Prove-that-pi-2-0-ln-1-sint-sint-dt-pi-2-8-1-2-arccos-2-

Question Number 196950 by Erico last updated on 05/Sep/23

Prove that {\int_{0}}^{\frac{π}{2}} \frac{\ln (1 + α \sin t)}{\sin t} d t = \frac{π^{2}}{8} - \frac{1}{2} {(\arccos α)}^{2}

Answered by Mathspace last updated on 06/Sep/23

f (x) = \int_{0}^{\frac{π}{2}} \frac{l n (1 + x s i n t)}{s i n t} d t

f^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x s i n t} (t a n (\frac{x}{2}) = y)

= \int_{0}^{1} \frac{2 d y}{(1 + y^{2}) (1 + x \frac{2 y}{1 + y^{2}})}

= \int_{0}^{1} \frac{2 d y}{1 + y^{2} + 2 x y} = \int_{0}^{1} \frac{2 d y}{y^{2} + 2 x y + x^{2} + 1 - x^{2}}

= \int_{0}^{1} \frac{2 d y}{{(y + x)}^{2} + 1 - x^{2}} d y (y + x = \sqrt{1 - x^{2}} z)

= \int_{\frac{x}{\sqrt{1 - x^{2}}}}^{\frac{1 + x}{\sqrt{1 - x^{2}}}} \frac{\sqrt{1 - x^{2}} d z}{(1 - x^{2}) (1 + z^{2})}

= \frac{2}{\sqrt{1 - x^{2}}} (a r c t a n (\frac{1 + x}{\sqrt{1 - x^{2}}}) - a r c t a n (\frac{x}{\sqrt{1 - x^{2}}})}

w e h a v e

t a n (a r c t a n a - a r c t a n b)

= \frac{a - b}{1 + a b} \Rightarrow

a r c t a n (\frac{1 + x}{\sqrt{1 - x^{2}}}) - a r c t a n (\frac{x}{\sqrt{1 - x^{2}}})

= a r c t a n (\frac{1 + x - x}{\sqrt{1 - x^{2}}})

= \frac{π}{2} - a r c t a n (\sqrt{1 - x^{2}}) \Rightarrow

f^{^{'}} (x) = \frac{2}{\sqrt{1 - x^{2}}} (\frac{π}{2} - a r c t a n \sqrt{1 - x^{2}})

= \frac{π}{\sqrt{1 - x^{2}}} - \frac{2 a r c t a n (\sqrt{1 - x^{2})}}{\sqrt{1 - x^{2}}}

\Rightarrow f (x) = π a r c t a n x

- 2 \int_{.}^{x} \frac{a r c t a n (\sqrt{1 - u^{2}})}{\sqrt{1 - u^{2}}} d u + c

t e s t t o f i n d

\int_{.}^{x} \frac{a r c t a n \sqrt{1 - u^{2}}}{\sqrt{1 - u^{2}}} d u

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

Commented by Mathspace last updated on 07/Sep/23

s o r r y f (x) = π a r c s i n x

- 2 \int^{x} \frac{a r c t a n \sqrt{1 - u^{2}}}{\sqrt{1 - u^{2}}} d u + c