find-the-value-of-0-pi-2-dx-1-tanx-2-

Question Number 63089 by mathmax by abdo last updated on 28/Jun/19

f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{d x}{1 + {(t a n x)}^{\sqrt{2}}} .

Commented by edafe ovwie last updated on 29/Jun/19

u s i n g

\int_{b}^{a} f (x) d x = \int_{b}^{a} f (a + b - x) d x

l e t

Missing \left or extra \right

Missing \left or extra \right

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

Missing \left or extra \right

Missing \left or extra \right

a d d 1 a n d 2

2 Ω = \int_{0}^{\frac{π}{2}} 1 d x

2 Ω = \frac{π}{2}

Ω = \frac{π}{4}

Missing \left or extra \right

Commented by mathmax by abdo last updated on 29/Jun/19

t h a n k y o u s i r .

Commented by mathmax by abdo last updated on 03/Jul/19

l e t A = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + {(t a n x)}^{\sqrt{2}}} c h a n g e m e n t x = \frac{π}{2} - t g i v e

A = - \int_{0}^{\frac{π}{2}} \frac{- d t}{1 + {(\frac{1}{t a n t})}^{\sqrt{2}}} = \int_{0}^{\frac{π}{2}} \frac{{(t a n t)}^{\sqrt{2}}}{1 + {(t a n t)}^{\sqrt{2}}} d t = \int_{0}^{\frac{π}{2}} \frac{1 + {(t a n t)}^{\sqrt{2}} - 1}{1 + {(t a n t)}^{\sqrt{2}}} d t

= \frac{π}{2} - A \Rightarrow 2 A = \frac{π}{2} \Rightarrow A = \frac{π}{4} .