\displaystyle \text{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \frac{1}{5+4\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{5+4\cos x}
\displaystyle \text{Putting } \cos x = \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle I = \int \frac{dx}{5+4\left(\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{5(1+\tan^2\frac{x}{2})+4(1-\tan^2\frac{x}{2})}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{5+5\tan^2\frac{x}{2}+4-4\tan^2\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{\tan^2\frac{x}{2}+9}
\displaystyle \text{Let } \tan\frac{x}{2} = t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx = dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx = 2\,dt
\displaystyle \therefore I = 2\int \frac{dt}{t^2+3^2}
\displaystyle = \frac{2}{3}\tan^{-1}\left(\frac{t}{3}\right) + C
\displaystyle = \frac{2}{3}\tan^{-1}\left(\frac{\tan\frac{x}{2}}{3}\right) + C

\displaystyle \textbf{Question 2: }~\int \frac{1}{5-4\sin x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{5-4\sin x}
\displaystyle \text{Putting } \sin x = \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{5-4\left(\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{5(1+\tan^2\frac{x}{2})-8\tan\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{5\tan^2\frac{x}{2}-8\tan\frac{x}{2}+5}
\displaystyle \text{Let } \tan\frac{x}{2} = t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx = dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx = 2\,dt
\displaystyle \therefore I = 2\int \frac{dt}{5t^2-8t+5}
\displaystyle = \frac{2}{5}\int \frac{dt}{t^2-\frac{8}{5}t+1}
\displaystyle = \frac{2}{5}\int \frac{dt}{t^2-\frac{8}{5}t+\left(\frac{4}{5}\right)^2-\left(\frac{4}{5}\right)^2+1}
\displaystyle = \frac{2}{5}\int \frac{dt}{\left(t-\frac{4}{5}\right)^2+\frac{9}{25}}
\displaystyle = \frac{2}{5}\cdot\frac{5}{3}\tan^{-1}\left(\frac{t-\frac{4}{5}}{\frac{3}{5}}\right)+C
\displaystyle = \frac{2}{3}\tan^{-1}\left(\frac{5t-4}{3}\right)+C
\displaystyle = \frac{2}{3}\tan^{-1}\left(\frac{5\tan\frac{x}{2}-4}{3}\right)+C

\displaystyle \textbf{Question 3: }~\int \frac{1}{1-2\sin x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{1-2\sin x}
\displaystyle \text{Putting } \sin x = \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{1-2\left(\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{(1+\tan^2\frac{x}{2})-4\tan\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{\tan^2\frac{x}{2}-4\tan\frac{x}{2}+1}
\displaystyle \text{Let } \tan\frac{x}{2} = t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx = dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx = 2\,dt
\displaystyle \therefore I = 2\int \frac{dt}{t^2-4t+1}
\displaystyle = 2\int \frac{dt}{t^2-4t+4-3}
\displaystyle = 2\int \frac{dt}{(t-2)^2-3}
\displaystyle = 2\int \frac{dt}{(t-2)^2-(\sqrt{3})^2}
\displaystyle = \frac{2}{2\sqrt{3}}\log\left|\frac{t-2-\sqrt{3}}{t-2+\sqrt{3}}\right| + C
\displaystyle = \frac{1}{\sqrt{3}}\log\left|\frac{\tan\frac{x}{2}-2-\sqrt{3}}{\tan\frac{x}{2}-2+\sqrt{3}}\right| + C

