Question-97173

Question Number 97173 by 675480065 last updated on 06/Jun/20

Answered by Sourav mridha last updated on 06/Jun/20

I = \frac{2}{π} \int_{- \frac{π}{4}}^{\frac{+ π}{4}} \frac{d x}{(1 + e^{s i n x}) (2 - c o s 2 x)} \dots (i)

a n d a l s o I = \frac{2}{π} \int_{+ \frac{π}{4}}^{- \frac{π}{4}} \frac{d (- x)}{(1 + e^{s i n (- x)}) (2 - c o s 2 (- x))}

= \frac{2}{π} \int_{- \frac{π}{4}}^{+ \frac{π}{4}} \frac{e^{s i n (x)}}{(1 + e^{s i n (x)}) (2 - c o s 2 x)} d x \dots . . (i i)

n o w (i) + (i i) w e g e t,

2 I = \frac{2}{π} \int_{- \frac{π}{4}}^{+ \frac{π}{4}} \frac{d x}{2 {s i n}^{2} (x) + 1} [e v e n f^{n}] s o

I = - \frac{2}{π} \int_{0}^{+ \frac{π}{4}} \frac{d (c o t x)}{{(\sqrt{3})}^{2} + {(c o t (x))}^{2}}

= - \frac{2}{π} . \frac{1}{\sqrt{3}} {[\tan^{- 1} (\frac{c o t x}{\sqrt{3}})]}_{0}^{+ \frac{π}{4}}

= - \frac{2}{π} . \frac{1}{\sqrt{3}} [\frac{π}{6} - \frac{π}{2}] = \frac{2}{3 \sqrt{3}}

s o n o w I^{2} = \frac{4}{27} \Rightarrow 27 I^{2} = 4

Commented by 675480065 last updated on 06/Jun/20

i dont understand from (i) + (ik)

Commented by Sourav mridha last updated on 07/Jun/20

t r y t o d o s t e p b y s t e p - -

a t f i r s t a d d (i) + (i i)

t h e n y o u w i l l s e e (1 + e^{s i n (x)}) i s v a n i s h .

n o w i t s r e m a i n (2 - c o s 2 x) a t d e n o m e n a t o r .

t h i s = \frac{1}{2 {s i n}^{2} (x) + 1} = \frac{{c o s e c}^{2} (x)}{3 + {c o t}^{2} (x)}

n o w g o e s o n a n d c h e a c k w i t h m y

{s o l}^{n} .