\displaystyle \text{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \frac{1}{4\cos^2 x+9\sin^2 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{4\cos^2 x + 9\sin^2 x}
\displaystyle \text{Dividing numerator and denominator by } \cos^2 x
\displaystyle \Rightarrow I = \int \frac{dx}{\cos^2 x(4 + 9\tan^2 x)}
\displaystyle = \int \frac{\sec^2 x\,dx}{4 + 9\tan^2 x}
\displaystyle \text{Let } \tan x = t
\displaystyle \Rightarrow \sec^2 x\,dx = dt
\displaystyle \therefore I = \int \frac{dt}{4 + 9t^2}
\displaystyle = \frac{1}{9}\int \frac{dt}{\frac{4}{9} + t^2}
\displaystyle = \frac{1}{9}\int \frac{dt}{\left(\frac{2}{3}\right)^2 + t^2}
\displaystyle = \frac{1}{9}\cdot\frac{3}{2}\tan^{-1}\left(\frac{t}{\frac{2}{3}}\right) + C
\displaystyle = \frac{1}{6}\tan^{-1}\left(\frac{3t}{2}\right) + C
\displaystyle = \frac{1}{6}\tan^{-1}\left(\frac{3\tan x}{2}\right) + C

\displaystyle \textbf{Question 2: }~\int \frac{1}{4\sin^2 x+5\cos^2 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{4\sin^2 x + 5\cos^2 x}
\displaystyle \text{Dividing numerator and denominator by } \cos^2 x
\displaystyle \Rightarrow I = \int \frac{\sec^2 x}{4\tan^2 x + 5}\,dx
\displaystyle \text{Let } \tan x = t
\displaystyle \Rightarrow \sec^2 x\,dx = dt
\displaystyle \therefore I = \int \frac{dt}{4t^2 + 5}
\displaystyle = \frac{1}{4}\int \frac{dt}{t^2 + \frac{5}{4}}
\displaystyle = \frac{1}{4}\int \frac{dt}{t^2 + \left(\frac{\sqrt{5}}{2}\right)^2}
\displaystyle = \frac{1}{4}\cdot\frac{2}{\sqrt{5}}\tan^{-1}\left(\frac{t}{\frac{\sqrt{5}}{2}}\right) + C
\displaystyle = \frac{1}{2\sqrt{5}}\tan^{-1}\left(\frac{2t}{\sqrt{5}}\right) + C
\displaystyle = \frac{1}{2\sqrt{5}}\tan^{-1}\left(\frac{2\tan x}{\sqrt{5}}\right) + C

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

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

\displaystyle \textbf{Question 5: }~\int \frac{1}{1+3\sin^2 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{1+3\sin^2 x}
\displaystyle \text{Dividing numerator and denominator by } \cos^2 x
\displaystyle \Rightarrow I = \int \frac{\sec^2 x}{\sec^2 x+3\tan^2 x}\,dx
\displaystyle = \int \frac{\sec^2 x}{1+\tan^2 x+3\tan^2 x}\,dx
\displaystyle = \int \frac{\sec^2 x}{1+4\tan^2 x}\,dx
\displaystyle = \int \frac{\sec^2 x}{1+(2\tan x)^2}\,dx
\displaystyle \text{Let } 2\tan x = t
\displaystyle \Rightarrow 2\sec^2 x\,dx = dt
\displaystyle \Rightarrow \sec^2 x\,dx = \frac{dt}{2}
\displaystyle \therefore I = \frac{1}{2}\int \frac{dt}{1+t^2}
\displaystyle = \frac{1}{2}\tan^{-1}(t) + C
\displaystyle = \frac{1}{2}\tan^{-1}(2\tan x) + C

\displaystyle \textbf{Question 6: }~\int \frac{1}{3+2\cos^2 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{3+2\cos^2 x}
\displaystyle \text{Dividing numerator and denominator by } \cos^2 x
\displaystyle \Rightarrow I = \int \frac{\sec^2 x}{3\sec^2 x+2}\,dx
\displaystyle = \int \frac{\sec^2 x}{3(1+\tan^2 x)+2}\,dx
\displaystyle = \int \frac{\sec^2 x}{3\tan^2 x+5}\,dx
\displaystyle = \int \frac{\sec^2 x}{(\sqrt{5})^2+(\sqrt{3}\tan x)^2}\,dx
\displaystyle \text{Let } \sqrt{3}\tan x = t
\displaystyle \Rightarrow \sqrt{3}\sec^2 x\,dx = dt
\displaystyle \Rightarrow \sec^2 x\,dx = \frac{dt}{\sqrt{3}}
\displaystyle \therefore I = \frac{1}{\sqrt{3}}\int \frac{dt}{(\sqrt{5})^2+t^2}
\displaystyle = \frac{1}{\sqrt{3}}\cdot\frac{1}{\sqrt{5}}\tan^{-1}\left(\frac{t}{\sqrt{5}}\right)+C
\displaystyle = \frac{1}{\sqrt{15}}\tan^{-1}\left(\frac{\sqrt{3}\tan x}{\sqrt{5}}\right)+C

