Evaluate : `int (sin x +cos x ) sqrt(9 +16 sin 2x) dx`

Answer

To evaluate the integral \( \int (\sin x + \cos x) \sqrt{9 + 16 \sin 2x} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int (\sin x + \cos x) \sqrt{9 + 16 \sin 2x} \, dx \]
We know that \( \sin 2x = 2 \sin x \cos x \), so we can rewrite the integral as:
\[ \int (\sin x + \cos x) \sqrt{9 + 32 \sin x \cos x} \, dx \]

### Step 2: Simplify the Expression Inside the Square Root
Next, we can express \( 9 + 32 \sin x \cos x \) in a different form. We can rewrite \( 32 \sin x \cos x \) as \( 16(2 \sin x \cos x) \):
\[ 9 + 32 \sin x \cos x = 9 + 16(2 \sin x \cos x) = 9 + 16 \sin 2x \]

### Step 3: Substitute for \( \sin x + \cos x \)
We can also express \( \sin x + \cos x \) in a different way. Let:
\[ t = \sin x - \cos x \]
Then, we have:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]

### Step 4: Change of Variables
Now, we will let \( t = \sin x - \cos x \). The derivative is:
\[ dt = (\cos x + \sin x) \, dx \]
Thus, we can express \( dx \) in terms of \( dt \):
\[ dx = \frac{dt}{\cos x + \sin x} \]

### Step 5: Rewrite the Integral
Now substituting \( t \) into the integral, we have:
\[ \int \sqrt{9 + 16 \sin 2x} \, dt \]

### Step 6: Factor Out Constants
We can factor out constants from the square root:
\[ \sqrt{9 + 16 \sin 2x} = \sqrt{25 - 16t^2} \]

### Step 7: Use the Integral Formula
We can apply the integral formula:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
In our case, \( a^2 = 25 \) and \( x = 4t \).

### Step 8: Final Expression
After applying the formula and simplifying, we get:
\[ \frac{t}{2} \sqrt{25 - 16t^2} + \frac{25}{2} \sin^{-1}\left(\frac{4t}{5}\right) + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \frac{1}{2}(\sin x - \cos x) \sqrt{25 - 16(\sin x - \cos x)^2} + \frac{25}{2} \sin^{-1}\left(\frac{4(\sin x - \cos x)}{5}\right) + C \]