`int _(0) ^(pi//2) (sin x +cos x)/ sqrt(1+sin 2x )" "dx=`

Options

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

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx, \]

we can simplify the expression under the integral.

### Step 1: Simplifying the denominator

First, we know that

\[ \sin 2x = 2 \sin x \cos x. \]

Thus, we can rewrite the term \(1 + \sin 2x\) as:

\[ 1 + \sin 2x = 1 + 2 \sin x \cos x. \]

Using the Pythagorean identity, we can express \(1\) as \(\sin^2 x + \cos^2 x\). Therefore, we have:

\[ 1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2. \]

### Step 2: Taking the square root

Now, taking the square root of both sides gives us:

\[ \sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2} = \sin x + \cos x, \]

for \(x\) in the interval \([0, \frac{\pi}{2}]\) since both \(\sin x\) and \(\cos x\) are non-negative in this range.

### Step 3: Substituting back into the integral

Now we can substitute this back into our integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx. \]

This simplifies to:

\[ I = \int_{0}^{\frac{\pi}{2}} 1 \, dx. \]

### Step 4: Evaluating the integral

The integral of \(1\) with respect to \(x\) over the interval \([0, \frac{\pi}{2}]\) is simply:

\[ I = x \bigg|_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \]

### Final Answer

Thus, the value of the integral is:

\[ I = \frac{\pi}{2}. \]

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