1) calculate f(θ) =∫_0 ^1 (√(t^2 +2sinθt +1))dt with 0≤θ≤(π/2) 2) calculate g(t) =∫_0 ^1 (√(t^2 +2(sinθ)t +1))dθ 3) find also h(θ) =∫


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Question Number 57666 by maxmathsup by imad last updated on 09/Apr/19

$1) c a l c u l a t e f (θ) = \int_{0}^{1} \sqrt{t^{2} + 2 s i n θ t + 1} d t w i t h 0 ⩽ θ ⩽ \frac{π}{2}$ $2) c a l c u l a t e g (t) = \int_{0}^{1} \sqrt{t^{2} + 2 (s i n θ) t + 1} d θ$ $3) f i n d a l s o h (θ) = \int_{0}^{1} \frac{t}{\sqrt{t^{2} + 2 (s i n θ) t + 1}} d t$
Commented by kaivan.ahmadi last updated on 10/Apr/19

$t h a n k s i r, i t w a s f a l s$
Commented by kaivan.ahmadi last updated on 10/Apr/19

$b u t i t s e q u a l t o$ ${(t + s i n θ)}^{2} + 1 - {s i n}^{2} θ = {(t + s i n θ)}^{2} + {c o s}^{2} θ$
Commented by maxmathsup by imad last updated on 10/Apr/19
$1) we have f(θ) =∫_0 ^1 (√(t^2 +2sinθ t +sin^2 θ +cos^2 ))dt =∫_0 ^1 (√((t+sinθ)^2 +cos^2 θ))dt changement t +sinθ =cosθ u give f(θ) = ∫_(tanθ) ^((1+sinθ)/(cosθ)) cosθ (√(1+u^2 ))cosθ du =cos^2 θ ∫_(tanθ) ^((1+sinθ)/(cosθ)) (√(1+u^2 ))du let find I =∫ (√(1+u^2 ))dy chang.u =shα give I =∫ chαch(α)dα=∫((1+ch(2α))/2)dα =(1/2)α +(1/4)sh(2α) +c =(α/2) +(1/2)sh(α)ch(α) +c =((argsh(u))/2) +(1/2) u(√(1+u^2 )) =(1/2)ln(u+(√(1+u^2 ))) +(u/2)(√(1+u^2 )) +c ⇒ f(θ) =((cos^2 (θ))/2)[ln(u+(√(1+u^2 )))+u(√(1+u^2 ))]_(tanθ) ^((1+sinθ)/(cosθ)) =(1/2) cos^2 θ{ ln(((1+sinθ)/(cosθ)) +(√(1+(((1+cosθ)/(sinθ)))^2 )))+((1+sinθ)/(cosθ))(√(1+(((1+sinθ)/(cosθ)))^2 )) −ln(tanθ +(√(1+tan^2 θ))) −tanθ (√(1+tan^2 θ))} .$
$1) w e h a v e f (θ) = \int_{0}^{1} \sqrt{t^{2} + 2 s i n θ t + {s i n}^{2} θ + {c o s}^{2}} d t$ $= \int_{0}^{1} \sqrt{{(t + s i n θ)}^{2} + {c o s}^{2} θ} d t c h a n g e m e n t t + s i n θ = c o s θ u g i v e$ $f (θ) = \int_{t a n θ}^{\frac{1 + s i n θ}{c o s θ}} c o s θ \sqrt{1 + u^{2}} c o s θ d u = {c o s}^{2} θ \int_{t a n θ}^{\frac{1 + s i n θ}{c o s θ}} \sqrt{1 + u^{2}} d u l e t f i n d$ $I = \int \sqrt{1 + u^{2}} d y c h a n g . u = s h α g i v e I = \int c h α c h (α) d α = \int \frac{1 + c h (2 α)}{2} d α$ $= \frac{1}{2} α + \frac{1}{4} s h (2 α) + c = \frac{α}{2} + \frac{1}{2} s h (α) c h (α) + c = \frac{a r g s h (u)}{2} + \frac{1}{2} u \sqrt{1 + u^{2}}$ $= \frac{1}{2} l n (u + \sqrt{1 + u^{2}}) + \frac{u}{2} \sqrt{1 + u^{2}} + c \Rightarrow$ $f (θ) = \frac{{c o s}^{2} (θ)}{2} {[l n (u + \sqrt{1 + u^{2}}) + u \sqrt{1 + u^{2}}]}_{t a n θ}^{\frac{1 + s i n θ}{c o s θ}}$ $= \frac{1}{2} {c o s}^{2} θ {l n (\frac{1 + s i n θ}{c o s θ} + \sqrt{1 + {(\frac{1 + c o s θ}{s i n θ})}^{2}}) + \frac{1 + s i n θ}{c o s θ} \sqrt{1 + {(\frac{1 + s i n θ}{c o s θ})}^{2}}$ $- l n (t a n θ + \sqrt{1 + {t a n}^{2} θ}) - t a n θ \sqrt{1 + {t a n}^{2} θ}} .$
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