nice-calculus-please-evaluate-0-pi-2-e-x-e-x-sin-x-cos-x-dx-2-0-pi-2-e-x-sin-x-cos-x-dx-0-

Question Number 116627 by mnjuly1970 last updated on 05/Oct/20

\dots n i c e c a l c u l u s \dots

p l e a s e e v a l u a t e ::

Φ = \frac{{(\int_{0}^{\frac{π}{2}} \frac{e^{x} + e^{- x}}{s i n (x) + c o s (x)} d x)}^{2}}{(\int_{0}^{\frac{π}{2}} \frac{e^{x}}{s i n (x) + c o s (x)} d x) (\int_{0}^{\frac{π}{2}} \frac{e^{- x}}{s i n (x) + c o s (x)} d x)} = ? ? ?

m . n . 1970

Answered by mindispower last updated on 05/Oct/20

\int_{0}^{\frac{π}{2}} \frac{e^{x}}{s i n (x) + c o s (x)} d x = J

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

= e^{\frac{π}{2}} \int_{0}^{\frac{π}{2}} \frac{e^{- x}}{s i n (x) + c o s (x)} = e^{\frac{π}{2}} I

Φ = \frac{{(J + I)}^{2}}{I . J} = \frac{{({I e}^{\frac{π}{2}} + I)}^{2}}{I . e^{\frac{π}{2}} I} = \frac{{(e^{\frac{π}{2}} + 1)}^{2}}{e^{\frac{π}{2}}}

= e^{\frac{π}{2}} + e^{- \frac{π}{2}} + 2 = 2 (c h (\frac{π}{2}) + 1)

Commented by mnjuly1970 last updated on 05/Oct/20

v o o d f o r y o u

Answered by Dwaipayan Shikari last updated on 05/Oct/20

I = \int_{0}^{\frac{π}{2}} \frac{e^{x} + e^{- x}}{s i n x + c o s x} d x

I_{1} = \int_{0}^{\frac{π}{2}} \frac{e^{x}}{s i n x + c o s x} d x = \int_{0}^{\frac{π}{2}} \frac{e^{\frac{π}{2} - x}}{s i n x + c o s x} d x = e^{\frac{π}{2}} \int_{0}^{\frac{π}{2}} \frac{e^{- x}}{s i n x + c o s x} d x

I_{2} = \int_{0}^{\frac{π}{2}} \frac{e^{- x}}{s i n x + c o s x} d x

I_{2} = e^{\frac{π}{2}} I_{1}

I_{1} + I_{2} = I (B y o b s e r v a t i o n)

Φ = \frac{{(I)}^{2}}{I_{1} I_{2}} = \frac{{(I_{1} + I_{2})}^{2}}{I_{1}^{2} e^{\frac{π}{2}}} = \frac{{(e^{\frac{π}{2}} + 1)}^{2}}{e^{\frac{π}{2}}} = i^{i} . {(\sqrt[i]{i} + 1)}^{2}

Commented by Dwaipayan Shikari last updated on 05/Oct/20

i^{i} = e^{- \frac{π}{2}} a n d \sqrt[i]{i} = i^{\frac{1}{i}} = i^{- i} = e^{\frac{π}{2}}