\displaystyle \text{Evaluate the following integrals:}
\displaystyle \textbf{Question 1: }~\int \frac{x}{\sqrt{x^4+a^4}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{x\,dx}{\sqrt{x^4+a^4}}
\displaystyle = \int \frac{x\,dx}{\sqrt{(x^2)^2+(a^2)^2}}
\displaystyle \text{Let } x^2=t
\displaystyle \Rightarrow 2x\,dx=dt
\displaystyle \Rightarrow x\,dx=\frac{dt}{2}
\displaystyle \text{Now, } \int \frac{x\,dx}{\sqrt{(x^2)^2+(a^2)^2}}
\displaystyle =\frac12\int \frac{dt}{\sqrt{t^2+a^4}}
\displaystyle =\frac12\log\left|t+\sqrt{t^2+a^4}\right|+C
\displaystyle =\frac12\log\left|x^2+\sqrt{x^4+a^4}\right|+C
\\
\displaystyle \textbf{Question 2: }~\int \frac{\sec^2 x}{\sqrt{4+\tan^2 x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{\sec^2 x\,dx}{\sqrt{4+\tan^2 x}}
\displaystyle \text{Let } \tan x=t
\displaystyle \Rightarrow \sec^2 x\,dx=dt
\displaystyle \text{Now, } \int \frac{\sec^2 x\,dx}{\sqrt{4+\tan^2 x}}
\displaystyle =\int \frac{dt}{\sqrt{4+t^2}}
\displaystyle =\log\left|t+\sqrt{4+t^2}\right|+C
\displaystyle =\log\left|\tan x+\sqrt{4+\tan^2 x}\right|+C
\\
\displaystyle \textbf{Question 3: }~\int \frac{e^x}{\sqrt{16-e^{2x}}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{e^x\,dx}{\sqrt{16-(e^x)^2}}
\displaystyle \text{Let } e^x=t
\displaystyle \Rightarrow e^x\,dx=dt
\displaystyle \text{Now, } \int \frac{e^x\,dx}{\sqrt{16-(e^x)^2}}
\displaystyle =\int \frac{dt}{\sqrt{16-t^2}}
\displaystyle =\int \frac{dt}{\sqrt{4^2-t^2}}
\displaystyle =\sin^{-1}\left(\frac{t}{4}\right)+C
\displaystyle =\sin^{-1}\left(\frac{e^x}{4}\right)+C
\\
\displaystyle \textbf{Question 4: }~\int \frac{\cos x}{\sqrt{4+\sin^2 x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{\cos x\,dx}{\sqrt{4+\sin^2 x}}
\displaystyle \text{Let } \sin x=t
\displaystyle \Rightarrow \cos x\,dx=dt
\displaystyle \text{Now, } \int \frac{\cos x\,dx}{\sqrt{4+\sin^2 x}}
\displaystyle =\int \frac{dt}{\sqrt{4+t^2}}
\displaystyle =\int \frac{dt}{\sqrt{2^2+t^2}}
\displaystyle =\log\left|t+\sqrt{4+t^2}\right|+C
\displaystyle =\log\left|\sin x+\sqrt{4+\sin^2 x}\right|+C
\\
\displaystyle \textbf{Question 5: }~\int \frac{\sin x}{\sqrt{4\cos^2 x-1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{\sin x\,dx}{\sqrt{4+\cos^2 x-1}}
\displaystyle \text{Let } \cos x=t
\displaystyle \Rightarrow -\sin x\,dx=dt
\displaystyle \Rightarrow \sin x\,dx=-dt
\displaystyle \text{Now, } \int \frac{\sin x\,dx}{\sqrt{4+\cos^2 x-1}}
\displaystyle =\int \frac{-dt}{\sqrt{t^2+3}}
\displaystyle =\int \frac{-dt}{\sqrt{4\left(t^2-\frac14\right)}}
\displaystyle =-\frac12\int \frac{dt}{\sqrt{t^2-\left(\frac12\right)^2}}
\displaystyle =-\frac12\log\left|t+\sqrt{t^2-\frac14}\right|+C
\displaystyle =-\frac12\log\left|\frac{2t+\sqrt{4t^2-1}}{2}\right|+C
\displaystyle =-\frac12\left[\log\left|2t+\sqrt{4t^2-1}\right|-\log 2\right]+C
\displaystyle =-\frac12\log\left|2t+\sqrt{4t^2-1}\right|+\frac{\log 2}{2}+C
\displaystyle \text{Let } C'=\frac{\log 2}{2}+C
\displaystyle =-\frac12\log\left|2\cos x+\sqrt{4\cos^2 x-1}\right|+C' latex \\ $
\displaystyle \textbf{Question 6: }~\int \frac{x}{\sqrt{4-x^4}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{x\,dx}{\sqrt{4-x^4}}
\displaystyle =\int \frac{x\,dx}{\sqrt{2^2-(x^2)^2}}
\displaystyle \text{Let } x^2=t
\displaystyle \Rightarrow 2x\,dx=dt
\displaystyle \Rightarrow x\,dx=\frac{dt}{2}
\displaystyle \text{Now, } \int \frac{x\,dx}{\sqrt{2^2-(x^2)^2}}
\displaystyle =\frac12\int \frac{dt}{\sqrt{2^2-t^2}}
\displaystyle =\frac12\sin^{-1}\left(\frac{t}{2}\right)+C
\displaystyle =\frac12\sin^{-1}\left(\frac{x^2}{2}\right)+C
\\
\displaystyle \textbf{Question 7: }~\int \frac{1}{x\sqrt{4-9(\log x)^2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{dx}{x\sqrt{4-9(\log x)^2}}
\displaystyle \text{Let } \log x=t
\displaystyle \Rightarrow \frac{1}{x}\,dx=dt
\displaystyle \text{Now, } \int \frac{dx}{x\sqrt{4-9(\log x)^2}}
\displaystyle =\int \frac{dt}{\sqrt{4-9t^2}}
\displaystyle =\int \frac{dt}{\sqrt{2^2-(3t)^2}}
\displaystyle =\frac13\sin^{-1}\left(\frac{3t}{2}\right)+C
\displaystyle =\frac13\sin^{-1}\left(\frac{3\log x}{2}\right)+C
\\
\displaystyle \textbf{Question 8: }~\int \frac{\sin 8x}{\sqrt{9+\sin^4 4x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{\sin 8x}{\sqrt{9+\sin^4 4x}}\,dx
\displaystyle =\int \frac{2\sin(4x)\cos(4x)}{\sqrt{9+(\sin^2 4x)^2}}\,dx
\displaystyle \text{Let } \sin^2(4x)=t
\displaystyle \Rightarrow 2\sin(4x)\cos(4x)\cdot 4\,dx=dt
\displaystyle \Rightarrow 2\sin(4x)\cos(4x)\,dx=\frac{dt}{4}
\displaystyle \text{Now, } \int \frac{2\sin(4x)\cos(4x)}{\sqrt{9+(\sin^2 4x)^2}}\,dx
\displaystyle =\frac14\int \frac{dt}{\sqrt{9+t^2}}
\displaystyle =\frac14\int \frac{dt}{\sqrt{3^2+t^2}}
\displaystyle =\frac14\log\left|t+\sqrt{9+t^2}\right|+C
\displaystyle =\frac14\log\left|\sin^2 4x+\sqrt{9+\sin^4 4x}\right|+C
\\
\displaystyle \textbf{Question 9: }~\int \frac{\cos 2x}{\sqrt{\sin^2 2x+8}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{\cos(2x)\,dx}{\sqrt{\sin^2 2x+8}}
\displaystyle \text{Let } \sin(2x)=t
\displaystyle \Rightarrow \cos(2x)\cdot 2\,dx=dt
\displaystyle \Rightarrow \cos(2x)\,dx=\frac{dt}{2}
\displaystyle \text{Now, } \int \frac{\cos(2x)\,dx}{\sqrt{\sin^2 2x+8}}
\displaystyle =\frac12\int \frac{dt}{\sqrt{t^2+8}}
\displaystyle =\frac12\int \frac{dt}{\sqrt{(2\sqrt2)^2+t^2}}
\displaystyle =\frac12\log\left|t+\sqrt{t^2+8}\right|+C
\displaystyle =\frac12\log\left|\sin(2x)+\sqrt{\sin^2(2x)+8}\right|+C
\\
\displaystyle \textbf{Question 10: }~\int \frac{\sin 2x}{\sqrt{\sin^4 x+4\sin^2 x-2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{\sin(2x)\,dx}{\sqrt{\sin^4 x+4\sin^2 x-2}}
\displaystyle \text{Let } \sin^2 x=t
\displaystyle \Rightarrow 2\sin x\cos x\,dx=dt
\displaystyle \Rightarrow \sin(2x)\,dx=dt
\displaystyle \text{Now, } \int \frac{\sin(2x)\,dx}{\sqrt{\sin^4 x+4\sin^2 x-2}}
\displaystyle =\int \frac{dt}{\sqrt{t^2+4t-2}}
\displaystyle =\int \frac{dt}{\sqrt{t^2+4t+4-4-2}}
\displaystyle =\int \frac{dt}{\sqrt{(t+2)^2-(\sqrt6)^2}}
\displaystyle =\log\left|t+2+\sqrt{(t+2)^2-6}\right|+C
\displaystyle =\log\left|t+2+\sqrt{t^2+4t-2}\right|+C
\displaystyle =\log\left|\sin^2 x+2+\sqrt{\sin^4 x+4\sin^2 x-2}\right|+C
\\
\displaystyle \textbf{Question 11: }~\int \frac{\sin 2x}{\sqrt{\cos^4 x-\sin^2 x+2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{\sin(2x)\,dx}{\sqrt{\cos^4 x-\sin^2 x+2}}
\displaystyle =\int \frac{2\sin x\cos x\,dx}{\sqrt{\cos^4 x-(1-\cos^2 x)+2}}
\displaystyle =\int \frac{2\sin x\cos x\,dx}{\sqrt{\cos^4 x+\cos^2 x+1}}
\displaystyle \text{Let } \cos^2 x=t
\displaystyle \Rightarrow 2\cos x(-\sin x)\,dx=dt
\displaystyle \Rightarrow \sin(2x)\,dx=-dt
\displaystyle \text{Now, } \int \frac{\sin(2x)\,dx}{\sqrt{\cos^4 x-\sin^2 x+2}}
\displaystyle =\int \frac{-dt}{\sqrt{t^2+t+1}}
\displaystyle =\int \frac{-dt}{\sqrt{t^2+t+\left(\frac12\right)^2-\left(\frac12\right)^2+1}}
\displaystyle =-\int \frac{dt}{\sqrt{\left(t+\frac12\right)^2+\frac34}}
\displaystyle =-\int \frac{dt}{\sqrt{\left(t+\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}}
\displaystyle =-\log\left|t+\frac12+\sqrt{\left(t+\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}\right|+C
\displaystyle =-\log\left|t+\frac12+\sqrt{t^2+t+1}\right|+C
\displaystyle =-\log\left|\cos^2 x+\frac12+\sqrt{\cos^4 x+\cos^2 x+1}\right|+C
\\
\displaystyle \textbf{Question 12: }~\int \frac{\cos x}{\sqrt{4-\sin^2 x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{\cos x\,dx}{\sqrt{4-\sin^2 x}}
\displaystyle \text{Let } \sin x=t
\displaystyle \Rightarrow \cos x\,dx=dt
\displaystyle \text{Now, } \int \frac{\cos x\,dx}{\sqrt{4-\sin^2 x}}
\displaystyle =\int \frac{dt}{\sqrt{4-t^2}}
\displaystyle =\int \frac{dt}{\sqrt{2^2-t^2}}
\displaystyle =\sin^{-1}\left(\frac{t}{2}\right)+C
\displaystyle =\sin^{-1}\left(\frac{\sin x}{2}\right)+C
\\
\displaystyle \textbf{Question 13: }~\int \frac{1}{x^{2/3}\sqrt{x^{2/3}-4}}\,dx.
