let f(t) = ∫_0 ^(π/2) ln( cosx +t sinx) 1) calculate f(0) 2) calculate f^′ (t) then find a simple form of f(t) 3) calculate ∫_0 ^(π/2) ln(cosx +2 sinx)dx 4) calculate ∫


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 40044 by abdo mathsup 649 cc last updated on 15/Jul/18

$l e t f (t) = \int_{0}^{\frac{π}{2}} l n (c o s x + t s i n x)$ $1) c a l c u l a t e f (0)$ $2) c a l c u l a t e f^{^{'}} (t) t h e n f i n d a s i m p l e f o r m o f f (t)$ $3) c a l c u l a t e \int_{0}^{\frac{π}{2}} l n (c o s x + 2 s i n x) d x$ $4) c a l c u l a t e \int_{0}^{\frac{π}{2}} l n (\sqrt{3} c o s x + s i n x) d x$
Commented by abdo mathsup 649 cc last updated on 15/Jul/18

$f (t) = \int_{0}^{\frac{π}{2}} l n (c o s x + t s i n x) d x .$
Commented by maxmathsup by imad last updated on 22/Jul/18
$1) we have f(t)= ∫_0 ^(π/2) ln(cosx+t sinx)dx ⇒f(0)=∫_0 ^(π/2) ln(cosx)dx=−(π/2)ln(2) (this value is proved ) 2)f^′ (t) =∫_0 ^(π/2) (∂/∂t){ln(cosx +tsinx)}dx =∫_0 ^(π/2) ((sinx)/(cosx+tsinx))dx hangement tan((x/2)) =u give f^′ (t) = ∫_0 ^1 (((2u)/(1+u^2 ))/(((1−u^2 )/(1+u^2 )) +t ((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) = ∫_0 ^1 ((4u du)/((1+u^2 ){1−u^2 +2tu})) = −∫_0 ^1 ((4udu)/((1+u^2 ){u^2 −2tu−1} )) let decompose F(u) = ((4u)/((u^2 +1)(u^2 −2tu −1))) roots of u^2 −2tu −1 Δ^′ = t^2 +1 ⇒u_1 = t+(√(1+t^2 )) and u_2 =t−(√(1+t^2 )) and F(u)= ((4u)/((u−u_1 )(u−u_2 )(1+u^2 ))) F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) + ((cu +d)/(u^2 +1)) a =lim_(u→u_1 ) (u−u_1 )F(u) = ((4u_1 )/((u_1 −u_2 )(1+u_1 ^2 ))) = ((4(t+(√(1+t^2 ))))/(2(√(1+t^2 ))( 1+(t+(√(1+t^2 )))^2 )) b =lim_(u→u_2 ) (u−u_2 )F(u) =((4u_2 )/((u_2 −u_1 )(1+u_2 ^2 ))) =((4(t−(√(1+t^2 ))))/(−2(√(1+t^2 )){1+(t−(√(1+t^2 )))^2 )) lim_(u→+∞) u F(u) =0 =a+b +c ⇒c =−a−b ⇒ F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) +(((−a−b)u +d)/(u^2 +1)) F(0) =0 =−(a/u_1 ) −(b/u_2 ) +d ⇒d = (a/u_1 ) +(b/u_2 ) ....be continued...$
$1) w e h a v e f (t) = \int_{0}^{\frac{π}{2}} l n (c o s x + t s i n x) d x \Rightarrow f (0) = \int_{0}^{\frac{π}{2}} l n (c o s x) d x = - \frac{π}{2} l n (2)$ $(t h i s v a l u e i s p r o v e d)$ $2) f^{^{'}} (t) = \int_{0}^{\frac{π}{2}} \frac{\partial}{\partial t} {l n (c o s x + t s i n x)} d x$ $= \int_{0}^{\frac{π}{2}} \frac{s i n x}{c o s x + t s i n x} d x h a n g e m e n t t a n (\frac{x}{2}) = u g i v e$ $f^{^{'}} (t) = \int_{0}^{1} \frac{\frac{2 u}{1 + u^{2}}}{\frac{1 - u^{2}}{1 + u^{2}} + t \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int_{0}^{1} \frac{4 u d u}{(1 + u^{2}) {1 - u^{2} + 2 t u}}$ $= - \int_{0}^{1} \frac{4 u d u}{(1 + u^{2}) {u^{2} - 2 t u - 1}} l e t d e c o m p o s e F (u) = \frac{4 u}{(u^{2} + 1) (u^{2} - 2 t u - 1)}$ $r o o t s o f u^{2} - 2 t u - 1$ $Δ^{^{'}} = t^{2} + 1 \Rightarrow u_{1} = t + \sqrt{1 + t^{2}} a n d u_{2} = t - \sqrt{1 + t^{2}} a n d F (u) = \frac{4 u}{(u - u_{1}) (u - u_{2}) (1 + u^{2})}$ $F (u) = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{c u + d}{u^{2} + 1}$ $a = {l i m}_{u \to u_{1}} (u - u_{1}) F (u) = \frac{4 u_{1}}{(u_{1} - u_{2}) (1 + u_{1}^{2})} = \frac{4 (t + \sqrt{1 + t^{2}})}{2 \sqrt{1 + t^{2}} (1 + {(t + \sqrt{1 + t^{2}})}^{2}}$ $b = {l i m}_{u \to u_{2}} (u - u_{2}) F (u) = \frac{4 u_{2}}{(u_{2} - u_{1}) (1 + u_{2}^{2})} = \frac{4 (t - \sqrt{1 + t^{2})}}{- 2 \sqrt{1 + t^{2}} {1 + {(t - \sqrt{1 + t^{2}})}^{2}}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + b + c \Rightarrow c = - a - b \Rightarrow$ $F (u) = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{(- a - b) u + d}{u^{2} + 1}$ $F (0) = 0 = - \frac{a}{u_{1}} - \frac{b}{u_{2}} + d \Rightarrow d = \frac{a}{u_{1}} + \frac{b}{u_{2}} . . . . b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum