find-the-exact-value-of-0-2pi-dx-1-sin-x-1-cos-x-

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

find the exact value of

\underset{0}{\int^{2 π}} \frac{d x}{\sqrt{1 + \sin x} + \sqrt{1 + \cos x}}

Commented by cortano1 last updated on 23/Mar/22

i g o t 3.962

i t t r u e ?

Commented by MJS_new last updated on 23/Mar/22

yes . but can you give the exact value ?

Answered by greogoury55 last updated on 23/Mar/22

\int_{0}^{2 π} \frac{d x}{\sqrt{1 + \sin x} + \sqrt{1 + \cos x}}

= 2 \int_{0}^{π} \frac{d x}{∣ \sin \frac{x}{2} + \cos \frac{x}{2} ∣ + \sqrt{2} ∣ \cos \frac{x}{2} ∣}

= 2 \int_{0}^{π} \frac{d t}{∣ \sin t + \cos t ∣ + \sqrt{2} ∣ \cos t ∣}

= \int_{0}^{\frac{π}{2}} \frac{2 d t}{\sin t + (1 + \sqrt{2}) \cos t} + \int_{\frac{π}{2}}^{\frac{3 π}{4}} \frac{2 d t}{\sin t + (1 - \sqrt{2}) \cos t}

- \int_{\frac{3 π}{4}}^{π} \frac{2 d t}{\sin t + (1 + \sqrt{2}) \cos t}

= {[\frac{4}{(1 + \sqrt{2}) \sqrt{4 - 2 \sqrt{2}}} \tanh^{- 1} (\frac{\tan \frac{x}{4} - \sqrt{2} + 1}{\sqrt{4 - 2 \sqrt{2}}})]}_{0}^{2 π}

Commented by MJS_new last updated on 23/Mar/22

I think this is wrong . I solved the integral at

question 167586 . but for the area we are not

allowed to ignore the absolute values

Commented by MJS_new last updated on 24/Mar/22

F (x) = \frac{4}{(1 + \sqrt{2}) \sqrt{4 - 2 \sqrt{2}}} \tanh^{- 1} (\frac{\tan \frac{x}{4} - \sqrt{2} + 1}{\sqrt{4 - 2 \sqrt{2}}})

F (x) is not defined for

x ⩾ 4 \arctan (- 1 + \sqrt{2} + \sqrt{4 - 2 \sqrt{2}}) \approx 3.927 < 2 π

Commented by greogoury55 last updated on 24/Mar/22

n o s i r . i t m y a n s w e r c o r r e c t b u t d i f f e r e n t

w a y w i t h y o u r s o l u t i o n