\displaystyle \text{Answer:}
\displaystyle =\int \frac{dx}{x^{2/3}\sqrt{(x^{1/3})^2-2^2}}
\displaystyle =\int \frac{dx}{x^{2/3}\sqrt{x^{2/3}-4}}
\displaystyle \text{Let } x^{1/3}=t
\displaystyle \Rightarrow \frac13 x^{-2/3}\,dx=dt
\displaystyle \Rightarrow \frac{1}{3x^{2/3}}\,dx=dt
\displaystyle \Rightarrow \frac{dx}{x^{2/3}}=3\,dt
\displaystyle \text{Now, } \int \frac{dx}{x^{2/3}\sqrt{x^{2/3}-4}}
\displaystyle =3\int \frac{dt}{\sqrt{t^2-2^2}}
\displaystyle =3\log\left|t+\sqrt{t^2-2^2}\right|+C
\displaystyle =3\log\left|x^{1/3}+\sqrt{x^{2/3}-4}\right|+C
\\
\displaystyle \textbf{Question 14: }~\int \frac{1}{\sqrt{(1-x^2)\{9+(\sin^{-1}x)^2\}}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{dx}{\sqrt{(1-x^2)\left(9+(\sin^{-1}x)^2\right)}}
\displaystyle \text{Let } \sin^{-1}x=t
\displaystyle \Rightarrow \frac{1}{\sqrt{1-x^2}}\,dx=dt
\displaystyle \text{Now, } \int \frac{dx}{\sqrt{(1-x^2)\left(9+(\sin^{-1}x)^2\right)}}
\displaystyle =\int \frac{dt}{\sqrt{9+t^2}}
\displaystyle =\int \frac{dt}{\sqrt{3^2+t^2}}
\displaystyle =\log\left|t+\sqrt{9+t^2}\right|+C
\displaystyle =\log\left|\sin^{-1}x+\sqrt{9+(\sin^{-1}x)^2}\right|+C
\\
\displaystyle \textbf{Question 15: }~\int \frac{\cos x}{\sqrt{\sin^2 x-2\sin x-3}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{\cos x\,dx}{\sqrt{\sin^2 x-2\sin x-3}}
\displaystyle \text{Let } \sin x=t
\displaystyle \Rightarrow \cos x\,dx=dt
\displaystyle \text{Now, } \int \frac{\cos x\,dx}{\sqrt{\sin^2 x-2\sin x-3}}
\displaystyle =\int \frac{dt}{\sqrt{t^2-2t-3}}
\displaystyle =\int \frac{dt}{\sqrt{t^2-2t+1-1-3}}
\displaystyle =\int \frac{dt}{\sqrt{(t-1)^2-2^2}}
\displaystyle =\log\left|t-1+\sqrt{(t-1)^2-2^2}\right|+C
\displaystyle =\log\left|t-1+\sqrt{t^2-2t-3}\right|+C
\displaystyle =\log\left|\sin x-1+\sqrt{\sin^2 x-2\sin x-3}\right|+C
\\
\displaystyle \textbf{Question 16: }~\int \sqrt{\mathrm{cosec x}-1}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \sqrt{\mathrm{cosec}\,x-1}\,dx
\displaystyle =\int \sqrt{\frac{1}{\sin x}-1}\,dx
\displaystyle =\int \sqrt{\frac{1-\sin x}{\sin x}}\,dx
\displaystyle =\int \sqrt{\frac{(1-\sin x)(1+\sin x)}{\sin x(1+\sin x)}}\,dx
\displaystyle =\int \sqrt{\frac{1-\sin^2 x}{\sin^2 x+\sin x}}\,dx
\displaystyle =\int \frac{\cos x\,dx}{\sqrt{\sin^2 x+\sin x}}
\displaystyle \text{Let } \sin x=t
\displaystyle \Rightarrow \cos x\,dx=dt
