Đáp án:

• Gợi ý:  

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

• Đặt \( t = \cos x \) và \( t = \sin x \), tương ứng:  

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

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

\( = -\frac{1}{2} \int \frac{1}{\sqrt{2}} \left[ \frac{1}{t - \frac{\sqrt{2}}{2}} - \frac{1}{t + \frac{\sqrt{2}}{2}} \right] \, dt \)

\( = -\frac{\sqrt{2}}{4} \left[ \ln \left| \frac{t - \frac{\sqrt{2}}{2}}{ t + \frac{\sqrt{2}}{2}} \right| \right] + C \)

\( = -\frac{\sqrt{2}}{4} \ln \left| \frac{2 \cos x - \sqrt{2}}{2 \cos x + \sqrt{2}} \right| + C \)

• \( \int \frac{\cos x}{1 - 2 \sin^2 x} \, dx \),  tương tự !