\displaystyle \textbf{Question 4: }~\int \frac{1}{4\cos x-1}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{4\cos x-1}
\displaystyle \text{Putting } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{4\left(\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)-1}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{4(1-\tan^2\frac{x}{2})-(1+\tan^2\frac{x}{2})}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{3-5\tan^2\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{3-5\tan^2\frac{x}{2}}
\displaystyle \text{Let } \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \therefore I = 2\int \frac{dt}{3-5t^2}
\displaystyle = \frac{2}{5}\int \frac{dt}{\frac{3}{5}-t^2}
\displaystyle = \frac{2}{5}\int \frac{dt}{\left(\frac{\sqrt{3}}{\sqrt{5}}\right)^2-t^2}
\displaystyle = \frac{2}{5}\cdot\frac{\sqrt{5}}{2\sqrt{3}}\log\left|\frac{\frac{\sqrt{3}}{\sqrt{5}}+t}{\frac{\sqrt{3}}{\sqrt{5}}-t}\right|+C
\displaystyle = \frac{1}{\sqrt{15}}\log\left|\frac{\sqrt{3}+\sqrt{5}\,t}{\sqrt{3}-\sqrt{5}\,t}\right|+C
\displaystyle = \frac{1}{\sqrt{15}}\log\left|\frac{\sqrt{3}+\sqrt{5}\tan\frac{x}{2}}{\sqrt{3}-\sqrt{5}\tan\frac{x}{2}}\right|+C

\displaystyle \textbf{Question 5: }~\int \frac{1}{1-\sin x+\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{1-\sin x+\cos x}
\displaystyle \text{Putting } \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \text{ and } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle = \int \frac{dx}{1-\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{(1+\tan^2\frac{x}{2})-2\tan\frac{x}{2}+(1-\tan^2\frac{x}{2})}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{2-2\tan\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{2-2\tan\frac{x}{2}}
\displaystyle = \frac{1}{2}\int \frac{\sec^2\frac{x}{2}\,dx}{1-\tan\frac{x}{2}}
\displaystyle \text{Let } 1-\tan\frac{x}{2}=t
\displaystyle \Rightarrow -\sec^2\frac{x}{2}\cdot\frac{1}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=-2\,dt
\displaystyle \therefore I = \frac{1}{2}\int \frac{-2\,dt}{t}
\displaystyle = -\int \frac{dt}{t}
\displaystyle = -\ln|t|+C
\displaystyle = -\ln\left|1-\tan\frac{x}{2}\right|+C

\displaystyle \textbf{Question 6: }~\int \frac{1}{3+2\sin x+\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{3+2\sin x+\cos x}
\displaystyle \text{Putting } \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \text{ and } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{3+2\left(\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)+\left(\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{3(1+\tan^2\frac{x}{2})+4\tan\frac{x}{2}+1-\tan^2\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{2\tan^2\frac{x}{2}+4\tan\frac{x}{2}+4}
\displaystyle = \frac{1}{2}\int \frac{\sec^2\frac{x}{2}\,dx}{\tan^2\frac{x}{2}+2\tan\frac{x}{2}+2}
\displaystyle \text{Let } \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \therefore I = \frac{1}{2}\int \frac{2\,dt}{t^2+2t+2}
\displaystyle = \int \frac{dt}{t^2+2t+1+1}
\displaystyle = \int \frac{dt}{(t+1)^2+1^2}
\displaystyle = \tan^{-1}(t+1)+C
\displaystyle = \tan^{-1}\left(1+\tan\frac{x}{2}\right)+C

\displaystyle \textbf{Question 7: }~\int \frac{1}{13+3\cos x+4\sin x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{13+3\cos x+4\sin x}
\displaystyle \text{Putting } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}} \text{ and } \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{13+3\left(\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)+4\left(\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{13(1+\tan^2\frac{x}{2})+3-3\tan^2\frac{x}{2}+8\tan\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{10\tan^2\frac{x}{2}+8\tan\frac{x}{2}+16}
\displaystyle \text{Let } \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \therefore I = \int \frac{2\,dt}{10t^2+8t+16}
\displaystyle = \int \frac{dt}{5t^2+4t+8}
\displaystyle = \frac{1}{5}\int \frac{dt}{t^2+\frac{4}{5}t+\frac{8}{5}}
\displaystyle = \frac{1}{5}\int \frac{dt}{t^2+\frac{4}{5}t+\left(\frac{2}{5}\right)^2-\left(\frac{2}{5}\right)^2+\frac{8}{5}}
\displaystyle = \frac{1}{5}\int \frac{dt}{\left(t+\frac{2}{5}\right)^2+\frac{36}{25}}
\displaystyle = \frac{1}{5}\cdot\frac{5}{6}\tan^{-1}\left(\frac{t+\frac{2}{5}}{\frac{6}{5}}\right)+C
\displaystyle = \frac{1}{6}\tan^{-1}\left(\frac{5t+2}{6}\right)+C
\displaystyle = \frac{1}{6}\tan^{-1}\left(\frac{5\tan\frac{x}{2}+2}{6}\right)+C

\displaystyle \textbf{Question 8: }~\int \frac{1}{\cos x-\sin x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Given } I = \int \frac{dx}{\cos x-\sin x}
\displaystyle \text{We know that } \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \text{ and } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}-\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{-2\tan\frac{x}{2}+1-\tan^2\frac{x}{2}}
\displaystyle \text{Replacing } 1+\tan^2\frac{x}{2} \text{ by } \sec^2\frac{x}{2}
\displaystyle \text{and letting } \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \therefore I = \int \frac{2\,dt}{-t^2-2t+1}
\displaystyle = -2\int \frac{dt}{t^2+2t-1}
\displaystyle = -2\int \frac{dt}{(t+1)^2-(\sqrt{2})^2}
\displaystyle = 2\int \frac{dt}{(\sqrt{2})^2-(t+1)^2}
\displaystyle \text{We know that } \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C
\displaystyle = \frac{2}{2\sqrt{2}}\log\left|\frac{\sqrt{2}+t+1}{\sqrt{2}-t-1}\right|+C
\displaystyle = \frac{1}{\sqrt{2}}\log\left|\frac{\sqrt{2}+t+1}{\sqrt{2}-t-1}\right|+C
\displaystyle = \frac{1}{\sqrt{2}}\log\left|\frac{\sqrt{2}+\tan\frac{x}{2}+1}{\sqrt{2}-\tan\frac{x}{2}-1}\right|+C

\displaystyle \textbf{Question 9: }~\int \frac{1}{\sin x+\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{\sin x+\cos x}
\displaystyle \text{Putting } \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \text{ and } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle = \int \frac{dx}{\frac{2tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{1-\tan^2\frac{x}{2}+2\tan\frac{x}{2}}
\displaystyle \text{Let } \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \therefore I = 2\int \frac{dt}{1-t^2+2t}
\displaystyle = -2\int \frac{dt}{t^2-2t-1}
\displaystyle = -2\int \frac{dt}{t^2-2t+1-2}
\displaystyle = 2\int \frac{dt}{(\sqrt{2})^2-(t-1)^2}
\displaystyle = 2\cdot\frac{1}{2\sqrt{2}}\log\left|\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}\right|+C
\displaystyle = \frac{1}{\sqrt{2}}\log\left|\frac{\sqrt{2}+\tan\frac{x}{2}-1}{\sqrt{2}-\tan\frac{x}{2}+1}\right|+C

\displaystyle \textbf{Question 10: }~\int \frac{1}{5-4\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{5-4\cos x}
\displaystyle \text{Putting } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{5-4\left(\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{5(1+\tan^2\frac{x}{2})-4(1-\tan^2\frac{x}{2})}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{9\tan^2\frac{x}{2}+1}
\displaystyle \text{Let } \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \therefore I = 2\int \frac{dt}{9t^2+1}
\displaystyle = \frac{2}{9}\int \frac{dt}{t^2+\frac{1}{9}}
\displaystyle = \frac{2}{9}\int \frac{dt}{t^2+\left(\frac{1}{3}\right)^2}
\displaystyle = \frac{2}{9}\cdot 3\tan^{-1}\left(\frac{t}{\frac{1}{3}}\right)+C
\displaystyle = \frac{2}{3}\tan^{-1}(3t)+C
\displaystyle = \frac{2}{3}\tan^{-1}\left(3\tan\frac{x}{2}\right)+C

