Calculate Integrals $ \int \sqrt{\sec 2x-1}\;dx$ and $ \int \sqrt{\sec 2x+1}\;dx$ Calculation of Integral of
$$\displaystyle \int \sqrt{\sec 2x-1}\;dx,\>\>\>\>\>\displaystyle \int \sqrt{\sec 2x+1}\;dx$$
$\bf{My\; Solution}::$ For $(a)::$
Let $$\displaystyle I = \int \sqrt{\sec 2x-1}\;dx = \int \sqrt{\frac{1-\cos 2x}{\cos 2x}}\;dx $$
$$\displaystyle = \int \sqrt{\frac{1-\cos 2x}{\cos 2x}\times \frac{1+\cos 2x}{1+\cos 2x}}dx = \int \frac{\sin (2x)}{\sqrt{\cos 2x\cdot (1+\cos 2x)}}dx$$
Now Let $\cos (2x) = t\;,$ Then $\displaystyle \sin (2x)dx = -2dt$
So Integral $$\displaystyle I =-2\int\frac{1}{\sqrt{t\cdot (1+t)}}dt=-2\int\frac{1}{\sqrt{(t+\frac{1}{2})^2+(\frac{1}{2})^2}}dt$$
So $\displaystyle I = -2\cdot \ln \left|\left(t+\frac{1}{2}\right)+\sqrt{t^2+t}\right|+\mathcal{C} = -2\cdot \ln \left|\left(\cos 2x+\frac{1}{2}\right)+\sqrt{\cos^2(2x)+\cos (2x)}\right|+\mathcal{C}$
Is There is any Method other then that,
If yes then plz explain here
Thanks

Integrate as follows
\begin{align}
&\int \sqrt{\sec 2x-1}\;dx \\=&
\int \frac{\sqrt2\sin x}{\sqrt{\cos 2x}}dx=- \int \frac{d(\sqrt2\cos x)}{\sqrt{2\cos^2 x -1}}dx
= -\cosh^{-1}(\sqrt2 \cos x)+C\\
\\
&\int \sqrt{\sec 2x+1}\;dx \\=&
\int \frac{\sqrt2\cos x}{\sqrt{\cos 2x}}dx=\int \frac{d(\sqrt2\sin x)}{\sqrt{1-2\sin^2 x}}dx
= \sin^{-1}(\sqrt2 \sin x)+C\\
\end{align}
