calculate Ω= ∫_0 ^( (π/2)) sin(x) (√( 1+^ sin(x)cos(x))) dx=?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
	��
	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

	$\sin (\frac{π}{2} - 2 x) = \cos 2 x = 2 t$ $1 - 2 \sin^{2} x = 2 t$ $- \sin 2 x d x = d t$ $\int = \sin x d x \sqrt{1 + \frac{\sin 2 x}{2}}$ $= \frac{1}{2 \sqrt{2}} \sqrt{\frac{1 - 2 t}{1 + 2 t}} \frac{\sqrt{2 + \sqrt{1 - 4 t^{2}}}}{\sqrt{1 - 4 t^{2}}} d t$ $. . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum