The value of `I=int_(0)^(pi//2)((sinx+cos)^(2))/(sqrt(1+sin2x))dx` is

Options

A. 3
B. 1
C. 2
D. 0

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{(\sin x + \cos x)^2}{\sqrt{1 + \sin 2x}} \, dx \), we will follow these steps:

### Step 1: Simplify the Denominator
We start by simplifying the denominator \( \sqrt{1 + \sin 2x} \).

Using the identity \( \sin 2x = 2 \sin x \cos x \), we can rewrite the denominator:
\[ 1 + \sin 2x = 1 + 2 \sin x \cos x \]
Now, we can express \( 1 \) as \( \sin^2 x + \cos^2 x \):
\[ 1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2 \]
Thus,
\[ \sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2} = \sin x + \cos x \]
(for \( x \) in the interval \( [0, \frac{\pi}{2}] \), \( \sin x + \cos x \) is non-negative).

### Step 2: Substitute Back into the Integral
Now we substitute this back into our integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{(\sin x + \cos x)^2}{\sin x + \cos x} \, dx \]
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx \]

### Step 3: Integrate
Now we can integrate:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin x \, dx + \int_{0}^{\frac{\pi}{2}} \cos x \, dx \]
Calculating each integral:
- The integral of \( \sin x \) is \( -\cos x \):
\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = [-\cos x]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = 0 + 1 = 1 \]
- The integral of \( \cos x \) is \( \sin x \):
\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]

### Step 4: Combine Results
Combining the results from both integrals:
\[ I = 1 + 1 = 2 \]

### Final Answer
Thus, the value of the integral \( I \) is:
\[ \boxed{2} \]