\displaystyle \textbf{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \frac{1}{1-\cot x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{1-\cot x}
\displaystyle = \int \frac{dx}{1-\frac{\cos x}{\sin x}}
\displaystyle = \int \frac{\sin x}{\sin x-\cos x}\,dx
\displaystyle = \frac{1}{2}\int \frac{2\sin x}{\sin x-\cos x}\,dx
\displaystyle = \frac{1}{2}\int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{\sin x-\cos x}\,dx
\displaystyle = \frac{1}{2}\int \frac{\sin x+\cos x}{\sin x-\cos x}\,dx + \frac{1}{2}\int dx
\displaystyle \text{Putting } \sin x-\cos x=t
\displaystyle \Rightarrow (\cos x+\sin x)\,dx=dt
\displaystyle \therefore I=\frac{1}{2}\int \frac{dt}{t}+\frac{1}{2}\int dx
\displaystyle =\frac{1}{2}\ln|t|+\frac{x}{2}+C
\displaystyle =\frac{x}{2}+\frac{1}{2}\ln|\sin x-\cos x|+C

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

\displaystyle \textbf{Question 3: }~\int \frac{3+2\cos x+4\sin x}{2\sin x+\cos x+3}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{3+2\cos x+4\sin x}{2\sin x+\cos x+3}\,dx
\displaystyle \text{Let } 3+2\cos x+4\sin x=A(2\sin x+\cos x+3)+B(2\cos x-\sin x)+C
\displaystyle \Rightarrow 3+2\cos x+4\sin x=(2A-B)\sin x+(A+2B)\cos x+3A+C
\displaystyle \text{Comparing coefficients of like terms}
\displaystyle 2A-B=4 \quad (1)
\displaystyle A+2B=2 \quad (2)
\displaystyle 3A+C=3 \quad (3)
\displaystyle \text{Multiplying (1) by }2 \text{ and adding to (2)}
\displaystyle 4A-2B+A+2B=8+2
\displaystyle 5A=10
\displaystyle \Rightarrow A=2
\displaystyle \text{Putting } A=2 \text{ in (1)}
\displaystyle 2\times2-B=4
\displaystyle \Rightarrow B=0
\displaystyle \text{Putting } A=2 \text{ in (3)}
\displaystyle 3\times2+C=3
\displaystyle \Rightarrow C=-3
\displaystyle \therefore I=\int \frac{2(2\sin x+\cos x+3)-3}{2\sin x+\cos x+3}\,dx
\displaystyle =2\int dx-3\int \frac{dx}{2\sin x+\cos x+3}
\displaystyle \text{Substituting } \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 =2\int dx-3\int \frac{dx}{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)+3}
\displaystyle =2\int dx-3\int \frac{(1+\tan^2\frac{x}{2})\,dx}{4\tan\frac{x}{2}+1-\tan^2\frac{x}{2}+3(1+\tan^2\frac{x}{2})}
\displaystyle =2\int dx-3\int \frac{\sec^2\frac{x}{2}\,dx}{2\tan^2\frac{x}{2}+4\tan\frac{x}{2}+4}
\displaystyle =2\int dx-\frac{3}{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=2\int dx-\frac{3}{2}\int \frac{2\,dt}{t^2+2t+2}
\displaystyle =2\int dx-3\int \frac{dt}{(t+1)^2+1}
\displaystyle =2x-3\tan^{-1}(t+1)+C
\displaystyle =2x-3\tan^{-1}\left(\tan\frac{x}{2}+1\right)+C

