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Question Number 167644 by MJS_new last updated on 22/Mar/22

$$\mathrm{find}\mathrm{the}\mathrm{exact}\mathrm{value}\mathrm{of}$$$$\underset{0}{\stackrel{2\pi }{\int }}\frac{dx}{\sqrt{1+\sin x}+\sqrt{1+\cos x}}$$

Commented by cortano1 last updated on 23/Mar/22

$$igot3.962$$$$ittrue?$$

Commented by MJS_new last updated on 23/Mar/22

$$\mathrm{yes}.\mathrm{but}\mathrm{can}\mathrm{you}\mathrm{give}\mathrm{the}\mathrm{exact}\mathrm{value}?$$

Answered by greogoury55 last updated on 23/Mar/22

$${\int }_0^{2\pi }\frac{dx}{\sqrt{1+\sin x}+\sqrt{1+\cos x}}$$$$=2{\int }_0^{\pi }\frac{dx}{\mid \sin \frac{x}{2}+\cos \frac{x}{2}\mid +\sqrt{2}\mid \cos \frac{x}{2}\mid }$$$$=2{\int }_0^{\pi }\frac{dt}{\mid \sin t+\cos t\mid +\sqrt{2}\mid \cos t\mid }$$$$={\int }_0^{\frac{\pi }{2}}\frac{2dt}{\sin t+(1+\sqrt{2})\cos t}+{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{4}}\frac{2dt}{\sin t+(1-\sqrt{2})\cos t}$$$$-{\int }_{\frac{3\pi }{4}}^{\pi }\frac{2dt}{\sin t+(1+\sqrt{2})\cos t}$$$$={\left[\frac{4}{(1+\sqrt{2})\sqrt{4-2\sqrt{2}}}\tanh {}^{-1}\left(\frac{\tan \frac{x}{4}-\sqrt{2}+1}{\sqrt{4-2\sqrt{2}}}\right)\right]}_0^{2\pi }$$

Commented by MJS_new last updated on 23/Mar/22

$$\mathrm{I}\mathrm{think}\mathrm{this}\mathrm{is}\mathrm{wrong}.\mathrm{I}\mathrm{solved}\mathrm{the}\mathrm{integral}\mathrm{at}$$$$\mathrm{question}167586.\mathrm{but}\mathrm{for}\mathrm{the}\mathrm{area}\mathrm{we}\mathrm{are}\mathrm{not}$$$$\mathrm{allowed}\mathrm{to}\mathrm{ignore}\mathrm{the}\mathrm{absolute}\mathrm{values}$$

Commented by MJS_new last updated on 24/Mar/22

$$F\left(x\right)=\frac{4}{(1+\sqrt{2})\sqrt{4-2\sqrt{2}}}\tanh {}^{-1}\left(\frac{\tan \frac{x}{4}-\sqrt{2}+1}{\sqrt{4-2\sqrt{2}}}\right)$$$$F\left(x\right)\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}$$$$x⩾4\arctan (-1+\sqrt{2}+\sqrt{4-2\sqrt{2}})\approx 3.927<2\pi$$

Commented by greogoury55 last updated on 24/Mar/22

$$nosir.itmyanswercorrectbutdifferent$$$$waywithyoursolution$$

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