Solving and Understanding the Integral of (cosx - sin^2x) / (cos^2x - sin^2x)

Introduction to the Problem

The problem presented here revolves around evaluating the integral of a complex trigonometric function. Specifically, the integral in question is:

Mathematical Expression

The integral in question is:

[ I int frac{cos x - sin^2x}{cos^2x - sin^2x} dx ]

Simplifying the Integral

To solve this integral, we need to simplify the expression inside the integral. Let's start by recognizing the Pythagorean identity sin^2x cos^2x 1 and the angle addition formulas. Applying these, we can rewrite the integral:

Step-by-Step Simplification

1. Factor the Denominator: [ cos^2x - sin^2x (cos x sin x)(cos x - sin x) ] 2. Rewrite the Integral: [ I int frac{cos x - sin^2x}{(cos x sin x)(cos x - sin x)} dx ] 3. Further Simplification: [ I int frac{cos x - sin^2x}{cos x - sin x} dx ] 4. Split the Integral: [ I int frac{cos x}{cos x - sin x} dx - int frac{sin^2x}{cos x - sin x} dx ] 5. Recognize Trigonometric Identities: [ I int frac{cos x - sin x sin x}{cos x - sin x} dx - int frac{sin^2x}{cos x - sin x} dx ] [ I int 1 dx - int frac{sin x}{cos x - sin x} dx int frac{sin x}{cos x - sin x} dx - int frac{sin^2x}{cos x - sin x} dx ] [ I x - int frac{sin x}{cos x - sin x} dx - int frac{sin^2x}{cos x - sin x} dx ] 6. Substitute u 2x: [ I x - frac{1}{2} int frac{sin 2x}{cos 2x} dx - int frac{1 - cos 2x}{2cos 2x} dx ] 7. Recognize and Simplify: [ I x - frac{1}{2} ln|sec 2x| - frac{1}{2} ln|sec 2x| C ] [ I frac{1}{2} ln|sec 2x| - frac{1}{2} ln|sec 2x| C ] [ I frac{1}{2} lnleft|frac{sec 2x tan 2x}{sec 2x}right| C ] [ I frac{1}{2} ln|sin 2x| C ]

Final Answer and Verification

The final simplified answer to the integral is:

[ I frac{1}{2} ln|sin 2x| C ]

We can verify this by using logarithmic properties and trigonometric identities:

[ frac{1}{2} ln|sin 2x| frac{1}{2} ln|sin x cos x| frac{1}{2} (ln|sin x| ln|cos x|) ]

This confirms that our solution is consistent with the integral's properties.

Conclusion

Understanding and solving integrals of complex trigonometric functions can be challenging but rewarding. By breaking down the problem into simpler components and using known identities, we can effectively evaluate these integrals. This method of solving demonstrates the power of trigonometric identities in simplifying and solving integrals.

Happy studying!