\displaystyle \textbf{Question 1: }~\text{Evaluate each of the followin the integrals: } \int x^4\,dx.

\displaystyle \text{(i): }\int x^4\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int x^4\,dx
\displaystyle = \frac{x^{4+1}}{4+1}+C
\displaystyle = \frac{x^5}{5}+C
\\

\displaystyle \text{(ii): } \int x^{5/4}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int x^{5/4}\,dx
\displaystyle = \frac{x^{5/4+1}}{5/4+1}+C
\displaystyle = \frac{x^{9/4}}{9/4}+C
\displaystyle = \frac{4}{9}x^{9/4}+C
\\

\displaystyle \text{(iii): }\int \frac{1}{x^5}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int x^{-5}\,dx
\displaystyle = \frac{x^{-5+1}}{-5+1}+C
\displaystyle = \frac{x^{-4}}{-4}+C
\displaystyle = -\frac{1}{4x^4}+C
\\

\displaystyle \text{(iv): }\int \frac{1}{x^{3/2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int x^{-3/2}\,dx
\displaystyle = \frac{x^{-3/2+1}}{-3/2+1}+C
\displaystyle = \frac{x^{-1/2}}{-1/2}+C
\displaystyle = -2x^{-1/2}+C
\displaystyle = -\frac{2}{\sqrt{x}}+C
\\

\displaystyle \text{(v): }\int 3^x\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int 3^x\,dx
\displaystyle = \frac{3^x}{\log 3}+C
\\

\displaystyle \text{(vi): }\int \frac{1}{\sqrt[3]{x^2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int x^{-2/3}\,dx
\displaystyle = \frac{x^{-2/3+1}}{-2/3+1}+C
\displaystyle = \frac{x^{1/3}}{1/3}+C
\displaystyle = 3x^{1/3}+C
\\

\displaystyle \text{(vii): }\int 3^{2\log_3 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \left(3^{\log_3 x}\right)^2\,dx
\displaystyle = \int x^2\,dx \quad (\because~3^{\log_3 x}=x)
\displaystyle = \frac{x^{2+1}}{2+1}+C
\displaystyle = \frac{x^3}{3}+C
\\

\displaystyle \text{(viii): }\int \log_x x\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int 1\,dx \quad (\because~\log_x x=1)
\displaystyle = x+C
\\

\displaystyle \textbf{Question 2: }

\displaystyle \text{(i): }~\text{Evaluate } \int \sqrt{\frac{1+\cos 2x}{2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \sqrt{\cos^2 x}\,dx \quad \left(\because~\frac{1+\cos 2x}{2}=\cos^2 x\right)
\displaystyle = \int \cos x\,dx
\displaystyle = \sin x + C
\\

\displaystyle \text{(ii): }~\text{Evaluate } \int \sqrt{\frac{1-\cos 2x}{2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \sqrt{\sin^2 x}\,dx \quad \left(\because~\frac{1-\cos 2x}{2}=\sin^2 x\right)
\displaystyle = \int \sin x\,dx
\displaystyle = -\cos x + C
\\

\displaystyle \textbf{Question 3: }~\text{Evaluate } \int \frac{e^{6\log_e x}-e^{5\log_e x}}{e^{4\log_e x}-e^{3\log_e x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{e^{\log_e x^6}-e^{\log_e x^5}}{e^{\log_e x^4}-e^{\log_e x^3}}\,dx
\displaystyle = \int \frac{x^6-x^5}{x^4-x^3}\,dx
\displaystyle = \int \frac{x^5(x-1)}{x^3(x-1)}\,dx
\displaystyle = \int x^2\,dx
\displaystyle = \frac{x^3}{3}+C
\\

\displaystyle \textbf{Question 4: }~\text{Evaluate } \int \frac{1}{a^x\,b^x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{1}{(ab)^x}\,dx
\displaystyle = \int (ab)^{-x}\,dx
\displaystyle = \int e^{-x\log(ab)}\,dx
\displaystyle = \frac{e^{-x\log(ab)}}{-\log(ab)}+C
\displaystyle = -\frac{(ab)^{-x}}{\log(ab)}+C
\displaystyle = -\frac{1}{(ab)^x\log(ab)}+C
\\

\displaystyle \textbf{Question 5(i): }~\text{Evaluate } \int \frac{\cos 2x + 2\sin^2 x}{\sin^2 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{\cos 2x}{\sin^2 x}\,dx+\int \frac{2\sin^2 x}{\sin^2 x}\,dx
\displaystyle = \int \frac{1-2\sin^2 x}{\sin^2 x}\,dx+\int 2\,dx \quad (\because~\cos 2x=1-2\sin^2 x)
\displaystyle = \int \left(\mathrm{cosec}^2 x-2\right)\,dx+\int 2\,dx
\displaystyle = \int \mathrm{cosec}^2 x\,dx
\displaystyle = -\cot x + C
\\

\displaystyle \textbf{Question 5(ii): }~\text{Evaluate } \int \frac{2\cos^2 x - \cos 2x}{\cos^2 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{2\cos^2 x}{\cos^2 x}\,dx-\int \frac{\cos 2x}{\cos^2 x}\,dx
\displaystyle = \int 2\,dx-\int \frac{2\cos^2 x-1}{\cos^2 x}\,dx \quad (\because~\cos 2x=2\cos^2 x-1)
\displaystyle = \int 2\,dx-\int \left(2-\sec^2 x\right)\,dx
\displaystyle = \int \sec^2 x\,dx
\displaystyle = \tan x + C
\\

\displaystyle \textbf{Question 6: }~\text{Evaluate } \int \frac{e^{\log \sqrt{x}}}{x}\,dx.
\displaystyle \text{Answer:}
\displaystyle = \int \frac{e^{\log(\sqrt{x})}}{x}\,dx
\displaystyle = \int \frac{\sqrt{x}}{x}\,dx \quad (\because~e^{\log t}=t)
\displaystyle = \int x^{-1/2}\,dx
\displaystyle = \frac{x^{-1/2+1}}{-1/2+1}+C
\displaystyle = \frac{x^{1/2}}{1/2}+C
\displaystyle = 2\sqrt{x}+C
\\

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