find-f-x-0-pi-4-ln-cost-xsint-dt-

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

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} = \dots .

b e c o n t i n u e d \dots .