\displaystyle \text{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \sin^4 x\,\cos^3 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sin^4 x \cos^3 x \, dx  
\displaystyle = \int \sin^4 x \cdot \cos^2 x \cdot \cos x \, dx  
\displaystyle = \int \sin^4 x \cdot (1 - \sin^2 x)\cos x \, dx  
\displaystyle \text{Let } \sin x = t \Rightarrow \cos x \, dx = dt  
\displaystyle = \int t^4 (1 - t^2)\, dt  
\displaystyle = \int (t^4 - t^6)\, dt  
\displaystyle = \frac{t^5}{5} - \frac{t^7}{7} + C
\displaystyle = \frac{\sin^5 x}{5} - \frac{\sin^7 x}{7} + C

\displaystyle \textbf{Question 2: }~\int \sin^5 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sin^5 x \, dx  
\displaystyle = \int \sin^4 x \cdot \sin x \, dx  
\displaystyle = \int (1 - \cos^2 x)^2 \sin x \, dx  
\displaystyle = \int (1 - 2\cos^2 x + \cos^4 x)\sin x \, dx  
\displaystyle \text{Let } \cos x = t \Rightarrow -\sin x \, dx = dt  
\displaystyle = -\int (1 - 2t^2 + t^4)\, dt  
\displaystyle = -\int (1 + t^4 - 2t^2)\, dt  
\displaystyle = -\left(t + \frac{t^5}{5} - \frac{2t^3}{3}\right) + C
\displaystyle = -t - \frac{t^5}{5} + \frac{2t^3}{3} + C
\displaystyle = -\cos x + \frac{2}{3}\cos^3 x - \frac{\cos^5 x}{5} + C

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

\displaystyle \int \cos^5 x \, dx  
\displaystyle = \int \cos^4 x \cdot \cos x \, dx  
\displaystyle = \int (1 - \sin^2 x)^2 \cos x \, dx  
\displaystyle \text{Let } \sin x = t \Rightarrow \cos x \, dx = dt  
\displaystyle = \int (1 - t^2)^2 \, dt  
\displaystyle = \int (1 - 2t^2 + t^4)\, dt  
\displaystyle = \int (1 + t^4 - 2t^2)\, dt  
\displaystyle = t + \frac{t^5}{5} - \frac{2t^3}{3} + C
\displaystyle = \sin x + \frac{\sin^5 x}{5} - \frac{2}{3}\sin^3 x + C

\displaystyle \textbf{Question 4: }~\int \sin^5 x\,\cos x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sin^5 x \cos x \, dx  
\displaystyle \text{Let } \sin x = t \Rightarrow \cos x \, dx = dt  
\displaystyle = \int t^5 \, dt  
\displaystyle = \frac{t^6}{6} + C
\displaystyle = \frac{\sin^6 x}{6} + C

\displaystyle \textbf{Question 5: }~\int \sin^3 x\,\cos^6 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sin^3 x \cos^6 x \, dx  
\displaystyle = \int \sin^2 x \cdot \cos^6 x \cdot \sin x \, dx  
\displaystyle = \int (1 - \cos^2 x)\cos^6 x \sin x \, dx  
\displaystyle \text{Let } \cos x = t \Rightarrow -\sin x \, dx = dt  
\displaystyle = -\int (1 - t^2)t^6 \, dt  
\displaystyle = \int (t^2 - 1)t^6 \, dt  
\displaystyle = \int (t^8 - t^6)\, dt  
\displaystyle = \frac{t^9}{9} - \frac{t^7}{7} + C
\displaystyle = \frac{\cos^9 x}{9} - \frac{\cos^7 x}{7} + C

