1)let 0<θ<(π/2) and A(θ) =∫_0 ^(π/2) (dx/(√(x^2 +2sinθ x +1))) calculate A(θ) 2) calculate ∫


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Question Number 53463 by maxmathsup by imad last updated on 22/Jan/19

$1) l e t 0 < θ < \frac{π}{2} a n d A (θ) = \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{x^{2} + 2 s i n θ x + 1}}$ $c a l c u l a t e A (θ)$ $2) c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{x^{2} + \sqrt{2} x + 1}}$
Commented bymaxmathsup by imad last updated on 23/Jan/19

$1) w e h a v e A (θ) = \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{x^{2} + 2 x s i n θ + {s i n}^{2} θ + {c o s}^{2} θ}}$ $= \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{{(x + s i n θ)}^{2} + {c o s}^{2} θ}} c h a n g e m e n t x + s i n θ = t c o s θ g i v e t = \frac{x + s i n θ}{c o s θ}$ $A (θ) = \int_{t a n θ}^{\frac{π}{2 c o s θ} + t a n θ} \frac{c o s θ d t}{c o s θ \sqrt{1 + t^{2}}} = \int_{t a n θ}^{\frac{π}{2 c o s θ} + t a n θ} \frac{d t}{\sqrt{1 + t^{2}}} = {[a r g s h (t)]}_{t a n θ}^{\frac{π}{2 c o s θ} + t a n θ}$ $= l n {(t + \sqrt{1 + t^{2}}]}_{t a n θ}^{\frac{π}{2 c o s θ} + t a n θ} = l n (\frac{π}{2 c o s θ} + t a n θ + \sqrt{1 + {(\frac{π}{2 c o s θ} + t a n θ)}^{2}})$ $- l n (t a n θ + \sqrt{1 + {t a n}^{2} θ})$ $2) \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{x^{2} + \sqrt{2} x + 1}} = A (\frac{π}{4}) = l n (\frac{π}{2 c o s (\frac{π}{4})} + t a n (\frac{π}{4}) + \sqrt{1 + (\frac{π}{2 c o s (\frac{π}{4})} + t a n {(\frac{π}{4})}^{2}}$ $- l n (t a n (\frac{π}{4}) + \sqrt{1 + {t a n}^{2} (\frac{π}{4})})$ $= l n (\frac{π}{\sqrt{2}} + 1 + \sqrt{1 + {(\frac{π}{\sqrt{2}} + 1)}^{2}}) - l n (1 + \sqrt{2}) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19
	$1)∫_0 ^(π/2) (dx/(√(x^2 +2xsinθ+1))) ∫_0 ^(π/2) (dx/(√((x+sinθ)^2 +cos^2 θ)))=∣ln{∣(x+sinθ)+(√((x+sinθ)^2 +cos^2 θ)) )∣_0 ^(π/2) =[ln{((π/2)+sinθ)+(√(((π/2)+sinθ)^2 +cos^2 θ)) }−ln{(0+sinθ)+(√((0+sinθ)^2 +cos^2 θ)) } =ln{((π/2)+sinθ)+(√((π^2 /4)+πsinθ+1)) }−ln{sinθ+1} using formula ∫(dx/(√(x^2 +a^2 )))=ln(x+(√(x^2 +a^2 )) ) 2)put θ=(π/4) for answer... ln{((π/4)+(1/(√2)))+(√((π^2 /4)+(π/(√2))+1)) }−ln{(1/(√2))+1} sir pls check...$
	$1) \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{x^{2} + 2 x s i n θ + 1}}$ $\int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{{(x + s i n θ)}^{2} + {c o s}^{2} θ}} =∣ l n {∣ (x + s i n θ) + \sqrt{{(x + s i n θ)}^{2} + {c o s}^{2} θ}) ∣_{0}^{\frac{π}{2}}$ $= [l n {(\frac{π}{2} + s i n θ) + \sqrt{{(\frac{π}{2} + s i n θ)}^{2} + {c o s}^{2} θ}} - l n {(0 + s i n θ) + \sqrt{{(0 + s i n θ)}^{2} + {c o s}^{2} θ}}$ $= l n {(\frac{π}{2} + s i n θ) + \sqrt{\frac{π^{2}}{4} + π s i n θ + 1}} - l n {s i n θ + 1}$ $u s i n g f o r m u l a$ $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = l n (x + \sqrt{x^{2} + a^{2}})$ $2) p u t θ = \frac{π}{4} f o r a n s w e r . . .$ $l n {(\frac{π}{4} + \frac{1}{\sqrt{2}}) + \sqrt{\frac{π^{2}}{4} + \frac{π}{\sqrt{2}} + 1}} - l n {\frac{1}{\sqrt{2}} + 1}$ $s i r p l s c h e c k . . .$
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