# Question #7d3ed

Dec 6, 2017

$\sqrt{2} \arcsin \left(\sqrt{2} \sin x\right) - \arctan \left(\sin \frac{x}{\sqrt{\cos 2 x}}\right) + C$

#### Explanation:

$\int \frac{\sqrt{\cos 2 x}}{\cos} x \cdot \mathrm{dx}$

=$\int \frac{\sqrt{1 - 2 {\left(\sin x\right)}^{2}}}{\cos x} ^ 2 \cdot \cos x \cdot \mathrm{dx}$

=$\int \frac{\sqrt{1 - 2 {\left(\sin x\right)}^{2}}}{1 - {\left(\sin x\right)}^{2}} \cdot \cos x \cdot \mathrm{dx}$

After using $\sqrt{2} \sin x = \sin y$ an $\sqrt{2} \cos x \cdot \mathrm{dx} = \cos y \cdot \mathrm{dy}$ substitution, this integral became

=$\int \frac{\sqrt{1 - {\left(\sin y\right)}^{2}}}{1 - {\left(\sin \frac{y}{\sqrt{2}}\right)}^{2}} \cdot \left(\frac{\cos y \cdot \mathrm{dy}}{\sqrt{2}}\right)$

=$\sqrt{2} \int \frac{\sqrt{{\left(\cos y\right)}^{2}}}{2 - {\left(\sin y\right)}^{2}} \cdot \cos y \cdot \mathrm{dy}$

=$\sqrt{2} \int \frac{{\left(\cos y\right)}^{2} \cdot \mathrm{dy}}{2 - {\left(\sin y\right)}^{2}}$

=$\sqrt{2} \int \frac{\mathrm{dy}}{2 {\left(\sec y\right)}^{2} - {\left(\tan y\right)}^{2}}$

=$\sqrt{2} \int \frac{\mathrm{dy}}{2 {\left(\tan y\right)}^{2} + 2 - {\left(\tan y\right)}^{2}}$

=$\sqrt{2} \int \frac{\mathrm{dy}}{{\left(\tan y\right)}^{2} + 2}$

=$\sqrt{2} \int \frac{\left[{\left(\tan y\right)}^{2} + 1\right] \cdot \mathrm{dy}}{\left[{\left(\tan y\right)}^{2} + 1\right] \left[{\left(\tan y\right)}^{2} + 2\right]}$

After using $z = \tan y$ and $\mathrm{dz} = \left[{\left(\tan y\right)}^{2} + 1\right] \cdot \mathrm{dy}$ transformation, it became

$\sqrt{2} \int \frac{\mathrm{dz}}{\left({z}^{2} + 1\right) \cdot \left({z}^{2} + 2\right)}$

=$\sqrt{2} \int \frac{\mathrm{dz}}{{z}^{2} + 1}$-$\sqrt{2} \int \frac{\mathrm{dz}}{{z}^{2} + 2}$

=$\sqrt{2} \arctan z - \arctan \left(\frac{z}{\sqrt{2}}\right) + C$

=$\sqrt{2} \arctan \left(\tan y\right) - \arctan \left(\tan \frac{y}{\sqrt{2}}\right) + C$

=$y \sqrt{2} - \arctan \left(\tan \frac{y}{\sqrt{2}}\right) + C$

After using $\sqrt{2} \sin x = \sin y$, $y = \arcsin \left(\sqrt{2} \sin x\right)$ and $\tan y = \frac{\sqrt{2} \sin x}{\sqrt{\cos 2 x}}$ inverse transforms, I found

$\int \frac{\sqrt{\cos 2 x}}{\cos} x \cdot \mathrm{dx}$

=$\sqrt{2} \arcsin \left(\sqrt{2} \sin x\right) - \arctan \left(\sin \frac{x}{\sqrt{\cos 2 x}}\right) + C$