\displaystyle \textbf{Question 4: }~\int \frac{1}{p+q\tan x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{dx}{p+q\tan x}
\displaystyle =\int \frac{dx}{p+\frac{q\sin x}{\cos x}}
\displaystyle =\int \frac{\cos x}{q\sin x+p\cos x}\,dx
\displaystyle \text{Let } \cos x=A(q\sin x+p\cos x)+B(q\cos x-p\sin x)
\displaystyle \Rightarrow \cos x=(Ap+Bq)\cos x+(Aq-Bp)\sin x
\displaystyle \text{Comparing coefficients of like terms}
\displaystyle Ap+Bq=1 \quad (1)
\displaystyle Aq-Bp=0 \quad (2)
\displaystyle \text{Multiplying (1) by }p\text{ and (2) by }q
\displaystyle Ap^2+Bpq=p
\displaystyle Aq^2-Bpq=0
\displaystyle \Rightarrow A(p^2+q^2)=p
\displaystyle \Rightarrow A=\frac{p}{p^2+q^2}
\displaystyle \text{Putting value of }A\text{ in (1)}
\displaystyle \frac{p^2}{p^2+q^2}+Bq=1
\displaystyle \Rightarrow Bq=\frac{q^2}{p^2+q^2}
\displaystyle \Rightarrow B=\frac{q}{p^2+q^2}
\displaystyle \therefore I=\int\left[\frac{p}{p^2+q^2}\frac{q\sin x+p\cos x}{q\sin x+p\cos x}+\frac{q}{p^2+q^2}\frac{q\cos x-p\sin x}{q\sin x+p\cos x}\right]dx
\displaystyle =\frac{p}{p^2+q^2}\int dx+\frac{q}{p^2+q^2}\int \frac{q\cos x-p\sin x}{q\sin x+p\cos x}\,dx
\displaystyle \text{Putting } q\sin x+p\cos x=t
\displaystyle \Rightarrow (q\cos x-p\sin x)\,dx=dt
\displaystyle \therefore I=\frac{p}{p^2+q^2}x+\frac{q}{p^2+q^2}\int \frac{dt}{t}
\displaystyle =\frac{p}{p^2+q^2}x+\frac{q}{p^2+q^2}\log|t|+C
\displaystyle =\frac{p}{p^2+q^2}x+\frac{q}{p^2+q^2}\log|q\sin x+p\cos x|+C

\displaystyle \textbf{Question 5: }~\int \frac{5\cos x+6}{2\cos x+\sin x+3}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{5\cos x+6}{2\cos x+\sin x+3}\,dx
\displaystyle \text{Let } 5\cos x+6=A(2\cos x+\sin x+3)+B(-2\sin x+\cos x)+C
\displaystyle \Rightarrow 5\cos x+6=(A-2B)\sin x+(2A+B)\cos x+3A+C
\displaystyle \text{Comparing coefficients of like terms}
\displaystyle A-2B=0 \quad (1)
\displaystyle 2A+B=5 \quad (2)
\displaystyle 3A+C=6 \quad (3)
\displaystyle \text{Multiplying (2) by }2\text{ and adding to (1)}
\displaystyle 4A+2B+A-2B=10
\displaystyle \Rightarrow 5A=10
\displaystyle \Rightarrow A=2
\displaystyle \text{Putting } A=2 \text{ in (1)}
\displaystyle 2-2B=0
\displaystyle \Rightarrow B=1
\displaystyle \text{Putting } A=2 \text{ in (3)}
\displaystyle 6+C=6
\displaystyle \Rightarrow C=0
\displaystyle \therefore I=\int \frac{2(2\cos x+\sin x+3)+(-2\sin x+\cos x)}{2\cos x+\sin x+3}\,dx
\displaystyle =2\int dx+\int \frac{-2\sin x+\cos x}{2\cos x+\sin x+3}\,dx
\displaystyle \text{Putting } 2\cos x+\sin x+3=t
\displaystyle \Rightarrow (-2\sin x+\cos x)\,dx=dt
\displaystyle \therefore I=2\int dx+\int \frac{dt}{t}
\displaystyle =2x+\ln|t|+C
\displaystyle =2x+\ln|2\cos x+\sin x+3|+C

