let-give-D-0-pi-2-0-1-2-find-the-value-of-D-dxdy-ycosx-1-

Question Number 28159 by abdo imad last updated on 21/Jan/18

l e t g i v e D = [0, \frac{π}{2}] \times [0, \frac{1}{2}] f i n d t h e v a l u e o f

\int \int_{D} \frac{d x d y}{y c o s x + 1} .

Commented by abdo imad last updated on 26/Jan/18

l e t p u t I = \int \int_{D} \frac{d x d y}{y c o s x + 1}

I = \int_{0}^{\frac{π}{2}} (\int_{0}^{\frac{1}{2}} \frac{d y}{y c o x + 1}) d x = \int_{0}^{\frac{π}{2}} {[l n (y c o s x + 1)]}_{y = 0}^{y = \frac{1}{2}} \frac{d x}{c o s x}

= \int_{0}^{\frac{π}{2}} \frac{l n (1 + \frac{1}{2} c o s x)}{c o s x} d x a n d b y f u b i n i

I = \int_{0}^{\frac{1}{2}} (\int_{0}^{\frac{π}{2}} \frac{d x}{y c o s x + 1}) d y t h e c h . t a n (\frac{x}{2}) = t g i v e

\int_{0}^{\frac{π}{2}} \frac{d x}{y c o s x + 1} = \int_{0}^{1} \frac{1}{1 + y \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

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

= \frac{2}{1 - y} \int_{0}^{1} \frac{d t}{t^{2} + \frac{1 + y}{1 - y}} (c h . t = \sqrt{\frac{1 + y}{1 - y}} u)

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

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

= \frac{2}{s i n θ} a r c t a n (t a n (\frac{θ}{2})) (c h . y = c o s θ)

= \frac{θ}{s i n θ} = \frac{a r c o s y}{\sqrt{1 - y^{2}}}

\Rightarrow I = \int_{0}^{\frac{1}{2}} \frac{a r c o s y}{\sqrt{1 - y^{2}}} d y b y p a r t s

I = {[- a r c o s y . a r c o s y]}_{0}^{\frac{1}{2}} - \int_{0}^{\frac{1}{2}} - a r c o s y \frac{- d y}{\sqrt{1 - y^{2}}}

2 I = {(\frac{π}{2})}^{2} - {(\frac{π}{3})}^{2} = \frac{π^{2}}{4} - \frac{π^{2}}{9} = \frac{5 π^{2}}{36}

a n d I = \frac{5 π^{2}}{72} .