\displaystyle \textbf{Question 11: }~\int \frac{1}{2+\sin x+\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{2+\sin x+\cos x}
\displaystyle \text{Putting } \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \text{ and } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{2+\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{2(1+\tan^2\frac{x}{2})+2\tan\frac{x}{2}+1-\tan^2\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{\tan^2\frac{x}{2}+2\tan\frac{x}{2}+3}
\displaystyle \text{Let } \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \therefore I = 2\int \frac{dt}{t^2+2t+3}
\displaystyle = 2\int \frac{dt}{(t+1)^2+2}
\displaystyle = 2\cdot\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t+1}{\sqrt{2}}\right)+C
\displaystyle = \sqrt{2}\tan^{-1}\left(\frac{\tan\frac{x}{2}+1}{\sqrt{2}}\right)+C

\displaystyle \textbf{Question 12: }~\int \frac{1}{\sin x+\sqrt{3}\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{\sin x+\sqrt{3}\cos x}
\displaystyle \text{Putting } \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \text{ and } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\sqrt{3}\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{2\tan\frac{x}{2}+\sqrt{3}-\sqrt{3}\tan^2\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{-\sqrt{3}\tan^2\frac{x}{2}+2\tan\frac{x}{2}+\sqrt{3}}
\displaystyle \text{Let } \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \therefore I = 2\int \frac{dt}{-\sqrt{3}t^2+2t+\sqrt{3}}
\displaystyle = -\frac{2}{\sqrt{3}}\int \frac{dt}{t^2-\frac{2}{\sqrt{3}}t-1}
\displaystyle = -\frac{2}{\sqrt{3}}\int \frac{dt}{t^2-\frac{2}{\sqrt{3}}t+\left(\frac{1}{\sqrt{3}}\right)^2-\left(\frac{1}{\sqrt{3}}\right)^2-1}
\displaystyle = -\frac{2}{\sqrt{3}}\int \frac{dt}{\left(t-\frac{1}{\sqrt{3}}\right)^2-\left(\frac{2}{\sqrt{3}}\right)^2}
\displaystyle = -\frac{2}{\sqrt{3}}\cdot\frac{1}{2\left(\frac{2}{\sqrt{3}}\right)}\log\left|\frac{t-\frac{1}{\sqrt{3}}-\frac{2}{\sqrt{3}}}{t-\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}}\right|+C
\displaystyle = -\frac{1}{2}\log\left|\frac{t-\sqrt{3}}{t+\frac{1}{\sqrt{3}}}\right|+C
\displaystyle = \frac{1}{2}\log\left|\frac{t+\frac{1}{\sqrt{3}}}{t-\sqrt{3}}\right|+C
\displaystyle = \frac{1}{2}\log\left|\frac{\sqrt{3}t+1}{\sqrt{3}t-3}\right|+C
\displaystyle = \frac{1}{2}\log\left|\frac{\sqrt{3}\tan\frac{x}{2}+1}{\sqrt{3}\tan\frac{x}{2}-3}\right|+C
\displaystyle = \frac{1}{2}\log\left|\frac{1+\sqrt{3}\tan\frac{x}{2}}{3-\sqrt{3}\tan\frac{x}{2}}\right|+C

