Step 1: Use trigonometric identity
Using
\[
\sin 4x = 4\sin x \cos x \cos 2x
\]
So,
\[
I=\int \frac{\sin x}{4\sin x \cos x \cos 2x}\,dx
\]
Cancelling $\sin x$,
\[
I=\frac{1}{4}\int \frac{1}{\cos x \cos 2x}\,dx
\]
Multiply numerator and denominator by $\cos x$:
\[
I=\frac{1}{4}\int \frac{\cos x}{\cos^2 x \cos 2x}\,dx
\]
Now use
\[
\cos^2 x = 1-\sin^2 x
\]
and
\[
\cos 2x = 1-2\sin^2 x
\]
Hence,
\[
I=\frac{1}{4}\int \frac{\cos x}{(1-\sin^2 x)(1-2\sin^2 x)}\,dx
\]
Step 2: Substitution
Let
\[
t=\sin x
\]
Then
\[
dt=\cos x\,dx
\]
So the integral becomes
\[
I=\frac{1}{4}\int \frac{1}{(1-t^2)(1-2t^2)}\,dt
\]
Step 3: Partial fraction decomposition
Write
\[
\frac{1}{(1-t^2)(1-2t^2)}
=
\frac{A}{1-t^2}
+
\frac{B}{1-2t^2}
\]
Multiplying throughout,
\[
1=A(1-2t^2)+B(1-t^2)
\]
Comparing coefficients,
\[
A=-1,\qquad B=2
\]
Therefore,
\[
\frac{1}{(1-t^2)(1-2t^2)}
=
\frac{-1}{1-t^2}
+
\frac{2}{1-2t^2}
\]
Thus,
\[
I=
\frac{1}{4}\int \left(
\frac{-1}{1-t^2}
+
\frac{2}{1-2t^2}
\right)\,dt
\]
\[
I=
-\frac{1}{4}\int \frac{1}{1-t^2}\,dt
+
\frac{1}{2}\int \frac{1}{1-2t^2}\,dt
\]
Step 4: Integrate each term
Using
\[
\int \frac{dx}{a^2-x^2}
=
\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C
\]
For the first integral:
\[
-\frac{1}{4}\int \frac{1}{1-t^2}\,dt
=
-\frac{1}{8}\log\left|\frac{1+t}{1-t}\right|
\]
For the second integral:
\[
\frac{1}{2}\int \frac{1}{1-2t^2}\,dt
=
\frac{1}{4\sqrt{2}}
\log\left|
\frac{1+\sqrt{2}t}{1-\sqrt{2}t}
\right|
\]
Step 5: Substitute back
Putting $t=\sin x$,
\[
I=
-\frac{1}{8}\log\left|
\frac{1+\sin x}{1-\sin x}
\right|
+
\frac{1}{4\sqrt{2}}
\log\left|
\frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}
\right|
+C
\]
Final Answer
\[
\boxed{
\int \frac{\sin x}{\sin 4x}\,dx
=
-\frac{1}{8}\log\left|
\frac{1+\sin x}{1-\sin x}
\right|
+
\frac{1}{4\sqrt{2}}
\log\left|
\frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}
\right|
+C
}
\]