\displaystyle \textbf{Question 6: }~\int \frac{2\sin x+3\cos x}{3\sin x+4\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{2\sin x+3\cos x}{3\sin x+4\cos x}\,dx
\displaystyle \text{Let } 2\sin x+3\cos x=A(3\sin x+4\cos x)+B(3\cos x-4\sin x)
\displaystyle \Rightarrow 2\sin x+3\cos x=(3A-4B)\sin x+(4A+3B)\cos x
\displaystyle \text{Comparing coefficients of like terms}
\displaystyle 3A-4B=2 \quad (1)
\displaystyle 4A+3B=3 \quad (2)
\displaystyle \text{Multiplying (1) by }3\text{ and (2) by }4\text{ and adding}
\displaystyle 9A-12B+16A+12B=6+12
\displaystyle 25A=18
\displaystyle \Rightarrow A=\frac{18}{25}
\displaystyle \text{Putting } A=\frac{18}{25} \text{ in (1)}
\displaystyle 3\times\frac{18}{25}-4B=2
\displaystyle \Rightarrow \frac{54}{25}-2=4B
\displaystyle \Rightarrow \frac{4}{25}=4B
\displaystyle \Rightarrow B=\frac{1}{25}
\displaystyle \therefore I=\int \frac{\frac{18}{25}(3\sin x+4\cos x)+\frac{1}{25}(3\cos x-4\sin x)}{3\sin x+4\cos x}\,dx
\displaystyle =\frac{18}{25}\int dx+\frac{1}{25}\int \frac{3\cos x-4\sin x}{3\sin x+4\cos x}\,dx
\displaystyle \text{Putting } 3\sin x+4\cos x=t
\displaystyle \Rightarrow (3\cos x-4\sin x)\,dx=dt
\displaystyle \therefore I=\frac{18}{25}\int dx+\frac{1}{25}\int \frac{dt}{t}
\displaystyle =\frac{18x}{25}+\frac{1}{25}\ln|t|+C
\displaystyle =\frac{18x}{25}+\frac{1}{25}\ln|3\sin x+4\cos x|+C

\displaystyle \textbf{Question 7: }~\int \frac{1}{3+4\cot x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{1}{3+4\cot x}\,dx
\displaystyle =\int \frac{1}{3+\frac{4\cos x}{\sin x}}\,dx
\displaystyle =\int \frac{\sin x}{3\sin x+4\cos x}\,dx
\displaystyle \text{Let } \sin x=A(3\sin x+4\cos x)+B(3\cos x-4\sin x)
\displaystyle \Rightarrow \sin x=(3A-4B)\sin x+(4A+3B)\cos x
\displaystyle \text{Comparing coefficients of like terms}
\displaystyle 3A-4B=1 \quad (1)
\displaystyle 4A+3B=0 \quad (2)
\displaystyle \text{Multiplying (1) by }3\text{ and (2) by }4\text{ and adding}
\displaystyle 9A-12B+16A+12B=3
\displaystyle 25A=3
\displaystyle \Rightarrow A=\frac{3}{25}
\displaystyle \text{Putting } A=\frac{3}{25} \text{ in (2)}
\displaystyle 4\times\frac{3}{25}+3B=0
\displaystyle \Rightarrow 3B=-\frac{12}{25}
\displaystyle \Rightarrow B=-\frac{4}{25}
\displaystyle \therefore I=\int \frac{\frac{3}{25}(3\sin x+4\cos x)-\frac{4}{25}(3\cos x-4\sin x)}{3\sin x+4\cos x}\,dx
\displaystyle =\frac{3}{25}\int dx-\frac{4}{25}\int \frac{3\cos x-4\sin x}{3\sin x+4\cos x}\,dx
\displaystyle \text{Putting } 3\sin x+4\cos x=t
\displaystyle \Rightarrow (3\cos x-4\sin x)\,dx=dt
\displaystyle \therefore I=\frac{3}{25}\int dx-\frac{4}{25}\int \frac{dt}{t}
\displaystyle =\frac{3x}{25}-\frac{4}{25}\ln|t|+C
\displaystyle =\frac{3x}{25}-\frac{4}{25}\ln|3\sin x+4\cos x|+C

