$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{Evaluate} \int \frac{1}{1-\cot \mathrm{x}}\mathrm{dx}. \\ \left(\mathrm{b}\right) \mathrm{Evaluate} \int \frac{3(\sin \mathrm{x}-2) \cos \mathrm{x}}{13-{\cos }^2\mathrm{x}-7 \sin \mathrm{x}}\mathrm{dx}. \end{gathered}$$

$$\begin{gathered} \mathrm{Let} \sin \mathrm{x}=A\frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x}-\cos \mathrm{x})+\mathrm{B}(\sin \mathrm{x}-\cos \mathrm{x}) \\ \Rightarrow \sin \mathrm{x}=\mathrm{A}(\cos \mathrm{x}+\sin \mathrm{x})+\mathrm{B}(\sin \mathrm{x}-\cos \mathrm{x}) ...\left(\mathrm{ii}\right) \end{gathered}$$

On equating the coefficient of  and cosx coefficients from both sides, we get

and $$\begin{gathered} \mathrm{A}+\mathrm{B}=1 ...\left(\mathrm{iii}\right) \\ \mathrm{A}-\mathrm{B}=0 ...\left(\mathrm{iv}\right) \end{gathered}$$

On adding Eqs. (iii) and (iv), we get

 $2\mathrm{A}=1\Rightarrow \mathrm{A}=\frac{1}{2}$

On putting the value of A in Eq. (iv), we get

 $\frac{1}{2}-\mathrm{B}=0 \Rightarrow B=\frac{1}{2}$

Now, from Eqs. (i) and (ii), we get

$$\begin{gathered} \mathrm{I}=\int \frac{{\displaystyle \frac{1}{2}}{\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}(\sin \mathrm{x}-\cos \mathrm{x})+{\displaystyle \frac{1}{2}}(\sin \mathrm{x}-\cos \mathrm{x})}{\sin \mathrm{x}-\cos \mathrm{x}}\mathrm{dx} \\ \Rightarrow \mathrm{I}=\int \frac{{\displaystyle \frac{1}{2}}(\cos \mathrm{x}+\sin \mathrm{x})+{\displaystyle \frac{1}{2}}(\sin \mathrm{x}-\cos \mathrm{x})}{\sin \mathrm{x}-\cos \mathrm{x}}\mathrm{dx} \\ =\frac{1}{2}\int \frac{\cos \mathrm{x}+\sin \mathrm{x}}{\sin \mathrm{x}-\cos \mathrm{x}}\mathrm{dx}+\frac{1}{2}\int \frac{\sin \mathrm{x}-\cos \mathrm{x}}{\sin \mathrm{x}-\cos \mathrm{x}}\mathrm{dx} \end{gathered}$$

$$\begin{gathered} \mathrm{Put} \mathrm{sinx}-\mathrm{cosx}=\mathrm{t} \mathrm{in} \mathrm{first} \mathrm{integral}. \\ \mathrm{Then}, (\cos \mathrm{x}+\sin \mathrm{x})\mathrm{dx}=\mathrm{dt} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{I}=\frac{1}{2}\int \frac{\mathrm{dt}}{\mathrm{t}}+\frac{1}{2}\int \mathrm{dx} \\ =\frac{1}{2}\log \left|\mathrm{t}\right|+\frac{1}{2}\mathrm{x}+\mathrm{C} \\ =\frac{1}{2}\log \left|\sin \mathrm{x}-\cos \mathrm{x}\right|+\frac{1}{2}\mathrm{x}+\mathrm{C} [\because \mathrm{t}=\sin \mathrm{x}-\cos \mathrm{x}] \end{gathered}$$

$$\begin{gathered} Given integral =\int \frac{3(\sin x-2)\cos x}{13-1+{\sin }^{2}x-7\sin x} dx \\ =\int \frac{3(\sin x-2)\cos x}{12+{\sin }^{2}x-7\sin x} dx\to \left(1\right) \\ put \sin x =t then \cos x dx =dt \\ then from \left(1\right) \\ I=\int \frac{3t-6}{t^2-7t+12} dt \\ Put 3t-6=A \frac{d}{dt} (t^2-7t+12)+B \\ Compare like terms on both sides \\ 2\mathrm{A}=3 \Rightarrow \mathrm{A}=\frac{3}{2} \\ \mathrm{and} -7\mathrm{A}+\mathrm{B}=-2 \\ \Rightarrow -7\times \frac{3}{2}+\mathrm{B}=-2 \\ \Rightarrow \mathrm{B}=\frac{21}{2}-2=\frac{21-4}{2}=\frac{17}{2} \\ \therefore \mathrm{I}=\int \frac{3\mathrm{t}-2}{{\mathrm{t}}^2-7\mathrm{t}+12}\mathrm{dt} \\ = \int \frac{{\displaystyle \frac{3}{2}}(2\mathrm{t}-7)}{{\mathrm{t}}^2-7\mathrm{t}+12}\mathrm{dt}+\int \frac{17/2}{{\mathrm{t}}^2-7\mathrm{t}+12}\mathrm{dt} \\ =\frac{3}{2}\int \frac{2\mathrm{t}-7}{{\mathrm{t}}^2-7\mathrm{t}+12}\mathrm{dt}+\frac{17}{2}\int \frac{1}{{\mathrm{t}}^2-7\mathrm{t}+12}\mathrm{dt} \\ ={\mathrm{I}}_1+{\mathrm{I}}_2 ...\left(\mathrm{ii}\right) \\ \mathrm{where}, {\mathrm{I}}_1=\frac{3}{2}\int \frac{2\mathrm{t}-7}{{\mathrm{t}}^2-7\mathrm{t}+12}\mathrm{dt} \\ \mathrm{Put} {\mathrm{t}}^2-7\mathrm{t}+12={\mathrm{t}}_1 \Rightarrow (2\mathrm{t}-7) \mathrm{dt}={\mathrm{dt}}_1 \\ \mathrm{Then} {\mathrm{I}}_1=\frac{3}{2}\int \frac{{\mathrm{dt}}_1}{{\mathrm{t}}_1}=\frac{3}{2}\log \left|{\mathrm{t}}_1\right|+{\mathrm{C}}_1 \\ =\frac{3}{2}\log |{\mathrm{t}}^2-7\mathrm{t}+12|+{\mathrm{C}}_1 \\ =\frac{3}{2} \log |{\sin }^2\mathrm{x}-7\sin \mathrm{x}+12|+{\mathrm{C}}_1 \\ \mathrm{and} {\mathrm{I}}_2=\frac{17}{2}\int \frac{1}{{\mathrm{t}}^2-7\mathrm{t}+12}\mathrm{dt} \\ =\frac{17}{2} \frac{1}{(\mathrm{t}-3)(\mathrm{t}-4)}\mathrm{dt} \\ =\frac{17}{2} [\log (\mathrm{t}-4)-\log (\mathrm{t}-3\left)\right] \\ =\frac{17}{2} \log \hspace{0.17em}\left|\frac{\mathrm{t}-4}{\mathrm{t}-3}\right| =\frac{17}{3}\log \left|\frac{\sin \mathrm{x}-4}{\sin \mathrm{x}-3}\right|+{\mathrm{C}}_2 \\ [\because \mathrm{t}=\sin \mathrm{x}] \end{gathered}$$