`int_(0)^(pi//2)(sin 2x)/(2+2sin^(2)x+cos^(2)x)dx=`

Options

A. `log((3)/(4))`
B. `-log((3)/(4))`
C. `2log ((4)/(3))`
D. `(pi)/(4)`

Answer

To solve the integral

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

we will simplify the integrand step by step.

### Step 1: Simplifying the Denominator
First, we rewrite the denominator:

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

Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), we can simplify this to:

\[ 2 + 2\sin^2 x + \cos^2 x = 2 + 1 + \sin^2 x = 3 + \sin^2 x. \]

Thus, we can rewrite our integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin 2x}{3 + \sin^2 x} \, dx. \]

### Step 2: Using the Identity for \(\sin 2x\)
Recall that \(\sin 2x = 2 \sin x \cos x\). We can substitute this into our integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{2 \sin x \cos x}{3 + \sin^2 x} \, dx. \]

### Step 3: Substitution
Let’s use the substitution \(u = \sin x\), which gives \(du = \cos x \, dx\). The limits change as follows: when \(x = 0\), \(u = 0\) and when \(x = \frac{\pi}{2}\), \(u = 1\). Thus, we have:

\[ I = \int_{0}^{1} \frac{2u}{3 + u^2} \, du. \]

### Step 4: Splitting the Integral
Now we can split the integral:

\[ I = 2 \int_{0}^{1} \frac{u}{3 + u^2} \, du. \]

### Step 5: Using Partial Fraction Decomposition
To evaluate the integral, we can use a simple substitution. Let \(v = 3 + u^2\), then \(dv = 2u \, du\) or \(du = \frac{dv}{2u}\). When \(u = 0\), \(v = 3\) and when \(u = 1\), \(v = 4\). The integral becomes:

\[ I = \int_{3}^{4} \frac{1}{v} \, dv = \ln |v| \bigg|_{3}^{4} = \ln 4 - \ln 3 = \ln \frac{4}{3}. \]

### Final Step: Multiply by 2
Since we have \(2\) from our earlier step, we multiply:

\[ I = 2 \ln \frac{4}{3}. \]

Thus, the final result is:

\[ I = 2 \ln \frac{4}{3}. \]