\displaystyle \textbf{Question 13: }~\int \frac{1}{\sqrt{3}\sin x+\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{\sqrt{3}\sin x+\cos x}
\displaystyle \text{Putting } \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} \text{ and } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{\sqrt{3}\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{2\sqrt{3}\tan\frac{x}{2}+1-\tan^2\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{-\tan^2\frac{x}{2}+2\sqrt{3}\tan\frac{x}{2}+1}
\displaystyle \text{Let } \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \therefore I = 2\int \frac{dt}{-t^2+2\sqrt{3}t+1}
\displaystyle = -2\int \frac{dt}{t^2-2\sqrt{3}t-1}
\displaystyle = -2\int \frac{dt}{t^2-2\sqrt{3}t+3-4}
\displaystyle = -2\int \frac{dt}{(t-\sqrt{3})^2-2^2}
\displaystyle = -2\cdot\frac{1}{2\cdot 2}\log\left|\frac{t-\sqrt{3}-2}{t-\sqrt{3}+2}\right|+C
\displaystyle = -\frac{1}{2}\log\left|\frac{t-\sqrt{3}-2}{t-\sqrt{3}+2}\right|+C
\displaystyle = \frac{1}{2}\log\left|\frac{t-\sqrt{3}+2}{t-\sqrt{3}-2}\right|+C
\displaystyle = \frac{1}{2}\log\left|\frac{\tan\frac{x}{2}+2-\sqrt{3}}{\tan\frac{x}{2}-2-\sqrt{3}}\right|+C

\displaystyle \textbf{Question 14: }~\int \frac{1}{\sin x-\sqrt{3}\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Given } I = \int \frac{dx}{\sin x-\sqrt{3}\cos x}
\displaystyle \text{Let } 1=r\cos\theta \text{ and } \sqrt{3}=r\sin\theta
\displaystyle r=\sqrt{1^2+(\sqrt{3})^2}=2
\displaystyle \tan\theta=\sqrt{3}\Rightarrow \theta=\frac{\pi}{3}
\displaystyle \Rightarrow \sin x-\sqrt{3}\cos x=r(\cos\theta\sin x-\sin\theta\cos x)
\displaystyle = r\sin(x-\theta)
\displaystyle \therefore I=\int \frac{dx}{r\sin(x-\theta)}
\displaystyle =\frac{1}{r}\int \mathrm{cosec}(x-\theta)\,dx
\displaystyle \text{We know that } \int \mathrm{cosec}\,u\,du=\log\left|\tan\frac{u}{2}\right|+C
\displaystyle \therefore I=\frac{1}{2}\log\left|\tan\left(\frac{x-\theta}{2}\right)\right|+C
\displaystyle =\frac{1}{2}\log\left|\tan\left(\frac{x}{2}-\frac{\pi}{6}\right)\right|+C

\displaystyle \textbf{Question 15: }~\int \frac{1}{5+7\cos x+\sin x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{5+7\cos x+\sin x}
\displaystyle \text{Putting } \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}} \text{ and } \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}
\displaystyle \Rightarrow I = \int \frac{dx}{5+7\left(\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)+\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}}
\displaystyle = \int \frac{(1+\tan^2\frac{x}{2})\,dx}{5(1+\tan^2\frac{x}{2})+7-7\tan^2\frac{x}{2}+2\tan\frac{x}{2}}
\displaystyle = \int \frac{\sec^2\frac{x}{2}\,dx}{-2\tan^2\frac{x}{2}+2\tan\frac{x}{2}+12}
\displaystyle \text{Let } \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt
\displaystyle \Rightarrow \sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \therefore I = \int \frac{2\,dt}{-2t^2+2t+12}
\displaystyle = \int \frac{dt}{-t^2+t+6}
\displaystyle = -\int \frac{dt}{t^2-t-6}
\displaystyle = -\int \frac{dt}{t^2-t+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2-6}
\displaystyle = -\int \frac{dt}{\left(t-\frac{1}{2}\right)^2-\left(\frac{5}{2}\right)^2}
\displaystyle = \int \frac{dt}{\left(\frac{5}{2}\right)^2-\left(t-\frac{1}{2}\right)^2}
\displaystyle = \frac{1}{2\left(\frac{5}{2}\right)}\log\left|\frac{\frac{5}{2}+t-\frac{1}{2}}{\frac{5}{2}-t+\frac{1}{2}}\right|+C
\displaystyle = \frac{1}{5}\log\left|\frac{t+2}{3-t}\right|+C
\displaystyle = \frac{1}{5}\log\left|\frac{2+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C

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