pi-2-pi-2-8-2-cosx-1-e-sinx-1-sin-4-x-dx-api-blog-3-2-2-then-find-a-b-

Question Number 205279 by gopikrishnan last updated on 14/Mar/24

{\int_{- π / 2}}^{π / 2} \frac{8 \sqrt{2} c o s x}{(1 + \overset{s i n x}{e}) (1 + s i \overset{4}{n x})} d x = a π + b l o g (3 + 2 \sqrt{2}) t h e n f i n d a + b

Answered by Berbere last updated on 14/Mar/24

x \to - x

Ω = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{8 \sqrt{2} c o s (x)}{(1 + e^{s i n (x)}) (1 + {s i n}^{4} (x))} d x = \int_{\frac{π}{2}}^{- \frac{π}{2}} - \frac{8 \sqrt{2} c o s (- x)}{(1 + e^{- s i n (x)}) (1 + {s i n}^{4} (x))} d x

= Ω

2 Ω = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{8 \sqrt{2} c o s (x)}{1 + {s i n}^{4} (x)}; s i n (x) = y

2 Ω = \int_{- 1}^{1} \frac{8 \sqrt{2} d y}{1 + y^{4}} \Rightarrow Ω = \int_{0}^{1} \frac{8 \sqrt{2} d y}{(y^{2} + 1 + y \sqrt{2}) (y^{2} + 1 - y \sqrt{2})}

\frac{8 \sqrt{2}}{(y^{4} + 1)} = \frac{a y + b}{y^{2} + 1 + y \sqrt{2}} + \frac{c y + d}{y^{2} + 1 - y \sqrt{2}}

a + c = 0, (- a \sqrt{2} + b + c \sqrt{2} + d) = 0

b + d = 8 \sqrt{2}; (a + c + d \sqrt{2} - b \sqrt{2}) = 0

a - c = 8

d - b = 0

d = b = 4 \sqrt{2}; a = 4, c = - 4

\frac{4 y + 4 \sqrt{2}}{y^{2} + y \sqrt{2} + 1} + \frac{- 4 y + 4 \sqrt{2}}{y^{2} + 1 - y \sqrt{2}}

\frac{2 (2 y + \sqrt{2})}{y^{2} + y \sqrt{2} + 1} + \frac{2 \sqrt{2}}{{(y + \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}} + \frac{- 2 (2 y - \sqrt{2})}{y^{2} + 1 - y \sqrt{2}} + \frac{2 \sqrt{2}}{{(y - \frac{1}{2})}^{2} + \frac{1}{2}}

\int \frac{8 \sqrt{2}}{y^{4} + 1} d y = 2 l n (\frac{y^{2} + y \sqrt{2} + 1}{y^{2} - y \sqrt{2} + 1}) + {4 \tan}^{- 1} (y \sqrt{2} + 1) + {4 \tan}^{- 1} (y \sqrt{2} - 1) + c

Ω = 2 l n (\frac{2 + \sqrt{2}}{2 - \sqrt{2}}) + 4 (\tan^{- 1} (\sqrt{2} + 1) + \tan^{- 1} (\sqrt{2} - 1) - \tan^{- 1} (1) - \tan^{- 1} (- 1))

= 2 l n (\frac{6 + 4 \sqrt{2}}{2}) + 4 (\tan^{- 1} (\sqrt{2} + 1) + \tan^{- 1} (\frac{1}{\sqrt{2} + 1}))

2 l n (3 + 2 \sqrt{2}) + 4 . \frac{π}{2} = 2 π + 2 l n (3 + 2 \sqrt{2)}

Commented by gopikrishnan last updated on 16/Mar/24

t h a n k u