\displaystyle \textbf{Question 6: }~\int \cos^7 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \cos^7 x \, dx  
\displaystyle = \int \cos^6 x \cdot \cos x \, dx  
\displaystyle = \int (\cos^2 x)^3 \cos x \, dx  
\displaystyle = \int (1 - \sin^2 x)^3 \cos x \, dx  
\displaystyle \text{Let } \sin x = t \Rightarrow \cos x \, dx = dt  
\displaystyle = \int (1 - t^2)^3 \, dt  
\displaystyle = \int (1 - 3t^2 + 3t^4 - t^6)\, dt  
\displaystyle = t - t^3 + \frac{3t^5}{5} - \frac{t^7}{7} + C
\displaystyle = \sin x - \sin^3 x + \frac{3}{5}\sin^5 x - \frac{1}{7}\sin^7 x + C

\displaystyle \textbf{Question 7: }~\int x\,\cos^3(x^2)\,\sin(x^2)\,dx.
\displaystyle \text{Answer:}

\displaystyle \int x \cos^3 x^2 \sin x^2 \, dx  
\displaystyle \text{Let } x^2 = t \Rightarrow 2x\,dx = dt \Rightarrow x\,dx = \frac{dt}{2}  
\displaystyle = \frac{1}{2}\int \cos^3 t \sin t \, dt  
\displaystyle \text{Let } \cos t = p \Rightarrow -\sin t\,dt = dp  
\displaystyle = -\frac{1}{2}\int p^3 \, dp  
\displaystyle = -\frac{1}{2}\cdot \frac{p^4}{4} + C
\displaystyle = -\frac{p^4}{8} + C
\displaystyle = -\frac{\cos^4 t}{8} + C
\displaystyle = -\frac{\cos^4 x^2}{8} + C

\displaystyle \textbf{Question 8: }~\int \sin^7 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sin^7 x \, dx  
\displaystyle = \int \sin^6 x \cdot \sin x \, dx  
\displaystyle = \int (\sin^2 x)^3 \sin x \, dx  
\displaystyle = \int (1 - \cos^2 x)^3 \sin x \, dx  
\displaystyle \text{Let } \cos x = t \Rightarrow -\sin x \, dx = dt  
\displaystyle = -\int (1 - t^2)^3 \, dt  
\displaystyle = -\int (1 - 3t^2 + 3t^4 - t^6)\, dt  
\displaystyle = -\left(t - t^3 + \frac{3t^5}{5} - \frac{t^7}{7}\right) + C
\displaystyle = -t + t^3 - \frac{3t^5}{5} + \frac{t^7}{7} + C
\displaystyle = -\cos x + \cos^3 x - \frac{3}{5}\cos^5 x + \frac{1}{7}\cos^7 x + C

\displaystyle \textbf{Question 9: }~\int \sin^3 x\,\cos^5 x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sin^3 x \cos^5 x \, dx  
\displaystyle = \int \sin^2 x \cdot \cos^5 x \cdot \sin x \, dx  
\displaystyle = \int (1 - \cos^2 x)\cos^5 x \sin x \, dx  
\displaystyle \text{Let } \cos x = t \Rightarrow -\sin x \, dx = dt  
\displaystyle = -\int (1 - t^2)t^5 \, dt  
\displaystyle = -\int (t^5 - t^7)\, dt  
\displaystyle = \int (t^7 - t^5)\, dt  
\displaystyle = \frac{t^8}{8} - \frac{t^6}{6} + C
\displaystyle = \frac{\cos^8 x}{8} - \frac{\cos^6 x}{6} + C

\displaystyle \textbf{Question 10: }~\int \frac{1}{\sin^4 x\,\cos^2 x}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{dx}{\sin^4 x \cos^2 x}  
\displaystyle = \int \frac{1}{\sin^4 x \cos^2 x}\, dx  
\displaystyle = \int \frac{\mathrm{cosec}^4 x}{\cot^2 x}\, dx  
\displaystyle = \int \frac{\mathrm{cosec}^4 x \cdot \mathrm{cosec}^2 x}{\cot^2 x}\, dx  
\displaystyle = \int \frac{(1 + \cot^2 x)^2 \mathrm{cosec}^2 x}{\cot^2 x}\, dx  
\displaystyle \text{Let } \cot x = t \Rightarrow -\mathrm{cosec}^2 x\, dx = dt  
\displaystyle = -\int \frac{(1 + t^2)^2}{t^2}\, dt  
\displaystyle = -\int \left(t^2 + 2 + \frac{1}{t^2}\right) dt  
\displaystyle = -\left(\frac{t^3}{3} + 2t - \frac{1}{t}\right) + C
\displaystyle = -\frac{t^3}{3} - 2t + \frac{1}{t} + C
\displaystyle = -\frac{1}{3}\cot^3 x - 2\cot x + \tan x + C

