∫_0 ^π (dx/((3+2cos x)^2 ))


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 63927 by aliesam last updated on 11/Jul/19

$\int_{0}^{π} \frac{d x}{{(3 + 2 c o s x)}^{2}}$
Commented by aliesam last updated on 12/Jul/19

$g o d b l e s s y o u s i r . . w e l l d o n e . .$
Commented by mathmax by abdo last updated on 12/Jul/19

$y o u a r e w e l c o m e .$
Commented by aliesam last updated on 11/Jul/19

$t h a n k y o u s i r$
Commented by aliesam last updated on 11/Jul/19

$b u t i t h i n k t h a t t h e s o l u t i o n i s \frac{3 π}{5 \sqrt{5}}$
Commented by mathmax by abdo last updated on 12/Jul/19
$let f(t)=∫_0 ^π (dx/(t+2cosx)) ⇒f^′ (t) =−∫_0 ^π (dx/((t+2cosx)^2 )) ⇒ ∫_0 ^π (dx/((t+2cosx)^2 )) =−f^′ (t) and ∫_0 ^π (dx/((3+2cosx)^2 )) =−f^′ (3) let calculate f(t) changement tan((x/2))=u give f(t)=∫_0 ^∞ ((2du)/((1+u^2 )( t +2((1−u^2 )/(1+u^2 ))))) =∫_0 ^∞ ((2du)/(t+tu^2 +2−2u^2 )) =∫_0 ^∞ ((2du)/((t−2)u^2 +t+2)) due to our case we take t>2 ⇒ f(t) =(2/((t−2)))∫_0 ^∞ (du/(u^2 +((t+2)/(t−2)))) =_(u=(√((t+2)/(t−2)))α) (2/(t−2)) ∫_0 ^∞ (1/(((t+2)/(t−2))(1+α^2 )))(√((t+2)/(t−2)))dα =(2/(√(t^2 −4))) (π/2) =(π/(√(t^2 −4))) ⇒f^′ (t) =π{(t^2 −4)^(−(1/2)) }^((1)) =−(π/2)(2t)(t^2 −4)^(−(3/2)) =((−πt)/((t^2 −4)(√(t^2 −4)))) ⇒ ∫_0 ^π (dx/((t+2cosx)^2 )) =((πt)/((t^2 −4)(√(t^2 −4)))) and ∫_0 ^π (dx/((3+2cosx)^2 )) =((3π)/((9−4)(√(9−4)))) =((3π)/(5(√5))) .$
$l e t f (t) = \int_{0}^{π} \frac{d x}{t + 2 c o s x} \Rightarrow f^{^{'}} (t) = - \int_{0}^{π} \frac{d x}{{(t + 2 c o s x)}^{2}} \Rightarrow$ $\int_{0}^{π} \frac{d x}{{(t + 2 c o s x)}^{2}} = - f^{^{'}} (t) a n d \int_{0}^{π} \frac{d x}{{(3 + 2 c o s x)}^{2}} = - f^{^{'}} (3)$ $l e t c a l c u l a t e f (t) c h a n g e m e n t t a n (\frac{x}{2}) = u g i v e$ $f (t) = \int_{0}^{\infty} \frac{2 d u}{(1 + u^{2}) (t + 2 \frac{1 - u^{2}}{1 + u^{2}})} = \int_{0}^{\infty} \frac{2 d u}{t + {t u}^{2} + 2 - 2 u^{2}}$ $= \int_{0}^{\infty} \frac{2 d u}{(t - 2) u^{2} + t + 2} d u e t o o u r c a s e w e t a k e t > 2 \Rightarrow$ $f (t) = \frac{2}{(t - 2)} \int_{0}^{\infty} \frac{d u}{u^{2} + \frac{t + 2}{t - 2}} =_{u = \sqrt{\frac{t + 2}{t - 2}} α} \frac{2}{t - 2} \int_{0}^{\infty} \frac{1}{\frac{t + 2}{t - 2} (1 + α^{2})} \sqrt{\frac{t + 2}{t - 2}} d α$ $= \frac{2}{\sqrt{t^{2} - 4}} \frac{π}{2} = \frac{π}{\sqrt{t^{2} - 4}} \Rightarrow f^{^{'}} (t) = π {{(t^{2} - 4)}^{- \frac{1}{2}}}^{(1)}$ $= - \frac{π}{2} (2 t) {(t^{2} - 4)}^{- \frac{3}{2}} = \frac{- π t}{(t^{2} - 4) \sqrt{t^{2} - 4}} \Rightarrow$ $\int_{0}^{π} \frac{d x}{{(t + 2 c o s x)}^{2}} = \frac{π t}{(t^{2} - 4) \sqrt{t^{2} - 4}} a n d$ $\int_{0}^{π} \frac{d x}{{(3 + 2 c o s x)}^{2}} = \frac{3 π}{(9 - 4) \sqrt{9 - 4}} = \frac{3 π}{5 \sqrt{5}} .$
Commented by aliesam last updated on 11/Jul/19

Commented by aliesam last updated on 11/Jul/19

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum