If $$\int {f\left( x \right)\sin \,x\,\cos \,x\,dx = \frac{1}{{2\left( {{b^2} - {a^2}} \right)}}{{\log }_e}\left( {f\left( x \right)} \right)} + C,\,b \ne \pm a,$$           then $${\left\{ {f\left( x \right)} \right\}^{ - 1}}$$  is equal to :

Solution :
$$\eqalign{ & \int {f\left( x \right)\sin \,x\,\cos \,x\,dx = \frac{1}{{2\left( {{b^2} - {a^2}} \right)}}{{\log }_e}\left( {f\left( x \right)} \right)} + C \cr & {\text{Therefore, }} \cr & f\left( x \right)\sin \,x\,\cos \,x = \frac{1}{{2\left( {{b^2} - {a^2}} \right)}}.\frac{1}{{f\left( x \right)}}f'\left( x \right)\,\,\,\left[ {{\text{by differentiating both the sides}}} \right] \cr & \Rightarrow 2\left( {{b^2} - {a^2}} \right)\sin \,x\,\cos \,x = \frac{{f'\left( x \right)}}{{{{\left( {f\left( x \right)} \right)}^2}}} \cr & \int {\left( {2{b^2}\sin \,x\,\cos \,x - 2{a^2}\sin \,x\,\cos \,x} \right)dx = } \int {\frac{{f'\left( x \right)}}{{{{\left( {f\left( x \right)} \right)}^2}}}} \,dx \cr & \left[ {{\text{by integrating both the sides}}} \right] \cr & \Rightarrow - {b^2}{\cos ^2}x - {a^2}{\sin ^2}x - C = - \frac{1}{{f\left( x \right)}} \cr} $$