prove-that-0-pi-2-ln-1-sinx-1-sinx-cosx-1-sinx-1-sinx-dx-8-

Question Number 114107 by mathdave last updated on 17/Sep/20

p r o v e t h a t

\int_{0}^{\frac{π}{2}} [\frac{\ln (\frac{1 - \sin x}{1 + \sin x}) \sqrt{\cos x}}{(1 + \sin x) \sqrt{1 - \sin x}}] d x = - 8

Commented by mathdave last updated on 17/Sep/20

Commented by Tawa11 last updated on 06/Sep/21

great sir

Answered by maths mind last updated on 17/Sep/20

= \int_{0}^{\frac{π}{2}} \frac{l n (\frac{1 - c o s (x)}{1 + c o s (x)}) \sqrt{s i n (x)}}{(1 + c o s (x)) \sqrt{1 - c o s (x)}} d x

1 + c o s (x) = 2 {c o s}^{2} (\frac{x}{2}) a n d 1 - c o s (x) = 2 {s i n}^{2} (\frac{x}{2}) \Rightarrow

= \int_{0}^{\frac{π}{2}} \frac{l n ({t g}^{2} (\frac{x}{2})) \sqrt{2 s i n (\frac{x}{2}) c o s (\frac{x}{2})}}{2 {c o s}^{2} (\frac{x}{2}) \sqrt{2 {s i n}^{2} (\frac{x}{2})}} d x

= 2 \int_{0}^{\frac{π}{2}} \frac{l n (t g (\frac{x}{2}))}{\sqrt{\frac{s i n (\frac{x}{2})}{c o s (\frac{x}{2})}}} \frac{d x}{2 {c o s}^{2} (\frac{x}{2})}

t g (\frac{x}{2}) = t \Rightarrow d t = \frac{d x}{2 {c o s}^{2} (\frac{x}{2})} w e g e t

= 2 \int_{0}^{1} \frac{l n (x)}{\sqrt{x}} d x = {[4 \sqrt{x} l n (x)]}_{0}^{1} - 4 \int_{0}^{1} \frac{d x}{\sqrt{x}} = - 4 {[2 \sqrt{x}]}_{0}^{1} = - 8