan^2 x}{1 + \tan^2 x} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We know that:
\[ 1 + \tan^2 x = \sec^2 x \]
Thus, we can rewrite the integrand:
\[ \frac{\tan^2 x}{1 + \tan^2 x} = \frac{\tan^2 x}{\sec^2 x} = \tan^2 x \cdot \cos^2 x \]

### Step 2: Rewrite \(\tan^2 x\)
Using the identity \(\tan x = \frac{\sin x}{\cos x}\), we have:
\[ \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \]
Substituting this into the integral gives:
\[ I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\cos^2 x} \cdot \cos^2 x \, dx = \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \]

### Step 3: Use the identity for \(\sin^2 x\)
We can use the identity:
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]
Thus, we rewrite the integral:
\[ I = \int_0^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx \]

### Step 4: Split the integral
This integral can be split into two parts:
\[ I = \frac{1}{2} \int_0^{\frac{\pi}{2}} (1 - \cos 2x) \, dx = \frac{1}{2} \left( \int_0^{\frac{\pi}{2}} 1 \, dx - \int_0^{\frac{\pi}{2}} \cos 2x \, dx \right) \]

### Step 5: Evaluate the integrals
1. The first integral:
\[ \int_0^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} \]

2. The second integral:
\[ \int_0^{\frac{\pi}{2}} \cos 2x \, dx = \left[ \frac{\sin 2x}{2} \right]_0^{\frac{\pi}{2}} = \frac{\sin(\pi)}{2} - \frac{\sin(0)}{2} = 0 \]

### Step 6: Combine the results
Putting it all together:
\[ I = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4} \]

Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{4}} \]