Evaluate
`int(sin x)/(sin 4x)dx`

Answer

To evaluate the integral

\[ I = \int \frac{\sin x}{\sin 4x} \, dx, \]

we will follow a series of steps to simplify and solve the integral.

### Step 1: Rewrite \(\sin 4x\)

Using the double angle formula, we know that

\[ \sin 4x = 2 \sin 2x \cos 2x. \]

Now, we can express \(\sin 2x\) as:

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

Thus,

\[ \sin 4x = 2 (2 \sin x \cos x) \cos 2x = 4 \sin x \cos x \cos 2x. \]

### Step 2: Substitute in the integral

Now substituting \(\sin 4x\) back into the integral, we have:

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

### Step 3: Simplify the integral

Now we can rewrite the integral:

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

### Step 4: Use the identity for \(\cos 2x\)

Recall that

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

Thus, we can express the integral as:

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

### Step 5: Substitute \(t = \sin x\)

Let \(t = \sin x\), then \(dt = \cos x \, dx\), or \(dx = \frac{dt}{\cos x}\). The integral becomes:

\[ I = \frac{1}{4} \int \frac{1}{\cos^2 x (1 - 2t^2)} \, dt. \]

### Step 6: Rewrite the integral

Using the identity \(\cos^2 x = 1 - t^2\), we can rewrite the integral:

\[ I = \frac{1}{4} \int \frac{1}{(1 - t^2)(1 - 2t^2)} \, dt. \]

### Step 7: Partial fraction decomposition

Now we can use partial fractions:

\[ \frac{1}{(1 - t^2)(1 - 2t^2)} = \frac{A}{1 - t^2} + \frac{B}{1 - 2t^2}. \]

Multiplying through by the denominator gives:

\[ 1 = A(1 - 2t^2) + B(1 - t^2). \]

Expanding and equating coefficients will give us values for \(A\) and \(B\).

### Step 8: Solve for \(A\) and \(B\)

Setting \(t = 0\):

\[ 1 = A + B \implies A + B = 1. \]

Setting \(t = 1\):

\[ 1 = A(1 - 2) + B(1 - 1) \implies -A = 1 \implies A = -1. \]

Thus, \(B = 2\). Therefore, we have:

\[ \frac{1}{(1 - t^2)(1 - 2t^2)} = \frac{-1}{1 - t^2} + \frac{2}{1 - 2t^2}. \]

### Step 9: Integrate each term

Now we can integrate:

\[ I = \frac{1}{4} \left( -\int \frac{1}{1 - t^2} \, dt + 2 \int \frac{1}{1 - 2t^2} \, dt \right). \]

The integrals can be solved using the standard integral forms:

\[ \int \frac{1}{1 - t^2} \, dt = \frac{1}{2} \log \left| \frac{1 + t}{1 - t} \right| + C, \]

and

\[ \int \frac{1}{1 - 2t^2} \, dt = \frac{1}{\sqrt{2}} \tanh^{-1}(\sqrt{2} t) + C. \]

### Step 10: Substitute back and simplify

After integrating and substituting back \(t = \sin x\), we will arrive at the final answer:

\[ I = \frac{1}{4} \left( -\frac{1}{2} \log \left| \frac{1 + \sin x}{1 - \sin x} \right| + 2 \cdot \frac{1}{\sqrt{2}} \tanh^{-1}(\sqrt{2} \sin x) \right) + C. \]

### Final Result

Thus, the final result of the integral is:

\[ I = \frac{1}{8} \log \left| \frac{1 + \sin x}{1 - \sin x} \right| + \frac{1}{2\sqrt{2}} \tanh^{-1}(\sqrt{2} \sin x) + C. \]