find f(x)=∫_0 ^(π/4) ln(sint +xcost)dt x real.


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Question Number 65133 by turbo msup by abdo last updated on 25/Jul/19

$f i n d f (x) = \int_{0}^{\frac{π}{4}} l n (s i n t + x c o s t) d t$ $x r e a l .$
Commented by mathmax by abdo last updated on 27/Jul/19

$w e h a v e f^{^{'}} (x) = \int_{0}^{\frac{π}{4}} \frac{c o s t}{s i n t + x c o s t} d t c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e$ $f^{^{'}} (x) = \int_{0}^{\sqrt{2} - 1} \frac{\frac{1 - u^{2}}{1 + u^{2}}}{\frac{2 u}{1 + u^{2}} + x \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = 2 \int_{0}^{\sqrt{2} - 1} \frac{1 - u^{2}}{(1 + u^{2}) (2 u + x - {x u}^{2})} d u$ $= 2 \int_{0}^{\sqrt{2} - 1} \frac{u^{2} - 1}{(u^{2} + 1) ({x u}^{2} - 2 u - x)} d u l e t d e c o m p o s e$ $F (u) = \frac{u^{2} - 1}{(u^{2} + 1) ({x u}^{2} - 2 u - x)}$ ${x u}^{2} - 2 u - x = 0 \to Δ^{^{'}} = 1 + x^{2} > 0 \Rightarrow u_{1} (x) = \frac{1 + \sqrt{1 + x^{2}}}{x}$ $u_{2} (x) = \frac{1 - \sqrt{1 + x^{2}}}{x} \Rightarrow F (u) = \frac{u^{2} - 1}{x (u^{2} + 1) (u - u_{1}) (u - u_{2})}$ $F (u) = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{c u + d}{u^{2} + 1}$ $a = {l i m}_{u \to u_{1}} (u - u_{1}) F (u) = \frac{u_{1}^{2} - 1}{x (u_{1}^{2} + 1) (u_{1} - u_{2})}$ $b = {l i m}_{u \to u_{2}} (u - u_{2}) F (u) = \frac{u_{2}^{2} - 1}{x (u_{2}^{2} + 1) (u_{2} - u_{1})}$ ${l i m}_{u \to + \infty} u F (u) = 0 = a + b + c \Rightarrow c = - a - b$ $F (0) = \frac{1}{x} = - \frac{a}{u_{1}} - \frac{b}{u_{2}} + d \Rightarrow d = \frac{1}{x} + \frac{a}{u_{1}} + \frac{b}{u_{2}} \Rightarrow$ $F (u) = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{- (a + b) u + \frac{1}{x} + \frac{a}{u_{1}} + \frac{b}{u_{2}}}{u^{2} + 1} \Rightarrow$ $f^{^{'}} (x) = 2 \int_{0}^{\sqrt{2} - 1} (\frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{- (a + b) u + λ_{x}}{u^{2} + 1}) d u$ $= 2 a {[l n ∣ u - u_{1} ∣]}_{0}^{\sqrt{2} - 1} + 2 b {[l n ∣ u - u_{2} ∣]}_{0}^{\sqrt{2} - 1} - (a + b) {[l n (u^{2} + 1)]}_{0}^{\sqrt{2} - 1}$ $+ 2 λ_{x} {[a r c t a n u]}_{0}^{\sqrt{2} - 1}$ $= 2 a l n ∣ \frac{\sqrt{2} - 1 - u_{1}}{u_{1}} ∣ + 2 b l n ∣ \frac{\sqrt{2} - 1 - u_{2}}{u_{2}} ∣ - (a + b) l n ({(\sqrt{2} - 1)}^{2} + 1)$ $+ 2 λ_{x} \frac{π}{8} \Rightarrow$ $f (x) = 2 a \int l n ∣ \frac{\sqrt{2} - 1 - u_{1} (x)}{u_{1} (x)} ∣ d x + 2 b \int l n ∣ \frac{\sqrt{2} - 1 - u_{2} (x)}{u_{2} (x)} ∣ d x$ $- (a + b) l n (4 - 2 \sqrt{2}) x + \frac{π}{4} \int λ_{x} d x + c . . . . b e c o n t i n u e d . . . .$
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