\displaystyle \text{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \tan^3 x~\sec^2 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \tan^3 x\,\sec^2 x\,dx

\displaystyle \text{Let } \tan x=t

\displaystyle \Rightarrow \sec^2 x\,dx=dt

\displaystyle \int \tan^3 x\,\sec^2 x\,dx=\int t^3\,dt

\displaystyle =\frac{t^4}{4}+C

\displaystyle =\frac{\tan^4 x}{4}+C

\displaystyle \textbf{Question 2: }~\int \tan x~\sec^4 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \tan x\,\sec^4 x\,dx

\displaystyle =\int \tan x\,\sec^2 x\,\sec^2 x\,dx

\displaystyle =\int \tan x\,(1+\tan^2 x)\,\sec^2 x\,dx

\displaystyle \text{Let } \tan x=t

\displaystyle \Rightarrow \sec^2 x\,dx=dt

\displaystyle \int \tan x\,(1+\tan^2 x)\,\sec^2 x\,dx=\int t(1+t^2)\,dt

\displaystyle =\int (t+t^3)\,dt

\displaystyle =\frac{t^2}{2}+\frac{t^4}{4}+C

\displaystyle =\frac{1}{2}\tan^2 x+\frac{1}{4}\tan^4 x+C

\displaystyle \textbf{Question 3: }~\int \tan^5 x~\sec^4 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \tan^5 x\,\sec^4 x\,dx

\displaystyle =\int \tan^5 x\,\sec^2 x\,\sec^2 x\,dx

\displaystyle =\int \tan^5 x\,(1+\tan^2 x)\,\sec^2 x\,dx

\displaystyle \text{Let } \tan x=t

\displaystyle \Rightarrow \sec^2 x\,dx=dt

\displaystyle \int \tan^5 x\,(1+\tan^2 x)\,\sec^2 x\,dx=\int t^5(1+t^2)\,dt

\displaystyle =\int (t^5+t^7)\,dt

\displaystyle =\frac{t^6}{6}+\frac{t^8}{8}+C

\displaystyle =\frac{\tan^6 x}{6}+\frac{\tan^8 x}{8}+C

\displaystyle \textbf{Question 4: }~\int \sec^6 x~\tan x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sec^6 x\,\tan x\,dx

\displaystyle =\int \sec^5 x\,\sec x\,\tan x\,dx

\displaystyle \text{Let } \sec x=t

\displaystyle \Rightarrow \sec x\,\tan x\,dx=dt

\displaystyle \int \sec^6 x\,\tan x\,dx=\int t^5\,dt

\displaystyle =\frac{t^6}{6}+C

\displaystyle =\frac{\sec^6 x}{6}+C

\displaystyle \textbf{Question 5: }~\int \tan^5 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \tan^5 x\,dx

\displaystyle =\int \tan^4 x\,\tan x\,dx

\displaystyle =\int (\sec^2 x-1)^2\,\tan x\,dx

\displaystyle =\int (\sec^4 x-2\sec^2 x+1)\,\tan x\,dx

\displaystyle =\int \sec^4 x\,\tan x\,dx-2\int \sec^2 x\,\tan x\,dx+\int \tan x\,dx

\displaystyle \text{Let } I_1=\int \sec^4 x\,\tan x\,dx

\displaystyle \text{and } I_2=\int \sec^2 x\,\tan x\,dx

\displaystyle \int \tan^5 x\,dx=I_1-2I_2+\int \tan x\,dx

\displaystyle \text{Now, } I_1=\int \sec^4 x\,\tan x\,dx

\displaystyle \text{Let } \tan x=t

\displaystyle \Rightarrow \sec^2 x\,dx=dt

\displaystyle I_1=\int (1+t^2)\,t\,dt

\displaystyle =\int (t+t^3)\,dt

\displaystyle =\frac{t^2}{2}+\frac{t^4}{4}+C_1

\displaystyle =\frac{\tan^2 x}{2}+\frac{\tan^4 x}{4}+C_1

\displaystyle I_2=\int \sec^2 x\,\tan x\,dx

\displaystyle \text{Let } \tan x=t

\displaystyle \Rightarrow \sec^2 x\,dx=dt

\displaystyle I_2=\int t\,dt=\frac{t^2}{2}+C_2

\displaystyle =\frac{\tan^2 x}{2}+C_2

\displaystyle \int \tan^5 x\,dx=\left(\frac{\tan^2 x}{2}+\frac{\tan^4 x}{4}\right)-2\left(\frac{\tan^2 x}{2}\right)+\log(\sec x)+C

\displaystyle =\frac{\tan^4 x}{4}-\frac{\tan^2 x}{2}+\log(\sec x)+C

\displaystyle \textbf{Question 6: }~\int \sqrt{\tan x}\,\sec^4 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sqrt{\tan x}\,\sec^4 x\,dx

\displaystyle =\int \sqrt{\tan x}\,\sec^2 x\,\sec^2 x\,dx

\displaystyle =\int \sqrt{\tan x}\,(1+\tan^2 x)\,\sec^2 x\,dx

\displaystyle \text{Let } \tan x=t

\displaystyle \Rightarrow \sec^2 x\,dx=dt

\displaystyle \int \sqrt{\tan x}\,(1+\tan^2 x)\,\sec^2 x\,dx=\int \sqrt{t}\,(1+t^2)\,dt

\displaystyle =\int (t^{\tfrac12}+t^{\tfrac52})\,dt

\displaystyle =\frac{2}{3}t^{\tfrac32}+\frac{2}{7}t^{\tfrac72}+C

\displaystyle =\frac{2}{3}\tan^{\tfrac32}x+\frac{2}{7}\tan^{\tfrac72}x+C

\displaystyle \textbf{Question 7: }~\int \sec^4 2x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sec^4 2x\,dx

\displaystyle =\int \sec^2 2x\,\sec^2 2x\,dx

\displaystyle =\int (1+\tan^2 2x)\,\sec^2 2x\,dx

\displaystyle \text{Let } \tan 2x=t

\displaystyle \Rightarrow 2\sec^2 2x\,dx=dt

\displaystyle \Rightarrow \sec^2 2x\,dx=\frac{dt}{2}

\displaystyle \int (1+\tan^2 2x)\,\sec^2 2x\,dx=\frac{1}{2}\int (1+t^2)\,dt

\displaystyle =\frac{1}{2}\left(t+\frac{t^3}{3}\right)+C

\displaystyle =\frac{t}{2}+\frac{t^3}{6}+C

\displaystyle =\frac{\tan(2x)}{2}+\frac{\tan^3(2x)}{6}+C

\displaystyle \textbf{Question 8: }~\int \mathrm{cosec}^4 3x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \mathrm{cosec}^4 3x\,dx

\displaystyle =\int \mathrm{cosec}^2 3x\cdot \mathrm{cosec}^2 3x\,dx

\displaystyle =\int (1+\cot^2 3x)\,\mathrm{cosec}^2 3x\,dx

\displaystyle \text{Let } \cot 3x=t

\displaystyle \Rightarrow -3\,\mathrm{cosec}^2 3x\,dx=dt

\displaystyle \Rightarrow \mathrm{cosec}^2 3x\,dx=-\frac{dt}{3}

\displaystyle \int (1+\cot^2 3x)\,\mathrm{cosec}^2 3x\,dx=-\frac{1}{3}\int (1+t^2)\,dt

\displaystyle =-\frac{1}{3}\left(t+\frac{t^3}{3}\right)+C

\displaystyle =-\frac{t}{3}-\frac{t^3}{9}+C

\displaystyle =-\frac{\cot 3x}{3}-\frac{\cot^3 3x}{9}+C

\displaystyle \textbf{Question 9: }~\int \cot^n x~\mathrm{cosec}^2 x~dx,\;n\neq -1.
\displaystyle \text{Answer:}

\displaystyle \int \cot^n x\,\mathrm{cosec}^2 x\,dx

\displaystyle \text{Let } \cot x=t

\displaystyle \Rightarrow -\mathrm{cosec}^2 x\,dx=dt

\displaystyle \Rightarrow \mathrm{cosec}^2 x\,dx=-dt

\displaystyle \int \cot^n x\,\mathrm{cosec}^2 x\,dx=-\int t^n\,dt

\displaystyle =-\frac{t^{n+1}}{n+1}+C

\displaystyle =-\frac{\cot^{\,n+1}x}{n+1}+C

\displaystyle \textbf{Question 10: }~\int \cot^5 x~\mathrm{cosec}^4 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \cot^5 x\,\mathrm{cosec}^4 x\,dx

\displaystyle =\int \cot^5 x\,\mathrm{cosec}^2 x\,\mathrm{cosec}^2 x\,dx

\displaystyle =\int \cot^5 x\,(1+\cot^2 x)\,\mathrm{cosec}^2 x\,dx

\displaystyle \text{Let } \cot x=t

\displaystyle \Rightarrow -\mathrm{cosec}^2 x\,dx=dt

\displaystyle \Rightarrow \mathrm{cosec}^2 x\,dx=-dt

\displaystyle \int \cot^5 x\,(1+\cot^2 x)\,\mathrm{cosec}^2 x\,dx=-\int t^5(1+t^2)\,dt

\displaystyle =-\int (t^5+t^7)\,dt

\displaystyle =-\left(\frac{t^6}{6}+\frac{t^8}{8}\right)+C

\displaystyle =-\left(\frac{\cot^6 x}{6}+\frac{\cot^8 x}{8}\right)+C

\displaystyle \textbf{Question 11: }~\int \cot^5 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \cot^5 x\,dx

\displaystyle =\int \cot^4 x\cdot \cot x\,dx

\displaystyle =\int (\mathrm{cosec}^2 x-1)^2\cot x\,dx

\displaystyle =\int (\mathrm{cosec}^4 x-2\mathrm{cosec}^2 x+1)\cot x\,dx

\displaystyle =\int \mathrm{cosec}^4 x\cot x\,dx-2\int \mathrm{cosec}^2 x\cot x\,dx+\int \cot x\,dx

\displaystyle =\int (1+\cot^2 x)\,\mathrm{cosec}^2 x\cdot \cot x\,dx-2\int \cot x\,\mathrm{cosec}^2 x\,dx+\int \cot x\,dx

\displaystyle =\int (\cot x+\cot^3 x)\,\mathrm{cosec}^2 x\,dx-2\int \cot x\,\mathrm{cosec}^2 x\,dx+\int \cot x\,dx

\displaystyle \text{Let } I_1=\int (\cot x+\cot^3 x)\,\mathrm{cosec}^2 x\,dx

\displaystyle \text{and } I_2=\int \cot x\,dx

\displaystyle \int \cot^5 x\,dx=I_1-2\int \cot x\,\mathrm{cosec}^2 x\,dx+I_2

\displaystyle \text{Now, } I_1=\int (\cot x+\cot^3 x)\,\mathrm{cosec}^2 x\,dx

\displaystyle \text{Let } \cot x=t

\displaystyle \Rightarrow -\mathrm{cosec}^2 x\,dx=dt

\displaystyle I_1=\int (t+t^3)(-dt)=-\int (t+t^3)\,dt

\displaystyle =-\left(\frac{t^2}{2}+\frac{t^4}{4}\right)+C_1

\displaystyle =-\frac{\cot^2 x}{2}-\frac{\cot^4 x}{4}+C_1

\displaystyle \int \cot x\,\mathrm{cosec}^2 x\,dx

\displaystyle \text{Let } \cot x=t

\displaystyle \Rightarrow -\mathrm{cosec}^2 x\,dx=dt

\displaystyle \int \cot x\,\mathrm{cosec}^2 x\,dx=-\int t\,dt=-\frac{t^2}{2}+C_2

\displaystyle =-\frac{\cot^2 x}{2}+C_2

\displaystyle I_2=\int \cot x\,dx=\log|\sin x|+C_3

\displaystyle \int \cot^5 x\,dx=\left(-\frac{\cot^2 x}{2}-\frac{\cot^4 x}{4}\right)-2\left(-\frac{\cot^2 x}{2}\right)+\log|\sin x|+C

\displaystyle =-\frac{\cot^4 x}{4}+\frac{\cot^2 x}{2}+\log|\sin x|+C

\displaystyle \textbf{Question 12: }~\int \cot^6 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \cot^6 x\,dx

\displaystyle =\int \cot^4 x(\mathrm{cosec}^2 x-1)\,dx

\displaystyle =\int \cot^4 x\,\mathrm{cosec}^2 x\,dx-\int \cot^4 x\,dx

\displaystyle =\int \cot^4 x\,\mathrm{cosec}^2 x\,dx-\int (\mathrm{cosec}^2 x-1)\cot^2 x\,dx

\displaystyle =\int \cot^4 x\,\mathrm{cosec}^2 x\,dx-\int \cot^2 x\,\mathrm{cosec}^2 x\,dx+\int \cot^2 x\,dx

\displaystyle =\int \cot^4 x\,\mathrm{cosec}^2 x\,dx-\int \cot^2 x\,\mathrm{cosec}^2 x\,dx+\int (\mathrm{cosec}^2 x-1)\,dx

\displaystyle \text{Let } I_1=\int \cot^4 x\,\mathrm{cosec}^2 x\,dx-\int \cot^2 x\,\mathrm{cosec}^2 x\,dx

\displaystyle \text{and } I_2=\int (\mathrm{cosec}^2 x-1)\,dx

\displaystyle \int \cot^6 x\,dx=I_1+I_2

\displaystyle \text{First we integrate } I_1

\displaystyle I_1=\int \cot^4 x\,\mathrm{cosec}^2 x\,dx-\int \cot^2 x\,\mathrm{cosec}^2 x\,dx

\displaystyle \text{Let } \cot x=t

\displaystyle \Rightarrow -\mathrm{cosec}^2 x\,dx=dt

\displaystyle \Rightarrow \mathrm{cosec}^2 x\,dx=-dt

\displaystyle I_1=\int t^4(-dt)-\int t^2(-dt)

\displaystyle =-\int t^4\,dt+\int t^2\,dt

\displaystyle =-\frac{t^5}{5}+\frac{t^3}{3}+C_1

\displaystyle =-\frac{\cot^5 x}{5}+\frac{\cot^3 x}{3}+C_1

\displaystyle \text{Now we integrate } I_2

\displaystyle I_2=\int (\mathrm{cosec}^2 x-1)\,dx

\displaystyle =-\cot x-x+C_2

\displaystyle \int \cot^6 x\,dx=-\frac{\cot^5 x}{5}+\frac{\cot^3 x}{3}-\cot x-x+C

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