Đáp án:

• \( \int \frac{\cos 3x}{\sin x} \, dx = \int \frac{4 \cos^3 x - 3 \cos x}{\sin x} \, dx \)

• \( = \int \sin x \frac{ (4 \cos^3 x - 3 \cos x)}{1 - \cos^2 x} \, dx \)

• Đặt \( t = \cos x \)

• \( I = -\int \frac{4t^3 - 3t}{1 - t^2} \, dt = \int \frac{4t^3 - 3t}{t^2 - 1} \, dt \)

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

\( = 2t^2 + \frac{1}{2} \ln |t^2 - 1| + C \)

\( = 2 \cos^2 x + \frac{1}{2} \ln |\cos^2 x - 1| + C \)