Chapter 8: Techniques of Integration - Practice Exercises - Page 517: 14

$$\frac{1}{5}\ln \left| {\frac{{\sin \theta - 2}}{{\sin \theta + 3}}} \right| + C$$

$$\eqalign{ & \int {\frac{{\cos \theta d\theta }}{{{{\sin }^2}\theta + \sin \theta - 6}}} \cr & {\text{let }}x = \sin \theta ,\,\,\,dx = \cos \theta d\theta \cr & {\text{use the substitution}} \cr & \int {\frac{{\cos \theta d\theta }}{{{{\sin }^2}\theta + \sin \theta - 6}}} = \int {\frac{{dx}}{{{x^2} + x - 6}}} \cr & \cr & {\text{Decompose the integrand into partial fractions.}} \cr & {\text{The form of the partial fraction decomposition is}} \cr & \frac{1}{{{x^2} + x - 6}} = \frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}} \cr & \frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}} = \frac{A}{{x + 3}} + \frac{B}{{x - 2}} \cr & {\text{Multiplying by }}\left( {x + 3} \right)\left( {x - 2} \right){\text{, we have}} \cr & 1 = A\left( {x - 2} \right) + B\left( {x + 3} \right) \cr & \cr & {\text{if we set }}x = - 3 \cr & 1 = A\left( { - 3 - 2} \right) + B\left( 0 \right) \cr & A = - \frac{1}{5} \cr & \cr & {\text{if we set }}x = 2 \cr & 1 = A\left( 0 \right) + B\left( {2 + 3} \right) \cr & B = \frac{1}{5} \cr & \cr & {\text{The decomposition of the integrand is}} \cr & \frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}} = \frac{{ - 1/5}}{{x + 3}} + \frac{{1/5}}{{x - 2}} \cr & \cr & \int {\frac{{dx}}{{{x^2} + x - 6}}} = - \frac{1}{5}\int {\frac{1}{{x + 3}}dx + \frac{1}{5}\int {\frac{1}{{x - 2}}} dx} \cr & \cr & {\text{Integrating}} \cr & \int {\frac{{dx}}{{{x^2} + x - 6}}} = - \frac{1}{5}\ln \left| {x + 3} \right| + \frac{1}{5}\ln \left| {x - 2} \right| + C \cr & \int {\frac{{dx}}{{{x^2} + x - 6}}} = \frac{1}{5}\ln \left| {\frac{{x - 2}}{{x + 3}}} \right| + C \cr & \cr & {\text{We know that }}x = \sin \theta ,\,\,\, then \cr & = \frac{1}{5}\ln \left| {\frac{{\sin \theta - 2}}{{\sin \theta + 3}}} \right| + C \cr} $$