find f(x) =∫_0 ^(π/4) ln(cost +xsint)dt


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Question Number 65445 by mathmax by abdo last updated on 30/Jul/19

$f i n d f (x) = \int_{0}^{\frac{π}{4}} l n (c o s t + x s i n t) d t$
Commented by Prithwish sen last updated on 30/Jul/19

$f (x) = \frac{π - 2 \sqrt{2}}{8} \ln ∣ 1 + x^{2} ∣ - \frac{1 - \sqrt{2}}{\sqrt{2}} \tan^{- 1} x + C$
Commented by mathmax by abdo last updated on 31/Jul/19
$we have f^′ (x) =∫_0 ^(π/4) ((sint)/(cost +xsint)) dt changement tan((t/2)) =u give f^′ (x) =∫_0 ^((√2)−1) (((2u)/(1+u^2 ))/(((1−u^2 )/(1+u^2 )) +x((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) =∫_0 ^((√2)−1) ((4u)/((1+u^2 )(1−u^2 +2xu)))du =−4 ∫_0 ^((√2)−1) ((udu)/((u^2 +1)(u^2 −2xu−1))) let decompose F(u) =(u/((u^2 +1)(u^2 −2xu−1))) u^2 −2xu−1 =0 →Δ^′ =x^2 +1 ⇒u_1 =x+(√(1+x^2 )) and u_2 =x−(√(1+x^2 )) F(u) =(u/((u^2 +1)(u−u_1 )(u−u_2 ))) =(a/(u−u_1 )) +(b/(u−u_2 )) +((cu +d)/(u^2 +1)) a =(u_1 /((u_1 ^2 +1)(u_1 −u_2 ))) =((x+(√(1+x^2 )))/((1+(x+(√(1+x^2 )))^2 2(√(1+x^2 )))) b =(u_2 /((1+u_2 ^2 )(u_2 −u_1 ))) =((x−(√(1+x^2 )))/((1+(x−(√(1+x^2 )))^2 (−2(√(1+x^2 ))))) lim_(u→+∞) uF(u) =0 =a+b +c ⇒c =−a−b ⇒ F(u) =(a/(u−u_1 ))+(b/(u−u_2 )) +(((−a−b)u +d)/(u^2 +1)) F(0)=0 =−(a/u_1 )−(b/u_2 ) +d ⇒d =(a/u_1 )+(b/u_2 ) ⇒ f^′ (x) =−4 ∫_0 ^((√2)−1) ((a/(u−u_1 )) +(b/(u−u_2 )) +((cu+d)/(u^2 +1)))du =−4 [aln∣u−u_1 ∣+bln∣u−u_2 ∣]_0 ^((√2)−1) −4 { (1/2)∫_0 ^((√2)−1) ((2cu +2d)/(u^2 +1))du} =−4{aln∣(√2)−1−u_1 ∣+bln∣(√2)−1−u_2 ∣−aln∣u_1 ∣−bln∣u_2 ∣} 2 { c[ln(u^2 +1)]_0 ^((√2)−1) +2d [arctanu]_0 ^((√2)−1) =.... be continued....$
$w e h a v e f^{^{'}} (x) = \int_{0}^{\frac{π}{4}} \frac{s i n t}{c o s t + x s i n t} d t c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e$ $f^{^{'}} (x) = \int_{0}^{\sqrt{2} - 1} \frac{\frac{2 u}{1 + u^{2}}}{\frac{1 - u^{2}}{1 + u^{2}} + x \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int_{0}^{\sqrt{2} - 1} \frac{4 u}{(1 + u^{2}) (1 - u^{2} + 2 x u)} d u$ $= - 4 \int_{0}^{\sqrt{2} - 1} \frac{u d u}{(u^{2} + 1) (u^{2} - 2 x u - 1)} l e t d e c o m p o s e$ $F (u) = \frac{u}{(u^{2} + 1) (u^{2} - 2 x u - 1)}$ $u^{2} - 2 x u - 1 = 0 \to Δ^{^{'}} = x^{2} + 1 \Rightarrow u_{1} = x + \sqrt{1 + x^{2}} a n d u_{2} = x - \sqrt{1 + x^{2}}$ $F (u) = \frac{u}{(u^{2} + 1) (u - u_{1}) (u - u_{2})} = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{c u + d}{u^{2} + 1}$ $a = \frac{u_{1}}{(u_{1}^{2} + 1) (u_{1} - u_{2})} = \frac{x + \sqrt{1 + x^{2}}}{(1 + {(x + \sqrt{1 + x^{2}})}^{2} 2 \sqrt{1 + x^{2}}}$ $b = \frac{u_{2}}{(1 + u_{2}^{2}) (u_{2} - u_{1})} = \frac{x - \sqrt{1 + x^{2}}}{(1 + {(x - \sqrt{1 + x^{2}})}^{2} (- 2 \sqrt{1 + x^{2}})}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + b + c \Rightarrow c = - a - b \Rightarrow$ $F (u) = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{(- a - b) u + d}{u^{2} + 1}$ $F (0) = 0 = - \frac{a}{u_{1}} - \frac{b}{u_{2}} + d \Rightarrow d = \frac{a}{u_{1}} + \frac{b}{u_{2}} \Rightarrow$ $f^{^{'}} (x) = - 4 \int_{0}^{\sqrt{2} - 1} (\frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{c u + d}{u^{2} + 1}) d u$ $= - 4 {[a l n ∣ u - u_{1} ∣ + b l n ∣ u - u_{2} ∣]}_{0}^{\sqrt{2} - 1} - 4 {\frac{1}{2} \int_{0}^{\sqrt{2} - 1} \frac{2 c u + 2 d}{u^{2} + 1} d u}$ $= - 4 {a l n ∣ \sqrt{2} - 1 - u_{1} ∣ + b l n ∣ \sqrt{2} - 1 - u_{2} ∣ - a l n ∣ u_{1} ∣ - b l n ∣ u_{2} ∣}$ $2 {c {[l n (u^{2} + 1)]}_{0}^{\sqrt{2} - 1} + 2 d {[a r c t a n u]}_{0}^{\sqrt{2} - 1} = . . . .$ $b e c o n t i n u e d . . . .$
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