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
Best Answer
Why not just perform this on computer?
$\int \sqrt{\sec{2 x}+1} dx = -8 \cos^4\left(\frac{x}{2}\right) \sqrt{\frac{-\left(1+\sqrt{2}\right) \sec (x)+\sqrt{2}+2}{\sec (x)+1}} \left(\left(1+\sqrt{2}\right) \sec (x)+\sqrt{2}+2\right) \sqrt{\frac{1}{\sec (x)+1}-\sqrt{2}+1} \sqrt{\frac{3-2 \sqrt{2}}{\sec (x)+1}-5 \sqrt{2}+7} \sqrt{\frac{3-2 \sqrt{2}}{\sec (x)+1}+\sqrt{2}-1} \sec (2 x) \sqrt{\sec (2 x)+1} \left(F\left(\sin ^{-1}\left(\left(1+\sqrt{2}\right) \tan \left(\frac{x}{2}\right)\right)|17-12 \sqrt{2}\right)+2 \Pi \left(-3+2 \sqrt{2};-\sin ^{-1}\left(\left(1+\sqrt{2}\right) \tan \left(\frac{x}{2}\right)\right)|17-12 \sqrt{2}\right)\right)$