\displaystyle \textbf{Question 11: }~\int \frac{1}{\sin^3 x\,\cos^5 x}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{dx}{\sin^3 x \cos^5 x}  
\displaystyle = \int \frac{1}{\sin^3 x \cos^5 x}\, dx  
\displaystyle = \int \frac{\sec^8 x}{\tan^3 x}\, dx  
\displaystyle = \int \frac{\sec^6 x \cdot \sec^2 x}{\tan^3 x}\, dx  
\displaystyle = \int \frac{(1+\tan^2 x)^3 \sec^2 x}{\tan^3 x}\, dx  
\displaystyle \text{Let } \tan x = t \Rightarrow \sec^2 x\, dx = dt  
\displaystyle = \int \frac{(1+t^2)^3}{t^3}\, dt  
\displaystyle = \int \left(\frac{1}{t^3} + 3t + 3t^3 + t^5\right) dt  
\displaystyle = -\frac{1}{2}t^{-2} + \frac{3}{2}t^2 + \frac{3}{4}t^4 + \frac{t^6}{6} + C
\displaystyle = -\frac{1}{2\tan^2 x} + \frac{3}{2}\tan^2 x + \frac{3}{4}\tan^4 x + \frac{\tan^6 x}{6} + C

\displaystyle \textbf{Question 12: }~\int \frac{1}{\sin^3 x\,\cos x}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{dx}{\sin^3 x \cos x}  
\displaystyle = \int \frac{1}{\sin^3 x \cos x}\, dx  
\displaystyle = \int \frac{\mathrm{cosec}^4 x}{\cot x}\, dx  
\displaystyle = \int \frac{\mathrm{cosec}^2 x \cdot \mathrm{cosec}^2 x}{\cot x}\, dx  
\displaystyle = \int \frac{(1+\cot^2 x)\mathrm{cosec}^2 x}{\cot x}\, dx  
\displaystyle \text{Let } \cot x = t \Rightarrow -\mathrm{cosec}^2 x\, dx = dt  
\displaystyle = -\int \frac{1+t^2}{t}\, dt  
\displaystyle = -\int \left(\frac{1}{t}+t\right) dt  
\displaystyle = -\left(\log |t| + \frac{t^2}{2}\right) + C
\displaystyle = -\log |\cot x| - \frac{\cot^2 x}{2} + C
\displaystyle = \log |\tan x| - \frac{1}{2\sin^2 x} + C

\displaystyle \textbf{Question 13: }~\int \frac{1}{\sin x\,\cos^3 x}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{dx}{\sin x \cos^3 x}  
\displaystyle = \int \frac{1}{\sin x \cos^3 x}\, dx  
\displaystyle = \int \frac{\sec^4 x}{\tan x}\, dx  
\displaystyle = \int \frac{\sec^2 x \cdot \sec^2 x}{\tan x}\, dx  
\displaystyle = \int \frac{(1+\tan^2 x)\sec^2 x}{\tan x}\, dx  
\displaystyle \text{Let } \tan x = t \Rightarrow \sec^2 x\, dx = dt  
\displaystyle = \int \frac{1+t^2}{t}\, dt  
\displaystyle = \int \left(\frac{1}{t}+t\right) dt  
\displaystyle = \log |t| + \frac{t^2}{2} + C
\displaystyle = \log |\tan x| + \frac{\tan^2 x}{2} + C

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