calculate-0-pi-2-sin-x-1-sin-x-cos-x-dx-

Question Number 197024 by mnjuly1970 last updated on 06/Sep/23

c a l c u l a t e

Ω = \int_{0}^{\frac{π}{2}} s i n (x) \sqrt{1 \overset{}{+} s i n (x) c o s (x)} d x = ?

Answered by Frix last updated on 06/Sep/23

\int \sin x \sqrt{1 + \sin x \cos x} d x \overset{t = x - \frac{π}{4}}{=}

= \int \frac{(\cos t + \sin t) \sqrt{1 + {2 \cos}^{2} t}}{2} d t \overset{u = \tan t}{=}

= \int \frac{(u + 1) \sqrt{u^{2} + 3}}{2 {(u^{2} + 1)}^{2}} d u \overset{v = \frac{u + \sqrt{u^{2} + 3}}{\sqrt{3}}}{=}

= \int \frac{\sqrt{3} (\sqrt{3} v - 1) (\sqrt{3} v + 3) {(v^{2} + 1)}^{2}}{{(3 v^{4} - 2 v^{2} + 3)}^{2}} d v \underset{Method}{\overset{{Ostrogradski}^{'} s}{=}}

= - \frac{\sqrt{3} v^{3} - v^{2} + \sqrt{3} v + 3}{2 (3 v^{4} - 2 v^{2} + 3)} + \int \frac{\sqrt{3} v^{2} + 6 v - \sqrt{3}}{2 (3 v^{4} - 2 v^{2} + 3)} d v =

= - \frac{\sqrt{3} v^{3} - v^{2} + \sqrt{3} v + 3}{2 (3 v^{4} - 2 v^{2} + 3)} + \frac{\sqrt{2}}{16} \ln \frac{3 v^{2} - 2 \sqrt{6} v + 3}{3 v^{2} + 2 \sqrt{6} v + 3} + \frac{3 \sqrt{2}}{8} (\tan^{- 1} (\sqrt{3} v - \sqrt{2}) - \tan^{- 1} (\sqrt{3} v + \sqrt{2}))

x \in [0, \frac{π}{2}] \Rightarrow v \in [\frac{\sqrt{3}}{3}, \sqrt{3}]

\Rightarrow

Ω = \frac{3 \sqrt{2} π}{16} + \frac{1}{2} - \frac{3 \sqrt{2}}{8} \tan^{- 1} \frac{\sqrt{2}}{4} \approx 1.15281482

Commented by mnjuly1970 last updated on 06/Sep/23

t h a n k y o u s o m u c h s i r F r i x

Commented by Frix last updated on 06/Sep/23

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Answered by universe last updated on 10/Sep/23

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

l e t

\sin x - \cos x = t \Rightarrow (\sin x + \cos x) d x = d t

I = \frac{1}{2 \sqrt{2}} \int_{- 1}^{1} \sqrt{3 - t^{2}} d t

I = \frac{1}{\sqrt{2}} \int_{0}^{1} \sqrt{{(\sqrt{3})}^{2} - t^{2}} d t

I = \frac{1}{\sqrt{2}} {[\frac{t}{2} \sqrt{3 - t^{2}} + \frac{3}{2} \sin^{- 1} \frac{t}{\sqrt{3}}]}_{0}^{1}

I = \frac{1}{2 \sqrt{2}} [\sqrt{2} + {3 \sin}^{- 1} \frac{1}{\sqrt{3}}]

Commented by mnjuly1970 last updated on 06/Sep/23

t h a n k y o u s o m u c h s i r

n i c e s o l u t i o n

Answered by ajfour last updated on 07/Sep/23

\sin (\frac{π}{2} - 2 x) = \cos 2 x = 2 t

1 - 2 \sin^{2} x = 2 t

- \sin 2 x d x = d t

\int = \sin x d x \sqrt{1 + \frac{\sin 2 x}{2}}

= \frac{1}{2 \sqrt{2}} \sqrt{\frac{1 - 2 t}{1 + 2 t}} \frac{\sqrt{2 + \sqrt{1 - 4 t^{2}}}}{\sqrt{1 - 4 t^{2}}} d t

\dots

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