\displaystyle \textbf{Question 7: }~\int \frac{1}{(\sin x-2\cos x)(2\sin x+\cos x)}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{(\sin x-2\cos x)(2\sin x+\cos x)}
\displaystyle \text{Dividing numerator and denominator by } \cos^2 x
\displaystyle \Rightarrow I = \int \frac{\sec^2 x}{\left(\frac{\sin x}{\cos x}-2\right)\left(\frac{2\sin x+\cos x}{\cos x}\right)}\,dx
\displaystyle = \int \frac{\sec^2 x}{(\tan x-2)(2\tan x+1)}\,dx
\displaystyle \text{Let } \tan x = t
\displaystyle \Rightarrow \sec^2 x\,dx = dt
\displaystyle \therefore I = \int \frac{dt}{(t-2)(2t+1)}
\displaystyle = \int \frac{dt}{2t^2-3t-2}
\displaystyle = \frac{1}{2}\int \frac{dt}{t^2-\frac{3}{2}t-1}
\displaystyle = \frac{1}{2}\int \frac{dt}{t^2-\frac{3}{2}t+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^2-1}
\displaystyle = \frac{1}{2}\int \frac{dt}{\left(t-\frac{3}{4}\right)^2-\frac{25}{16}}
\displaystyle = \frac{1}{2}\int \frac{dt}{\left(t-\frac{3}{4}\right)^2-\left(\frac{5}{4}\right)^2}
\displaystyle = \frac{1}{2}\cdot\frac{1}{2\left(\frac{5}{4}\right)}\log\left|\frac{t-\frac{3}{4}-\frac{5}{4}}{t-\frac{3}{4}+\frac{5}{4}}\right| + C
\displaystyle = \frac{1}{5}\log\left|\frac{t-2}{t+\frac{1}{2}}\right| + C
\displaystyle = \frac{1}{5}\log\left|\frac{t-2}{2t+1}\right| + C
\displaystyle = \frac{1}{5}\log\left|\frac{\tan x-2}{2\tan x+1}\right| + C

\displaystyle \textbf{Question 8: }~\int \frac{\sin 2x}{\sin^4 x+\cos^4 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{\sin(2x)}{\sin^4 x + \cos^4 x}\,dx
\displaystyle = \int \frac{2\sin x\cos x}{\sin^4 x + \cos^4 x}\,dx
\displaystyle \text{Dividing numerator and denominator by } \cos^4 x
\displaystyle \Rightarrow I = \int \frac{\frac{2\sin x\cos x}{\cos^4 x}}{\tan^4 x + 1}\,dx
\displaystyle = \int \frac{2\tan x\sec^2 x}{(\tan^2 x)^2 + 1}\,dx
\displaystyle \text{Let } \tan^2 x = t
\displaystyle \Rightarrow 2\tan x\sec^2 x\,dx = dt
\displaystyle \therefore I = \int \frac{dt}{t^2 + 1}
\displaystyle = \tan^{-1}(t) + C
\displaystyle = \tan^{-1}(\tan^2 x) + C

\displaystyle \textbf{Question 9: }~\int \frac{1}{\cos x(\sin x+2\cos x)}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{\cos x(\sin x+2\cos x)}
\displaystyle \text{Dividing numerator and denominator by } \cos^2 x
\displaystyle \Rightarrow I = \int \frac{\sec^2 x}{\left(\frac{\cos x}{\cos x}\right)\left(\frac{\sin x+2\cos x}{\cos x}\right)}\,dx
\displaystyle = \int \frac{\sec^2 x}{\tan x+2}\,dx
\displaystyle \text{Let } \tan x+2 = t
\displaystyle \Rightarrow \sec^2 x\,dx = dt
\displaystyle \therefore I = \int \frac{dt}{t}
\displaystyle = \ln|t| + C
\displaystyle = \ln|\tan x+2| + C

\displaystyle \textbf{Question 10: }~\int \frac{1}{\sin^2 x+\sin 2x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{dx}{\sin^2 x + \sin(2x)}
\displaystyle = \int \frac{dx}{\sin^2 x + 2\sin x\cos x}
\displaystyle \text{Dividing numerator and denominator by } \cos^2 x
\displaystyle \Rightarrow I = \int \frac{\sec^2 x}{\tan^2 x + 2\tan x}\,dx
\displaystyle \text{Let } \tan x = t
\displaystyle \Rightarrow \sec^2 x\,dx = dt
\displaystyle \therefore I = \int \frac{dt}{t^2 + 2t}
\displaystyle = \int \frac{dt}{t^2 + 2t + 1 - 1}
\displaystyle = \int \frac{dt}{(t+1)^2 - 1^2}
\displaystyle = \frac{1}{2}\log\left|\frac{t+1-1}{t+1+1}\right| + C
\displaystyle = \frac{1}{2}\log\left|\frac{t}{t+2}\right| + C
\displaystyle = \frac{1}{2}\log\left|\frac{\tan x}{\tan x+2}\right| + C

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

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