let f(x) =∫_0 ^(π/4) ln(1−x^2 cosθ)dθ with ∣x∣<1 1) find a explicit form of f(x) 2) calculate ∫


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Question Number 49806 by maxmathsup by imad last updated on 10/Dec/18

$l e t f (x) = \int_{0}^{\frac{π}{4}} l n (1 - x^{2} c o s θ) d θ w i t h ∣ x ∣< 1$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{\frac{π}{4}} l n (1 - \frac{1}{4} c o s θ) d θ .$
Commented by Abdo msup. last updated on 12/Dec/18
$1) we have f^′ (x)=∫_0 ^(π/4) ((−2xcosθ)/(1−x^2 cosθ)) dθ =(2/x) ∫_0 ^(π/4) ((1−x^2 cosθ −1)/(1−x^2 cosθ)) dθ =(π/(2x)) −(2/x) ∫_0 ^(π/4) (dθ/(1−x^2 cosθ)) let find ∫_0 ^(π/4) (dθ/(1−x^2 cosθ)) changement tan((θ/2))=t ⇒ ∫_0 ^(π/4) (dθ/(1−x^2 cosθ)) =∫_0 ^((√2)−1) (1/(1−x^2 ((1−t^2 )/(1+t^2 )))) ((2t)/(1+t^2 )) = ∫_0 ^((√2)−1) ((2dt)/(1+t^2 −x^2 +x^2 t^2 )) =∫_0 ^((√2)−1) ((2dt)/(1−x^2 +(1+x^2 )t^2 )) =(2/(1−x^2 )) ∫_0 ^((√2)−1) (dt/(1+((1+x^2 )/(1−x^2 ))t^2 ))let suppose ∣x∣<1 =_((√(((1+x^2 )/(1−x^2 ))t))=u) (2/(1−x^2 )) ∫_0 ^(((√2)−1)(√((1+x^2 )/(1−x^2 )))) (1/(1+u^2 )) (√((1−x^2 )/(1+x^2 )))du =(2/(√(1−x^4 ))) arctan{((√2)−1)(√((1+x^2 )/(1−x^2 )))} ⇒ f^′ (x)= (π/(2x)) −(1/(x(√(1−x^4 )))) arctan{((√2)−1)(√((1+x^2 )/(1−x^2 )))} ⇒ f(x)=(π/2)ln∣x∣ −∫_1 ^x (1/(t(√(1−t^4 )))) arctan{((√2)−1)(√((1+t^2 )/(1−t^2 )))} +λ$
$1) w e h a v e f^{^{'}} (x) = \int_{0}^{\frac{π}{4}} \frac{- 2 x c o s θ}{1 - x^{2} c o s θ} d θ$ $= \frac{2}{x} \int_{0}^{\frac{π}{4}} \frac{1 - x^{2} c o s θ - 1}{1 - x^{2} c o s θ} d θ$ $= \frac{π}{2 x} - \frac{2}{x} \int_{0}^{\frac{π}{4}} \frac{d θ}{1 - x^{2} c o s θ} l e t f i n d \int_{0}^{\frac{π}{4}} \frac{d θ}{1 - x^{2} c o s θ}$ $c h a n g e m e n t t a n (\frac{θ}{2}) = t \Rightarrow$ $\int_{0}^{\frac{π}{4}} \frac{d θ}{1 - x^{2} c o s θ} = \int_{0}^{\sqrt{2} - 1} \frac{1}{1 - x^{2} \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 t}{1 + t^{2}}$ $= \int_{0}^{\sqrt{2} - 1} \frac{2 d t}{1 + t^{2} - x^{2} + x^{2} t^{2}} = \int_{0}^{\sqrt{2} - 1} \frac{2 d t}{1 - x^{2} + (1 + x^{2}) t^{2}}$ $= \frac{2}{1 - x^{2}} \int_{0}^{\sqrt{2} - 1} \frac{d t}{1 + \frac{1 + x^{2}}{1 - x^{2}} t^{2}} l e t s u p p o s e ∣ x ∣< 1$ $=_{\sqrt{\frac{1 + x^{2}}{1 - x^{2}} t} = u} \frac{2}{1 - x^{2}} \int_{0}^{(\sqrt{2} - 1) \sqrt{\frac{1 + x^{2}}{1 - x^{2}}}} \frac{1}{1 + u^{2}} \sqrt{\frac{1 - x^{2}}{1 + x^{2}}} d u$ $= \frac{2}{\sqrt{1 - x^{4}}} a r c t a n {(\sqrt{2} - 1) \sqrt{\frac{1 + x^{2}}{1 - x^{2}}}} \Rightarrow$ $f^{^{'}} (x) = \frac{π}{2 x} - \frac{1}{x \sqrt{1 - x^{4}}} a r c t a n {(\sqrt{2} - 1) \sqrt{\frac{1 + x^{2}}{1 - x^{2}}}} \Rightarrow$ $f (x) = \frac{π}{2} l n ∣ x ∣ - \int_{1}^{x} \frac{1}{t \sqrt{1 - t^{4}}} a r c t a n {(\sqrt{2} - 1) \sqrt{\frac{1 + t^{2}}{1 - t^{2}}}} + λ$
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