let-f-x-0-pi-4-ln-1-x-2-cos-d-with-x-lt-1-1-find-a-explicit-form-of-f-x-2-calculate-0-pi-4-ln-1-1-4-cos-d-

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) 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