let f(θ) =∫_0 ^(π/4) (dx/(1+sinθ sinx)) with 0<θ<(π/2) 1) explicite f(θ) 2) calculate ∫


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Question Number 78273 by msup trace by abdo last updated on 15/Jan/20

$l e t f (θ) = \int_{0}^{\frac{π}{4}} \frac{d x}{1 + s i n θ s i n x}$ $w i t h 0 < θ < \frac{π}{2}$ $1) e x p l i c i t e f (θ)$ $2) c a l c u l a t e \int_{0}^{\frac{π}{4}} \frac{d x}{{(1 + s i n θ s i n x)}^{2}}$
Commented by mathmax by abdo last updated on 18/Jan/20
$1) changement tan((x/2))=u ⇒f(θ)=∫_0 ^((√2)−1) (1/(1+sinθ×((2u)/(1+u^2 ))))((2du)/(1+u^2 )) =∫_0 ^((√2)−1) ((2du)/(1+u^2 +2sinθ u)) =∫_0 ^((√2)−1) ((2du)/(u^2 +2sinθ u +sin^2 θ +cos^2 θ)) =∫_0 ^((√2)−1) ((2du)/((u+sinθ)^2 +cos^2 θ)) =_(u+sinθ =(cosθ)t) 2 ∫_(tanθ) ^((((√2)−1)/(cosθ))+tanθ) ((cosθ dt)/(cos^2 θ(1+t^2 ))) =(2/(cosθ))[arctant]_(tanθ) ^((((√2)−1)/(cosθ))+tanθ) ⇒f(θ) =(2/(cosθ)){ arctan((((√2)−1)/(cosθ))+tanθ)−θ}$
$1) c h a n g e m e n t t a n (\frac{x}{2}) = u \Rightarrow f (θ) = \int_{0}^{\sqrt{2} - 1} \frac{1}{1 + s i n θ \times \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= \int_{0}^{\sqrt{2} - 1} \frac{2 d u}{1 + u^{2} + 2 s i n θ u} = \int_{0}^{\sqrt{2} - 1} \frac{2 d u}{u^{2} + 2 s i n θ u + {s i n}^{2} θ + {c o s}^{2} θ}$ $= \int_{0}^{\sqrt{2} - 1} \frac{2 d u}{{(u + s i n θ)}^{2} + {c o s}^{2} θ} =_{u + s i n θ = (c o s θ) t} 2 \int_{t a n θ}^{\frac{\sqrt{2} - 1}{c o s θ} + t a n θ} \frac{c o s θ d t}{{c o s}^{2} θ (1 + t^{2})}$ $= \frac{2}{c o s θ} {[a r c t a n t]}_{t a n θ}^{\frac{\sqrt{2} - 1}{c o s θ} + t a n θ} \Rightarrow f (θ) = \frac{2}{c o s θ} {a r c t a n (\frac{\sqrt{2} - 1}{c o s θ} + t a n θ) - θ}$
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