Integration Techniques for Complex Trigonometric Functions: Solving sin(x)/(1 - sin(x)cos(x))

Integration of complex trigonometric functions can often require the application of multiple integration techniques. One such example is int (sin x) / (1 - sin(x)cos(x)) dx. In this article, we will explore the integral of this particular function and walk through the steps of the solution process.

Overview of the Method

The integral of sin(x) / (1 - sin(x)cos(x)) can be solved using a substitution method. This article will detail the integration using trigonometric identities and substitution, focusing on the key steps involved in simplifying the integral.

Step-by-Step Solution

We are going to evaluate the integral [ I int frac{sin x}{1 - sin xcos x} dx ] by using substitutions. Let's start by rewriting the integral:

[ I int frac{sin x}{1 - sin xcos x} dx ][ int frac{sin x(1 - sin xcos x)}{1 - sin^2 xcos^2 x} dx ][ int frac{sin x dx}{1 - sin^2 xcos^2 x} - int frac{sin^2 xcos x dx}{1 - sin^2 xcos^2 x} ]

We will consider each of these integrals separately.

Integral of the First Term

Consider ( J int frac{sin x dx}{1 - sin^2 xcos^2 x} ).

[ J -int frac{d cos x}{1 - (1 - cos^2 x)cos^2 x} -int frac{dy}{1 - y^2y^4} ]where ( y cos x ).[ -int frac{d y / y^2}{(y^2 - 1)y^4} frac{1}{2} int frac{(1 - y^{-2}) - (1 - y^{-2})}{(y^2 - 1)y^4} dy ][ frac{1}{2} left[ int frac{d(y - y^{-1})}{(y - y^{-1})^2 - 3} - int frac{d(y y^{-1})}{(y y^{-1})^2 - 1} right] ][ frac{1}{2 sqrt{3}} ln left| frac{(y - y^{-1}) - sqrt{3}}{(y - y^{-1}) sqrt{3}} right| - tan^{-1} (y - y^{-1}) C_1 ][ frac{1}{2 sqrt{3}} ln left| frac{cos^2 x - y - y^{-1}}{cos^2 x y y^{-1}} right| - tan^{-1} left( frac{cos^2 x - 1}{cos x} right) C_1 ]

Integral of the Second Term

Consider ( K int frac{sin^2 x cos x dx}{1 - sin^2 xcos^2 x} ). where ( z sin x ).

[ K int frac{z^2 dz}{1 - z^2 z^4} cdot frac{1}{1 - z^2} int frac{dz / z^2}{(z^2 - 1)(z^2 - 1)} int frac{d z / z^2}{(z - z^{-1})^2 - 3} ][ frac{1}{2} left[ tan^{-1} (z - z^{-1}) - frac{1}{2 sqrt{3}} ln left| frac{(z - z^{-1}) - sqrt{3}}{(z - z^{-1}) sqrt{3}} right| right] C_2 ][ frac{1}{2 sqrt{3}} left[ sqrt{3} tan^{-1} left( frac{sin^2 x - 1}{sin x} right) - ln left| frac{sin^2 x - sqrt{3} sin x - 1}{sin^2 x sqrt{3} sin x - 1} right| right] C_2 ]

Now, combining the results from both integrals, we get:

[ I frac{1}{2 sqrt{3}} left[ ln left| frac{cos^2 x - sqrt{3} cos x - 1}{cos^2 x sqrt{3} cos x - 1} right| right. ][ left. - sqrt{3} tan^{-1} left( frac{sin x}{cos x} right) sqrt{3} tan^{-1} left( frac{cos x}{sin x} right) right] C ]

This final expression represents the solution of the given integral using trigonometric substitution and identities.

Conclusion

In conclusion, solving integrals involving complex trigonometric functions requires a combination of algebraic manipulation, trigonometric identities, and substitution. Understanding these techniques not only helps in solving specific integrals but also in developing a strong foundation for tackling more advanced problems in calculus and beyond.

Related Keywords

Trigonometric integration Integral substitution Trigonometric identities