let-I-0-pi-2-cosx-1-cosx-sinx-dx-and-J-0-pi-2-sinx-1-cosx-sinx-dx-prove-that-I-J-then-calculate-I-and-J-

Question Number 36942 by maxmathsup by imad last updated on 07/Jun/18

l e t I = \int_{0}^{\frac{π}{2}} \frac{c o s x}{\sqrt{1 + c o s x s i n x}} d x a n d J = \int_{0}^{\frac{π}{2}} \frac{s i n x}{\sqrt{1 + c o s x s i n x}} d x

p r o v e t h a t I = J t h e n c a l c u l a t e I a n d J .

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18

I = \int_{0}^{\frac{Π}{2}} \frac{c o s x}{\sqrt{1 + c o s x s i n x}} d x

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

= \int_{0}^{\frac{Π}{2}} \frac{s i n x}{\sqrt{1 + s i n x c o s x}} d x

= J (p r o v d)

I + J = \int_{0}^{\frac{Π}{2}} \frac{c o s x + s i n x}{\sqrt{1 + s i n x c o s x}} d x

= \sqrt{2} \int_{0}^{\frac{Π}{2}} \frac{c o s x + s i n x}{\sqrt{1 + 1 + 2 s i n x c o s x}} d x

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

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

\sqrt{} 2 \int_{0}^{\frac{Π}{2}} \frac{d (s i n x - c o s x)}{\sqrt{3 - {(s i n x - c o s x)}^{2}}} [\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = {s i n}^{- 1} (\frac{x}{a})

= \sqrt{2} \times ∣ {s i n}^{- 1} (\frac{s i n x - c o s x}{\sqrt{3}}) ∣_{0}^{\frac{Π}{2}}

= \sqrt{2} \times {{s i n}^{- 1} (\frac{1}{\sqrt{3}}) - {s i n}^{- 1} (\frac{- 1}{\sqrt{3}})}

= 2 \sqrt{2} {s i n}^{- 1} (\frac{1}{\sqrt{3}})

I + j = 2 I = 2 \sqrt{2} {s i n}^{- 1} (\frac{1}{\sqrt{3}})

I = J = \sqrt{2} {s i n}^{- 1} (\frac{1}{\sqrt{3}})