\displaystyle \text{Now, } \int \frac{\cos x\,dx}{\sqrt{\sin^2 x+\sin x}}
\displaystyle =\int \frac{dt}{\sqrt{t^2+t}}
\displaystyle =\int \frac{dt}{\sqrt{t^2+t+\left(\frac12\right)^2-\left(\frac12\right)^2}}
\displaystyle =\int \frac{dt}{\sqrt{\left(t+\frac12\right)^2-\left(\frac12\right)^2}}
\displaystyle =\log\left|t+\frac12+\sqrt{\left(t+\frac12\right)^2-\left(\frac12\right)^2}\right|+C
\displaystyle =\log\left|t+\frac12+\sqrt{t^2+t}\right|+C
\displaystyle =\log\left|\sin x+\frac12+\sqrt{\sin^2 x+\sin x}\right|+C
\\
\displaystyle \textbf{Question 17: }~\int \frac{\sin x-\cos x}{\sqrt{\sin 2x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{\sin x-\cos x}{\sqrt{\sin 2x}}\,dx
\displaystyle =\int \frac{\sin x-\cos x}{\sqrt{1+\sin 2x-1}}\,dx
\displaystyle =\int \frac{\sin x-\cos x}{\sqrt{\sin^2 x+\cos^2 x+2\sin x\cos x-1}}\,dx
\displaystyle =\int \frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^2-1}}\,dx
\displaystyle \text{Let } \sin x+\cos x=t
\displaystyle \Rightarrow (\cos x-\sin x)\,dx=dt
\displaystyle \Rightarrow (\sin x-\cos x)\,dx=-dt
\displaystyle \text{Now, } \int \frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^2-1}}\,dx
\displaystyle =-\int \frac{dt}{\sqrt{t^2-1^2}}
\displaystyle =-\log\left|t+\sqrt{t^2-1}\right|+C
\displaystyle =-\log\left|\sin x+\cos x+\sqrt{(\sin x+\cos x)^2-1}\right|+C
\displaystyle =-\log\left|\sin x+\cos x+\sqrt{\sin^2 x+\cos^2 x+2\sin x\cos x-1}\right|+C
\displaystyle =-\log\left|\sin x+\cos x+\sqrt{\sin 2x}\right|+C
\\
\displaystyle \textbf{Question 18: }~\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}\,dx
\displaystyle =\int \frac{\cos x-\sin x}{\sqrt{9-1-\sin 2x}}\,dx
\displaystyle =\int \frac{\cos x-\sin x}{\sqrt{9-\sin^2 x-\cos^2 x-2\sin x\cos x}}\,dx
\displaystyle =\int \frac{\cos x-\sin x}{\sqrt{9-(\sin x+\cos x)^2}}\,dx
\displaystyle \text{Let } \sin x+\cos x=t
\displaystyle \text{On differentiating both sides, we get}
\displaystyle (\cos x-\sin x)\,dx=dt
\displaystyle \therefore I=\int \frac{dt}{\sqrt{3^2-t^2}}
\displaystyle =\sin^{-1}\left(\frac{t}{3}\right)+C
\displaystyle =\sin^{-1}\left(\frac{\sin x+\cos x}{3}\right)+C
\displaystyle \text{Hence, } \int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}\,dx=\sin^{-1}\left(\frac{\sin x+\cos x}{3}\right)+C

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