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Question Number 94718 by john santu last updated on 20/May/20

$$\int \sqrt{\tan \mathrm{x}}\mathrm{dx}=$$$$\int \frac{\sqrt{\tan \mathrm{x}}+\sqrt{\cot \mathrm{x}}}{2}\mathrm{dx}+\int \frac{\sqrt{\tan \mathrm{x}}-\sqrt{\cot \mathrm{x}}}{2}\mathrm{dx}$$$$=\frac{1}{\sqrt{2}}\int \frac{\sin \mathrm{x}+\cos \mathrm{x}}{\sqrt{\sin 2\mathrm{x}}}\mathrm{dx}+\frac{1}{\sqrt{2}}\int \frac{\sin \mathrm{x}-\cos \mathrm{x}}{\sqrt{\sin 2\mathrm{x}}}\mathrm{dx}$$$$=\frac{1}{\sqrt{2}}\int \frac{\sin \mathrm{x}+\cos \mathrm{x}}{\sqrt{1-{(\sin \mathrm{x}-\cos \mathrm{x})}^2}}\mathrm{dx}+$$$$\frac{1}{\sqrt{2}}\int \frac{\sin \mathrm{x}-\cos \mathrm{x}}{\sqrt{{(\sin \mathrm{x}+\cos \mathrm{x})}^2-1}}\mathrm{dx}$$$$=\frac{1}{\sqrt{2}}\int \frac{\mathrm{dt}}{\sqrt{1-{\mathrm{t}}^2}}+\frac{1}{\sqrt{2}}\int \frac{-\mathrm{du}}{\sqrt{{\mathrm{u}}^2-1}}$$$$=\frac{1}{\sqrt{2}}{\sin }^{-1}\left(\mathrm{t}\right)-\frac{1}{\sqrt{2}}\ln (\mathrm{u}+\sqrt{{\mathrm{u}}^2-1})+\mathrm{c}$$$$=\frac{1}{\sqrt{2}}{\sin }^{-1}(\sin \mathrm{x}-\cos \mathrm{x})-$$$$\frac{1}{\sqrt{2}}\ln (\sin \mathrm{x}+\cos \mathrm{x}+\sqrt{\sin 2\mathrm{x}})+\mathrm{c}$$$$\mathrm{where}\mathrm{t}=\sin \mathrm{x}-\cos \mathrm{x};$$$$\mathrm{u}=\sin \mathrm{x}+\cos \mathrm{x}$$

Commented by i jagooll last updated on 20/May/20

$$\mathrm{waww}\mathrm{simple}$$

Commented by MJS last updated on 20/May/20

$$\mathrm{nice}\mathrm{but}\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{easiest}\mathrm{way}\mathrm{to}\mathrm{solve}\mathrm{is}$$$$t=\sqrt{\tan x}\mathrm{or}t=\sqrt{\cot x};\mathrm{they}\mathrm{lead}\mathrm{to}$$$$2\int \frac{t^2}{t^4+1}dt\mathrm{or}-2\int \frac{dt}{t^4+1}$$$$$$

Commented by peter frank last updated on 20/May/20

$$\mathrm{good}$$

Commented by peter frank last updated on 20/May/20

$$\mathrm{good}$$

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