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

Answer

To evaluate the integral

\[ \int \frac{\sin x + \cos x}{9 + 16 \sin 2x} \, dx, \]

we will follow a systematic approach.

### Step 1: Substitute for \(\sin x + \cos x\)

Let

\[ t = \sin x - \cos x. \]

Then, differentiating both sides with respect to \(x\):

\[ dt = \cos x \, dx + \sin x \, dx = (\cos x + \sin x) \, dx. \]

This implies:

\[ dx = \frac{dt}{\cos x + \sin x}. \]

### Step 2: Express \(\sin x + \cos x\) in terms of \(t\)

We can express \(\sin x + \cos x\) using the identity:

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

From our substitution, we have:

\[ \sin^2 x + \cos^2 x = 1, \]
\[ (\sin x - \cos x)^2 = t^2 \implies \sin^2 x + \cos^2 x - 2\sin x \cos x = t^2. \]

Thus,

\[ 1 - 2\sin x \cos x = t^2 \implies \sin x \cos x = \frac{1 - t^2}{2}. \]

### Step 3: Rewrite \(\sin 2x\)

Using the double angle identity:

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

### Step 4: Substitute into the integral

Now we substitute \(\sin 2x\) into the integral:

\[ \int \frac{\sin x + \cos x}{9 + 16(1 - t^2)} \, dx = \int \frac{\sin x + \cos x}{9 + 16 - 16t^2} \, dx = \int \frac{\sin x + \cos x}{25 - 16t^2} \, dx. \]

### Step 5: Substitute \(dx\)

Now substituting \(dx\):

\[ \int \frac{(\sin x + \cos x)}{25 - 16t^2} \cdot \frac{dt}{\sin x + \cos x} = \int \frac{dt}{25 - 16t^2}. \]

### Step 6: Solve the integral

This integral can be solved using the formula for the integral of the form \(\int \frac{dt}{a^2 - t^2}\):

\[ \int \frac{dt}{a^2 - t^2} = \frac{1}{2a} \ln \left| \frac{a + t}{a - t} \right| + C. \]

Here, \(a^2 = 25\) so \(a = 5\):

\[ \int \frac{dt}{25 - 16t^2} = \frac{1}{10} \ln \left| \frac{5 + 4t}{5 - 4t} \right| + C. \]

### Step 7: Substitute back for \(t\)

Now substituting back \(t = \sin x - \cos x\):

\[ = \frac{1}{10} \ln \left| \frac{5 + 4(\sin x - \cos x)}{5 - 4(\sin x - \cos x)} \right| + C. \]

### Final Answer

Thus, the evaluated integral is:

\[ \int \frac{\sin x + \cos x}{9 + 16 \sin 2x} \, dx = \frac{1}{10} \ln \left| \frac{5 + 4(\sin x - \cos x)}{5 - 4(\sin x - \cos x)} \right| + C. \]