\displaystyle \textbf{Question 8: }~\int \frac{2\tan x+3}{3\tan x+4}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{2\tan x+3}{3\tan x+4}\,dx
\displaystyle =\int \frac{\frac{2\sin x}{\cos x}+3}{\frac{3\sin x}{\cos x}+4}\,dx
\displaystyle =\int \frac{2\sin x+3\cos x}{3\sin x+4\cos x}\,dx
\displaystyle \text{Let } 2\sin x+3\cos x=A(3\sin x+4\cos x)+B(3\cos x-4\sin x)
\displaystyle \Rightarrow 2\sin x+3\cos x=(3A-4B)\sin x+(4A+3B)\cos x
\displaystyle \text{Equating coefficients of like terms}
\displaystyle 3A-4B=2 \quad (1)
\displaystyle 4A+3B=3 \quad (2)
\displaystyle \text{Multiplying (1) by }3\text{ and (2) by }4\text{ and adding}
\displaystyle 9A-12B+16A+12B=6+12
\displaystyle 25A=18
\displaystyle \Rightarrow A=\frac{18}{25}
\displaystyle \text{Putting } A=\frac{18}{25} \text{ in (1)}
\displaystyle 3\times\frac{18}{25}-4B=2
\displaystyle \Rightarrow \frac{54}{25}-2=4B
\displaystyle \Rightarrow \frac{4}{25}=4B
\displaystyle \Rightarrow B=\frac{1}{25}
\displaystyle \therefore I=\int \frac{\frac{18}{25}(3\sin x+4\cos x)+\frac{1}{25}(3\cos x-4\sin x)}{3\sin x+4\cos x}\,dx
\displaystyle =\frac{18}{25}\int dx+\frac{1}{25}\int \frac{3\cos x-4\sin x}{3\sin x+4\cos x}\,dx
\displaystyle \text{Putting } 3\sin x+4\cos x=t
\displaystyle \Rightarrow (3\cos x-4\sin x)\,dx=dt
\displaystyle \therefore I=\frac{18}{25}\int dx+\frac{1}{25}\int \frac{dt}{t}
\displaystyle =\frac{18x}{25}+\frac{1}{25}\ln|t|+C
\displaystyle =\frac{18x}{25}+\frac{1}{25}\ln|3\sin x+4\cos x|+C

\displaystyle \textbf{Question 9: }~\int \frac{1}{4+3\tan x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{dx}{4+3\tan x}
\displaystyle =\int \frac{dx}{4+\frac{3\sin x}{\cos x}}
\displaystyle =\int \frac{\cos x}{4\cos x+3\sin x}\,dx
\displaystyle \text{Consider, } \cos x=A(4\cos x+3\sin x)+B\frac{d}{dx}(4\cos x+3\sin x)
\displaystyle \Rightarrow \cos x=A(4\cos x+3\sin x)+B(-4\sin x+3\cos x)
\displaystyle \Rightarrow \cos x=(4A+3B)\cos x+(3A-4B)\sin x
\displaystyle \text{Equating coefficients of like terms}
\displaystyle 4A+3B=1 \quad (1)
\displaystyle 3A-4B=0 \quad (2)
\displaystyle \text{Solving (1) and (2), we get}
\displaystyle A=\frac{4}{25},\; B=\frac{3}{25}
\displaystyle \therefore I=\int \frac{\frac{4}{25}(4\cos x+3\sin x)+\frac{3}{25}(-4\sin x+3\cos x)}{4\cos x+3\sin x}\,dx
\displaystyle =\frac{4}{25}\int dx+\frac{3}{25}\int \frac{-4\sin x+3\cos x}{4\cos x+3\sin x}\,dx
\displaystyle \text{Let } 4\cos x+3\sin x=t
\displaystyle \Rightarrow (-4\sin x+3\cos x)\,dx=dt
\displaystyle \therefore I=\frac{4}{25}\int dx+\frac{3}{25}\int \frac{dt}{t}
\displaystyle =\frac{4x}{25}+\frac{3}{25}\ln|t|+C
\displaystyle =\frac{4x}{25}+\frac{3}{25}\ln|4\cos x+3\sin x|+C

