find f(a)= ∫_0 ^a arctan((√(a^2 −x^2 )))dx


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Question Number 36940 by maxmathsup by imad last updated on 07/Jun/18

$f i n d f (a) = \int_{0}^{a} a r c t a n (\sqrt{a^{2} - x^{2}}) d x$
Commented by math khazana by abdo last updated on 08/Jun/18
$changement (√(a^2 −x^2 ))=t givea^2 −x^2 =t^2 ⇒ x^2 =a^2 −t^2 ⇒x=(√(a^2 −t^2 ))⇒dx = ((−2tdt)/(2(√(a^2 −t^2 )))) f(a) = ∫_0 ^a arctan(t) ((t dt)/(√(a^2 −t^2 ))) by parts f(a) =[−arctan(t)(√(a^2 −t^2 ))]_0 ^a −∫_0 ^a ((−(√(a^2 −t^2 )))/(1+t^2 )) dt = ∫_0 ^a ((√(a^2 −t^2 ))/(1+t^2 )) dt changement t= a sinα give f(a) = ∫_0 ^(π/2) ((∣a∣cosα)/(1+a^2 sin^2 α)) acosα dα =a∣a∣ ∫_0 ^(π/2) ((cos^2 α)/(1+a^2 sin^2 α))dα but ((cos^2 α)/(1+a^2 sin^2 α)) = ((1/(1+tan^2 α))/(1+a^2 (1−(1/(1+tan^2 α))))) = (1/((1+tan^2 α)(1+a^2 tan^2 α).(1/(1+tan^2 α)))) = (1/(1 +a^2 tan^2 α)) ⇒ f(a)=a∣a∣ ∫_0 ^(π/2) (dα/(1+a^2 tan^2 α)) =_(atan(α)=x) a∣a∣ ∫_0 ^(+∞) (1/(1+x^2 )) (1/(a( 1+(x^2 /a^2 ))))dx =∣a∣ ∫_0 ^∞ ((a^2 dx)/((1+x^2 )( x^2 +a^2 ))) case1 a>0 anda≠1⇒f(a) =(a^3 /2) ∫_(−∞) ^(+∞) (dx/((x^2 +1)(x^2 +a^2 ))) let ϕ(z) = (1/((z^2 +1)(z^2 +a^2 ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,i)+Res(ϕ,ia)} ϕ(z)= (1/((z−i)(z+i)(z−ia)(z+ia))) Res(ϕ,i)= (1/(2i(a^2 −1))) Res(ϕ^� ,ia) = (1/(2ia(1−a^2 ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/(2i(a^2 −1))) +(1/(2ia(1−a^2 )))} =(π/(a^2 −1)){ 1−(1/a)}= ((π(a−1))/(a(a^2 −1))) =(π/(a(a+1))) f(a)= (a^3 /2) .(π/(a(a+1))) = ((πa^2 )/(2(a+1))) case 2 a<0 and a≠−1 we folow the same method...$
$c h a n g e m e n t \sqrt{a^{2} - x^{2}} = t {g i v e a}^{2} - x^{2} = t^{2} \Rightarrow$ $x^{2} = a^{2} - t^{2} \Rightarrow x = \sqrt{a^{2} - t^{2}} \Rightarrow d x = \frac{- 2 t d t}{2 \sqrt{a^{2} - t^{2}}}$ $f (a) = \int_{0}^{a} a r c t a n (t) \frac{t d t}{\sqrt{a^{2} - t^{2}}} b y p a r t s$ $f (a) = {[- a r c t a n (t) \sqrt{a^{2} - t^{2}}]}_{0}^{a} - \int_{0}^{a} \frac{- \sqrt{a^{2} - t^{2}}}{1 + t^{2}} d t$ $= \int_{0}^{a} \frac{\sqrt{a^{2} - t^{2}}}{1 + t^{2}} d t c h a n g e m e n t t = a s i n α g i v e$ $f (a) = \int_{0}^{\frac{π}{2}} \frac{∣ a ∣ c o s α}{1 + a^{2} {s i n}^{2} α} a c o s α d α$ $= a ∣ a ∣ \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} α}{1 + a^{2} {s i n}^{2} α} d α$ $b u t \frac{{c o s}^{2} α}{1 + a^{2} {s i n}^{2} α} = \frac{\frac{1}{1 + {t a n}^{2} α}}{1 + a^{2} (1 - \frac{1}{1 + {t a n}^{2} α})}$ $= \frac{1}{(1 + {t a n}^{2} α) (1 + a^{2} {t a n}^{2} α) . \frac{1}{1 + {t a n}^{2} α}}$ $= \frac{1}{1 + a^{2} {t a n}^{2} α} \Rightarrow f (a) = a ∣ a ∣ \int_{0}^{\frac{π}{2}} \frac{d α}{1 + a^{2} {t a n}^{2} α}$ $=_{a t a n (α) = x} a ∣ a ∣ \int_{0}^{+ \infty} \frac{1}{1 + x^{2}} \frac{1}{a (1 + \frac{x^{2}}{a^{2}})} d x$ $=∣ a ∣ \int_{0}^{\infty} \frac{a^{2} d x}{(1 + x^{2}) (x^{2} + a^{2})}$ $c a s e 1 a > 0 a n d a \neq 1 \Rightarrow f (a) = \frac{a^{3}}{2} \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) (x^{2} + a^{2})}$ $l e t φ (z) = \frac{1}{(z^{2} + 1) (z^{2} + a^{2})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, i a)}$ $φ (z) = \frac{1}{(z - i) (z + i) (z - i a) (z + i a)}$ $R e s (φ, i) = \frac{1}{2 i (a^{2} - 1)}$ $R e s (\hat{φ}, i a) = \frac{1}{2 i a (1 - a^{2})} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{2 i (a^{2} - 1)} + \frac{1}{2 i a (1 - a^{2})}}$ $= \frac{π}{a^{2} - 1} {1 - \frac{1}{a}} = \frac{π (a - 1)}{a (a^{2} - 1)} = \frac{π}{a (a + 1)}$ $f (a) = \frac{a^{3}}{2} . \frac{π}{a (a + 1)} = \frac{π a^{2}}{2 (a + 1)}$ $c a s e 2 a < 0 a n d a \neq - 1 w e f o l o w t h e s a m e m e t h o d . . .$
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