calculate f(a)=∫_0 ^π (dx/(1−a cosx)) a from R . 2) application calculate ∫


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Question Number 35687 by prof Abdo imad last updated on 22/May/18

$c a l c u l a t e f (a) = \int_{0}^{π} \frac{d x}{1 - a c o s x} a f r o m R .$ $2) a p p l i c a t i o n c a l c u l a t e \int_{0}^{π} \frac{d x}{1 - 2 c o s x}$
Commented by prof Abdo imad last updated on 22/May/18
$vhangement tan((x/2))=t give f(a) = ∫_0 ^∞ (1/(1−a ((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^∞ ((2dt)/(1+t^2 −a(1−t^2 ))) = ∫_0 ^∞ ((2dt)/(1−a +(1+a)t^2 )) = (1/(1−a))∫_0 ^∞ ((2dt)/(1+((1+a)/(1−a))t^2 )) case 1 ((1+a)/(1−a))>0 and a≠1 ⇒ (((1+a)(1−a))/((1−a)^2 ))>0 ⇒ 1−a^2 >0 ⇒ ∣a∣<1 we get f(a) = (2/(1−a))∫_0 ^∞ (dt/(1+((√((1+a)/(1−a)))t)^2 )) =_(u= (√((1+a)/(1−a)))t) (2/(1−a))∫_0 ^∞ (1/(1+u^2 )) ((√(1−a))/(√(1+a))) du =(2/((√(1−a))(√(1+a)))) .(π/2) = (π/(√(1−a^2 ))) case2 ((1+a)/(1−a))<0 ⇒ ((1+a)/(a−1))>0 and f(a) = (1/(1−a))∫_0 ^∞ ((2dt)/(1 −((a+1)/(a−1))t^2 )) =_((√((a+1)/(a−1))) t =u) (2/(1−a)) ∫_0 ^∞ (1/(1 −u^2 )) ((√(a−1))/(√(a+1))) du = −(2/(√(a^2 −1))) ∫_0 ^∞ (du/(1−u^2 )) = (1/(√(a^2 −1))) ∫_0 ^∞ { (1/(u−1)) −(1/(u+1))}du = (1/(√(a^2 −1))) [ln∣((u−1)/(u+1))∣]_0 ^(+∞) =0 .$
$v h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $f (a) = \int_{0}^{\infty} \frac{1}{1 - a \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int_{0}^{\infty} \frac{2 d t}{1 + t^{2} - a (1 - t^{2})} = \int_{0}^{\infty} \frac{2 d t}{1 - a + (1 + a) t^{2}}$ $= \frac{1}{1 - a} \int_{0}^{\infty} \frac{2 d t}{1 + \frac{1 + a}{1 - a} t^{2}}$ $c a s e 1 \frac{1 + a}{1 - a} > 0 a n d a \neq 1 \Rightarrow \frac{(1 + a) (1 - a)}{{(1 - a)}^{2}} > 0 \Rightarrow$ $1 - a^{2} > 0 \Rightarrow ∣ a ∣< 1 w e g e t$ $f (a) = \frac{2}{1 - a} \int_{0}^{\infty} \frac{d t}{1 + {(\sqrt{\frac{1 + a}{1 - a}} t)}^{2}}$ $=_{u = \sqrt{\frac{1 + a}{1 - a}} t} \frac{2}{1 - a} \int_{0}^{\infty} \frac{1}{1 + u^{2}} \frac{\sqrt{1 - a}}{\sqrt{1 + a}} d u$ $= \frac{2}{\sqrt{1 - a} \sqrt{1 + a}} . \frac{π}{2} = \frac{π}{\sqrt{1 - a^{2}}}$ $c a s e 2 \frac{1 + a}{1 - a} < 0 \Rightarrow \frac{1 + a}{a - 1} > 0 a n d$ $f (a) = \frac{1}{1 - a} \int_{0}^{\infty} \frac{2 d t}{1 - \frac{a + 1}{a - 1} t^{2}}$ $=_{\sqrt{\frac{a + 1}{a - 1}} t = u} \frac{2}{1 - a} \int_{0}^{\infty} \frac{1}{1 - u^{2}} \frac{\sqrt{a - 1}}{\sqrt{a + 1}} d u$ $= - \frac{2}{\sqrt{a^{2} - 1}} \int_{0}^{\infty} \frac{d u}{1 - u^{2}}$ $= \frac{1}{\sqrt{a^{2} - 1}} \int_{0}^{\infty} {\frac{1}{u - 1} - \frac{1}{u + 1}} d u$ $= \frac{1}{\sqrt{a^{2} - 1}} {[l n ∣ \frac{u - 1}{u + 1} ∣]}_{0}^{+ \infty} = 0 .$
Commented by prof Abdo imad last updated on 22/May/18

$2) w e h a v e a = 2 a n d ∣ a ∣> 1 \Rightarrow$ $\int_{0}^{π} \frac{d x}{1 - 2 c o s x} = 0$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18

	$\int_{0}^{Π} \frac{d x}{1 - a c o s x}$ $= \frac{1}{a} \int_{0}^{Π} \frac{d x}{\frac{1}{a} - c o s x}$ $l e t k = 1 / a$ $= k \int_{0}^{Π} \frac{d x}{k - \frac{1 - {t a n}^{2} (\frac{x}{2})}{1 + {t a n}^{2} (\frac{x}{2})}}$ $= k \int_{0}^{Π} \frac{(1 + {t a n}^{2} \frac{x}{2})}{k + {k t a n}^{2} \frac{x}{2} - 1 + {t a n}^{2} \frac{x}{2}} d x$ $t = t a n \frac{x}{2} d t = {s e c}^{2} \frac{x}{2} \times \frac{1}{2} d x$ $= k \int_{0}^{\infty} \frac{2 d t}{(k - 1) + (k + 1) t^{2}}$ $= 2 k \int_{0}^{\infty} \frac{d t}{(k + 1) (\frac{k - 1}{k + 1} + t^{2})}$ $= \frac{2 k}{k + 1} \int_{0}^{\infty} \frac{d t}{\frac{k - 1}{k + 1} + t^{2}}$ $= \frac{2 k}{k + 1} \times \frac{1}{\sqrt{\frac{k - 1}{k + 1}}} \times ∣ {t a n}^{- 1} (\frac{t}{\sqrt{\frac{k - 1}{k + 1}}}) ∣_{0}^{\infty}$ $= \frac{2 k}{k + 1} \times \frac{\sqrt{k + 1}}{\sqrt{k - 1}} \times (\frac{Π}{2})$ $= \frac{2}{1 + \frac{1}{k}} \times \frac{\sqrt{1 + \frac{1}{k}}}{\sqrt{1 - \frac{1}{k}}} \times (\frac{Π}{2})$ $= \frac{2}{1 + a} \times \frac{\sqrt{1 + a}}{\sqrt{1 - a}} \times (\frac{Π}{2})$
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