\displaystyle \textbf{Question 10: }~\int \frac{8\cot x+1}{3\cot x+2}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{8\cot x+1}{3\cot x+2}\,dx
\displaystyle =\int \frac{\frac{8\cos x}{\sin x}+1}{\frac{3\cos x}{\sin x}+2}\,dx
\displaystyle =\int \frac{8\cos x+\sin x}{3\cos x+2\sin x}\,dx
\displaystyle \text{Now, let } 8\cos x+\sin x=A(3\cos x+2\sin x)+B(-3\sin x+2\cos x)
\displaystyle \Rightarrow 8\cos x+\sin x=3A\cos x+2A\sin x-3B\sin x+2B\cos x
\displaystyle \Rightarrow 8\cos x+\sin x=(3A+2B)\cos x+(2A-3B)\sin x
\displaystyle \text{Equating coefficients of like terms}
\displaystyle 2A-3B=1 \quad (1)
\displaystyle 3A+2B=8 \quad (2)
\displaystyle \text{Solving (1) and (2), we get}
\displaystyle A=2,\; B=1
\displaystyle \therefore I=\int \frac{2(3\cos x+2\sin x)+1(-3\sin x+2\cos x)}{3\cos x+2\sin x}\,dx
\displaystyle =2\int \frac{3\cos x+2\sin x}{3\cos x+2\sin x}\,dx+\int \frac{-3\sin x+2\cos x}{3\cos x+2\sin x}\,dx
\displaystyle =2\int dx+\int \frac{-3\sin x+2\cos x}{3\cos x+2\sin x}\,dx
\displaystyle \text{Putting } 3\cos x+2\sin x=t
\displaystyle \Rightarrow (-3\sin x+2\cos x)\,dx=dt
\displaystyle \therefore I=2\int dx+\int \frac{dt}{t}
\displaystyle =2x+\ln|t|+C
\displaystyle =2x+\ln|3\cos x+2\sin x|+C

\displaystyle \textbf{Question 11: }~\int \frac{4\sin x+5\cos x}{5\sin x+4\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{4\sin x+5\cos x}{5\sin x+4\cos x}\,dx
\displaystyle \text{Let } 4\sin x+5\cos x=A(5\sin x+4\cos x)+B(5\cos x-4\sin x)
\displaystyle \Rightarrow 4\sin x+5\cos x=(5A-4B)\sin x+(4A+5B)\cos x
\displaystyle \text{By equating the coefficients of like terms}
\displaystyle 5A-4B=4 \quad (1)
\displaystyle 4A+5B=5 \quad (2)
\displaystyle \text{Multiplying (1) by }5\text{ and (2) by }4\text{ and adding}
\displaystyle 25A-20B+16A+20B=20+20
\displaystyle 41A=40
\displaystyle \Rightarrow A=\frac{40}{41}
\displaystyle \text{Putting } A=\frac{40}{41} \text{ in (2)}
\displaystyle \frac{160}{41}+5B=5
\displaystyle \Rightarrow 5B=\frac{45}{41}
\displaystyle \Rightarrow B=\frac{9}{41}
\displaystyle \therefore I=\int \frac{\frac{40}{41}(5\sin x+4\cos x)+\frac{9}{41}(5\cos x-4\sin x)}{5\sin x+4\cos x}\,dx
\displaystyle =\frac{40}{41}\int dx+\frac{9}{41}\int \frac{5\cos x-4\sin x}{5\sin x+4\cos x}\,dx
\displaystyle \text{Putting } 5\sin x+4\cos x=t
\displaystyle \Rightarrow (5\cos x-4\sin x)\,dx=dt
\displaystyle \therefore I=\frac{40}{41}\int dx+\frac{9}{41}\int \frac{dt}{t}
\displaystyle =\frac{40x}{41}+\frac{9}{41}\ln|t|+C
\displaystyle =\frac{40x}{41}+\frac{9}{41}\ln|5\sin x+4\cos x|+C

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