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Question Number 183109 by cortano1 last updated on 20/Dec/22

$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sqrt{\sin x+1}}{\sqrt{\cos x+1}}dx=?$$

Answered by universe last updated on 20/Dec/22

$$$$$${\int }_0^{\pi /2}\frac{\cos (\frac{\pi }{4}-\frac{x}{2})}{\cos \frac{x}{2}}dx$$$${\int }_0^{\pi /2}\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}}dx$$$${\int }_0^{\pi /2}(1+tan\frac{x}{2})dx$$$$$$

Commented by MJS_new last updated on 21/Dec/22

$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sqrt{1+\sin x}}{\sqrt{1+\cos x}}dx\approx 1.60$$$$\underset{0}{\stackrel{\pi /2}{\int }}(1+\tan \frac{x}{2})dx\approx 2.26$$

Answered by MJS_new last updated on 21/Dec/22

$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sqrt{1+\sin x}}{\sqrt{1+\cos x}}dx=$$$$[t=\tan \frac{x}{2}\to dx=\frac{2dt}{t^2+1}]$$$$=\sqrt{2}\underset{0}{\stackrel{1}{\int }}\frac{t+1}{t^2+1}dt=$$$$=\sqrt{2}{[\frac{1}{2}\ln (t^2+1)+\arctan t]}_0^1=$$$$=\frac{\pi \sqrt{2}}{4}+\frac{\sqrt{2}}{2}\ln 2$$

Commented by cortano1 last updated on 21/Dec/22

$$yess....$$

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