Solve the following integration
`int(sin 2x+sin 5x-sin 3x)/(cos x+1-2 sin^(2)2x)dx`

Answer

To solve the integral

\[ I = \int \frac{\sin 2x + \sin 5x - \sin 3x}{\cos x + 1 - 2 \sin^2 2x} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Denominator
We know that

\[ 1 - 2 \sin^2 \theta = \cos 2\theta. \]

Thus, we can rewrite the denominator:

\[ \cos x + 1 - 2 \sin^2 2x = \cos x + 1 - \cos 4x. \]

### Step 2: Rewrite the Integral
Now, we can express the integral as:

\[ I = \int \frac{\sin 2x + \sin 5x - \sin 3x}{\cos x + 1 - \cos 4x} \, dx. \]

### Step 3: Use Sine Difference Formula
We can use the sine difference formula:

\[ \sin a - \sin b = 2 \cos\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right). \]

Applying this to \(\sin 5x - \sin 3x\):

Let \(a = 5x\) and \(b = 3x\):

\[ \sin 5x - \sin 3x = 2 \cos\left(4x\right) \sin\left(x\right). \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{\sin 2x + 2 \cos 4x \sin x}{\cos x + 1 - \cos 4x} \, dx. \]

### Step 4: Factor Out Common Terms
Now, we can factor out \(2 \sin x\):

\[ I = \int \frac{2 \sin x \left(\cos 4x + \sin 2x\right)}{\cos x + 1 - \cos 4x} \, dx. \]

### Step 5: Cancel Common Terms
Notice that \(\cos x + 1 - \cos 4x\) can be simplified. Let's denote \(D = \cos x + 1 - \cos 4x\).

Thus, we have:

\[ I = 2 \int \sin x \, dx. \]

### Step 6: Integrate
The integral of \(\sin x\) is:

\[ \int \sin x \, dx = -\cos x + C. \]

So,

\[ I = 2(-\cos x) + C = -2\cos x + C. \]

### Final Answer
Thus, the final result of the integral is:

\[ I = -2\cos x + C. \]

---