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Question Number 151568 by tabata last updated on 21/Aug/21

$$\int \sqrt{sec\left(x\right)+tan\left(x\right)}dx$$$$$$$$howcanitsolve$$

Answered by puissant last updated on 22/Aug/21

$$=\int \sqrt{\frac{1}{cosx}+\frac{sinx}{cosx}}dx$$$$t=tan\frac{x}{2}\to dx=\frac{2}{1+t^2}dt$$$$K=\int \sqrt{\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}}\times \frac{2}{1+t^2}dt$$$$=\int \sqrt{\frac{{(1+t)}^2}{(1-t)(1+t)}}\times \frac{2}{1+t^2}dt$$$$=2\int \sqrt{\frac{1+t}{1-t}}\times \frac{1}{1+t^2}dt$$$$u^2=\frac{1+t}{1-t}\Rightarrow u^2-u^{2}t=1+t\Rightarrow t(1+u^2)=u^2-1$$$$\Rightarrow t=\frac{u^2-1}{u^2+1}\to dt=\frac{2u(u^2+1)-2u(u^2-1)}{{(u^2+1)}^2}du$$$$\to dt=\frac{4u}{{(u^2+1)}^2}du$$$$K=2\int \frac{4u^2}{{(u^2+1)}^2}\times \frac{1}{1+{\left(\frac{u^2-1}{u^2+1}\right)}^2}du$$$$K=2\int \frac{4u^2}{{(u^2+1)}^2}\times \frac{{(u^2+1)}^2}{{(u^2+1)}^2+{(u^2-1)}^2}du$$$$K=8\int \frac{u^2}{u^4+2u^2+1+u^4-2u^2+1}du$$$$K=8\int \frac{u^2}{2(u^4+1)}du$$$$K=4\int \frac{u^2}{u^4+1}du$$$$tobecontinued...$$

Commented by peter frank last updated on 22/Aug/21

$$4\int .\frac{1}{2}.\frac{{2\mathrm{u}}^2}{1+{\mathrm{u}}^4}=2\int \frac{{2\mathrm{u}}^2}{1+{\mathrm{u}}^4}$$$$2\int \frac{{2\mathrm{u}}^2}{1+{\mathrm{u}}^4}=4\int \frac{({\mathrm{u}}^2+1)+({\mathrm{u}}^2-1)}{1+{\mathrm{u}}^4}\mathrm{du}$$$$4[\int \frac{1+{\mathrm{u}}^2}{1+{\mathrm{u}}^4}\mathrm{du}+\int \frac{1-{\mathrm{u}}^2}{1+{\mathrm{u}}^4}\mathrm{du}]$$$$4[\int \frac{1+\frac{1}{{\mathrm{u}}^2}}{{(\mathrm{u}-\frac{1}{\mathrm{u}})}^2+2}+\int \frac{1-\frac{1}{{\mathrm{u}}^2}}{{(\mathrm{u}+\frac{1}{\mathrm{u}})}^2+2}$$$$\mathrm{t}=\mathrm{u}-\frac{1}{\mathrm{u}}$$$$\mathrm{dt}=(1+\frac{1}{{\mathrm{u}}^2})\mathrm{du}$$$$\mathrm{x}=\mathrm{u}+\frac{1}{\mathrm{u}}$$$$\mathrm{dx}=(1-\frac{1}{{\mathrm{u}}^2})\mathrm{du}$$$$4[\int \frac{1}{2}.\frac{\mathrm{dt}}{{\mathrm{t}}^2+2}+\int \frac{1}{2}.\int \frac{\mathrm{dx}}{{\mathrm{x}}^2-2}$$$$2[\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\mathrm{t}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln \left(\frac{\mathrm{x}-\sqrt{2}}{\mathrm{x}+\sqrt{2}}\right)$$$$$$

Answered by peter frank last updated on 22/Aug/21

$$\int \sqrt{\frac{1}{\cos \mathrm{x}}+\frac{\sin \mathrm{x}}{\cos \mathrm{x}}}\mathrm{dx}$$$$\int \sqrt{\frac{1+\sin \mathrm{x}}{\cos \mathrm{x}}}\mathrm{dx}$$$$\int \sqrt{\frac{{(\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2})}^2}{\cos {}^2\frac{\mathrm{x}}{2}-\sin {}^2\frac{\mathrm{x}}{2}}}\mathrm{dx}$$$$\int \sqrt{\frac{{(\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2})}^2}{(\cos \frac{\mathrm{x}}{2}+\sin \frac{\mathrm{x}}{2})(\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2})}}\mathrm{dx}$$$$\int \sqrt{\frac{(\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2})}{(\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2})}}\mathrm{dx}$$$$\int \sec {}^2\frac{\mathrm{x}}{2}.\sqrt{\frac{(\tan \frac{\mathrm{x}}{2}+1)}{(1-\tan \frac{\mathrm{x}}{2})}}\mathrm{dx}$$$$\mathrm{u}=\tan \frac{\mathrm{x}}{2}$$$$\mathrm{du}=\frac{1}{2}\sec {}^2\frac{\mathrm{x}}{2}\mathrm{dx}$$$$\int \sqrt{\frac{\mathrm{u}+1}{1-\mathrm{u}}}\mathrm{du}.2$$$$\int 2\sqrt{\frac{\mathrm{u}+1}{1-\mathrm{u}}.\frac{\mathrm{u}+1}{\mathrm{u}-1}}\mathrm{du}.$$$$\int 2.\frac{\mathrm{u}+1}{\sqrt{1-{\mathrm{u}}^2}}\mathrm{du}.$$$$$$$$$$

Commented by talminator2856791 last updated on 22/Aug/21

$$\mathrm{the}\mathrm{font}\mathrm{is}\mathrm{so}\mathrm{big}!\mathrm{hahahahahah}$$

Answered by MJS_new last updated on 22/Aug/21

$$\int \sqrt{\sec x+\tan x}dx=$$$$[t=\tan \frac{x}{2}\to dx=\frac{2dt}{1+t^2}]$$$$=2\int \frac{\sqrt{1+t}}{(1+t^2)\sqrt{1-t}}dt=$$$$[u=\frac{\sqrt{1-t}}{\sqrt{1+t}}\to dt=-\sqrt{(1-t){(1+t)}^3}du]$$$$=-4\int \frac{du}{u^4+1}=$$$$=\frac{1}{\sqrt{2}}\int \frac{2u-\sqrt{2}}{u^2-\sqrt{2}u+1}du-\int \frac{du}{u^2-\sqrt{2}u+1}-$$$$-\frac{1}{\sqrt{2}}\int \frac{2u+\sqrt{2}}{u^2+\sqrt{2}u+1}du-\int \frac{du}{u^2+\sqrt{2}u+1}=$$$$=\frac{\sqrt{2}}{2}\ln (u^2-\sqrt{2}u+1)-\sqrt{2}\arctan (\sqrt{2}u-1)-$$$$+\frac{\sqrt{2}}{2}\ln (u^2+\sqrt{2}u+1)-\sqrt{2}\arctan (\sqrt{2}u+1)$$$$...$$

Commented by peter frank last updated on 22/Aug/21

$$\mathrm{thank}\mathrm{you}$$

Answered by Ar Brandon last updated on 22/Aug/21

$$I=\int \sqrt{\sec x+\tan x}dx=\int \sqrt{\frac{1+\sin x}{\cos x}}dx$$$$=\int \sqrt{\frac{{(\cos \frac{x}{2}+\sin \frac{x}{2})}^2}{\cos x}}dx=\int \frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\sqrt{\cos x}}dx$$$$=\int \frac{\cos \frac{x}{2}}{\sqrt{1-{2\sin }^2\frac{x}{2}}}dx+\int \frac{\sin \frac{x}{2}}{\sqrt{{2\cos }^2\frac{x}{2}-1}}dx$$$$=\sqrt{2}\arcsin \left(\sqrt{2}\sin \frac{x}{2}\right)-\sqrt{2}\mathrm{argcosh}\left(\sqrt{2}\cos \frac{x}{2}\right)+C$$$$=\sqrt{2}\arcsin \left(\sqrt{2}\sin \frac{x}{2}\right)-\sqrt{2}\ln \mid \sqrt{2}\cos \frac{x}{2}+\sqrt{{2\cos }^2\frac{x}{2}-1}\mid +C$$

Commented by peter frank last updated on 22/Aug/21

$$\mathrm{thank}\mathrm{you}$$

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