... nice calculus... calculate:: φ=^(???) ∫_0 ^( (π/4)) (dx/((sin(x)+cos(x)+(√2) )^2 ))


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Question Number 129180 by mnjuly1970 last updated on 13/Jan/21

$. . . n i c e c a l c u l u s . . .$ $c a l c u l a t e ::$ $ϕ \overset{? ? ?}{=} \int_{0}^{\frac{π}{4}} \frac{d x}{{(s i n (x) + c o s (x) + \sqrt{2})}^{2}}$
Commented by Dwaipayan Shikari last updated on 13/Jan/21

$I n t e g r a l {d o e s n}^{'} t C o n v e r g e$
Commented by mnjuly1970 last updated on 13/Jan/21

$t h a n k y o u f o r y o u r a t t a n t i o n .$ $i c h a n g e d t h e i n t e r v a l . . .$
	Answered by Dwaipayan Shikari last updated on 13/Jan/21

	$\frac{1}{2} \int_{0}^{\frac{π}{4}} \frac{d x}{{(c o s (\frac{π}{4} - x) + 1)}^{2}} = \frac{1}{2} \int_{0}^{\frac{π}{4}} \frac{1}{{(1 + c o s x)}^{2}} d x t a n \frac{x}{2} = t$ $= \int_{0}^{\sqrt{2} - 1} \frac{1}{{(1 + \frac{1 - t^{2}}{1 + t^{2}})}^{2}} . \frac{1}{1 + t^{2}} d t$ $= \int_{0}^{\sqrt{2} - 1} \frac{1 + t^{2}}{4} d t = \frac{\sqrt{2} - 1}{4} + \frac{{(\sqrt{2} - 1)}^{3}}{12} = \frac{2 \sqrt{2}}{3} - \frac{5}{6}$
	Commented by mnjuly1970 last updated on 13/Jan/21

	$v e r y n i c e a s a l w a y s . . t h x . . .$
	Answered by MJS_new last updated on 13/Jan/21

	$\int \frac{d x}{{(\sin x + \cos x + \sqrt{2})}^{2}} =$ $[t = x + \frac{5 π}{4} \to d x = d t]$ $= \frac{1}{2} \int \frac{d t}{{(1 - \sin t)}^{2}} =$ $[u = \tan \frac{t}{2} \to d t = \frac{2 d u}{u^{2} + 1}]$ $= \int \frac{u^{2} + 1}{{(u - 1)}^{4}} d u = - \frac{3 u^{2} - 3 u + 2}{3 {(u - 1)}^{3}}$ $\Rightarrow answer is - \frac{5}{6} + \frac{2 \sqrt{2}}{3}$
	Commented by mnjuly1970 last updated on 13/Jan/21

	$t h a n k s a l o t m r m j s . g r a t e f u l . .$
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