\(\int{\sqrt{\tan^2{x}+2}\;dx}\)

[Grigorios Kostakos]

Evaluate \(\displaystyle\int{\sqrt{\tan^2{x}+2}\;dx}\,.\)

[jacks]

\(I=\displaystyle\int{\sqrt{\tan^2{x}+2}\;dx}\,.\)

Put \(\tan{x}=\sqrt{2}\,\tan{t}\quad\Rightarrow\quad \sec^2{x}\,dx=\sqrt{2}\, \sec^2{t}\,dt\,.\)

So \(dx=\dfrac{\sqrt{2}}{\sin^2{t}+1}\,dt\,.\)

So our integral is \(I=\displaystyle\int{\frac{2}{\cos{t}\,(1+\sin^2{t})}\,dt}=\int{\frac{2\cos{t}}{(1-\sin^2{t})\,(1+\sin^2{t})}\;dt}\,.\)

Now let \(\sin{t}=u\quad\Rightarrow\quad \cos{t}\,dt=du\,.\) So \begin{align*}
I&=\displaystyle\int{\Bigl({\frac{1}{1-u^2}+\frac{1}{1+u^2}}\Bigr)\;du}\\
&=\frac{1}{2}\,\ln\left|{\frac{1+u}{1-u}}\right|+\tan^{-1}(u)+c\\
&=\frac{1}{2}\,\ln\left|{\frac{1+\sin{t}}{1-\sin{t}}}\right|+\tan^{-1}(\sin{t})+c\,.
\end{align*} So \(\displaystyle\int{\sqrt{\tan^2{x}+2}\;dx}=\frac{1}{2}\,\ln\left|{\frac{\sqrt{\tan^2{x}+2}+\tan{x}}{\sqrt{\tan^2{x}+2}-\tan{x}}}\right|+\tan^{-1}\Bigl({\frac{\tan{x}}{\sqrt{\tan^2{x}+2}}}\Bigr)+c\,.\)

[Grigorios Kostakos]

2nd solution: \begin{align*}
\displaystyle\int{\sqrt{\tan^2{x}+2}\;dx}&=\int{\sqrt{\tan^2{x}+1+1}\;dx}\\
&=\int{\sqrt{\frac{1}{\cos^2{x}}+1}\;dx}\\
&=\displaystyle\int{\frac{\sqrt{1+\cos^2{x}}}{\sqrt{\cos^2{x}}}\,dx}\\
&=\int{\frac{\sqrt{2-\sin^2{x}}}{\cos{x}}\,dx}\\
&=\int{\frac{\sqrt{2-\sin^2{x}}}{\cos^2{x}}\,\cos{x}\,dx}\\
&=\displaystyle\int{\frac{\sqrt{2-\sin^2{x}}}{1-\sin^2{x}}\,\cos{x}\,dx}\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{{\begin{subarray}{c}
{t\,=\,\sin{x}} \\
{dt\,=\,\cos{x}\,dx}
\end{subarray}}}\,\int{\frac{\sqrt{2-t^2}}{1-t^2}\,dt}\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{{\begin{subarray}{c}
{t\,=\,\sqrt{2}\,\sin{\theta}} \\
{dt\,=\,\sqrt{2}\,\cos{\theta}\,d\theta}
\end{subarray}}}\displaystyle\int{\frac{\sqrt{2-2\sin^2{\theta}}}{1-2\,\sin^2{\theta}}\,\sqrt{2}\,\cos{\theta}\,d\theta}\\
&=\int{\frac{\sqrt{2}\,\sqrt{\cos^2{\theta}}}{1-2\,\sin^2{\theta}}\,\sqrt{2}\,\cos{\theta}\,d\theta}\\
&=\int{\frac{2\,\cos^2{\theta}}{1-2\,\sin^2{\theta}}\,d\theta}\\
&=\displaystyle\int{\frac{2\,\cos^2{\theta}+1-1}{1-2\,\sin^2{\theta}}\,d\theta}\\
&=\int{\frac{2\,\cos^2{\theta}+1}{1-2\,\sin^2{\theta}}\,d\theta}-\int{\frac{1}{1-2\,\sin^2{\theta}}\,d\theta}\\
&=\displaystyle\int{\frac{\cos({2\theta})}{\cos({2\theta})}\,d\theta}-\int{\frac{1}{\cos({2\theta})}\,d\theta}\\
&=\int{d\theta}-\frac{1}{2}\int{\frac{1}{\cos({2\theta})}\,d({2\theta})}\\
&\stackrel{(*)}{=\!=}\displaystyle\theta-\frac{1}{2}\,\log\bigl|{\tfrac{1+\sin({2\theta})}{\cos({2\theta})}}\bigr|+c\\
&=\theta-\frac{1}{2}\,\log\bigl|{\tfrac{1+2\sin{\theta}\cos{\theta}}{1-2\sin^2{\theta}}}\bigr|+c\\
&\stackrel{t\,=\,\sqrt{2}\,\sin{\theta}}{=\!=\!=\!=\!=\!=}\displaystyle\arcsin\bigl({\tfrac{t}{\sqrt{2}}}\bigr)-\frac{1}{2}\,\log\bigl|{\tfrac{1+t\,\sqrt{2-t^2}}{1-t^2}}\bigr|+c\\
&\stackrel{t\,=\,\sin{x}}{=\!=\!=\!=}\arcsin\bigl({\tfrac{\sin{x}}{\sqrt{2}}}\bigr)-\frac{1}{2}\,\log\Bigl|{\tfrac{1+\sin{x}\,\sqrt{2-\sin^2{x}}}{1-\sin^2{x}}}\Bigr|+c\\
&=\displaystyle\arcsin\bigl({\tfrac{\sin{x}}{\sqrt{2}}}\bigr)-\frac{1}{2}\,\log\Bigl|{\tfrac{1+\sin{x}\,\sqrt{2-\sin^2{x}}}{\cos^2{x}}\,}\Bigr|+c\,.
\end{align*}

\((*)\;\displaystyle\int{\frac{1}{\cos{x}}\,dx}=\int{\frac{1}{\frac{1+\sin{x}}{\cos{x}}}\,\frac{1+\sin{x}}{\cos^2{x}}\,dx}=\int{\frac{1}{\frac{1+\sin{x}}{\cos{x}}}\left({\frac{1+\sin{x}}{\cos{x}}}\right)'\,dx}=\log\bigl|{\tfrac{1+\sin{x}}{\cos{x}}}\bigr|+c\,.\)