find-0-pi-4-dx-2cosx-cos-2x-

Question Number 37784 by prof Abdo imad last updated on 17/Jun/18

f i n d \int_{0}^{\frac{π}{4}} \frac{d x}{2 c o s x + c o s (2 x)}

Commented by math khazana by abdo last updated on 18/Jun/18

I = \int_{0}^{\frac{π}{4}} \frac{d x}{2 c o s x + 2 {c o s}^{2} x - 1}

= \int_{0}^{\frac{π}{4}} \frac{d x}{2 {c o s}^{2} x + 2 c o s x - 1} l e t d e c o m p o s e

F (x) = \frac{1}{2 x^{2} + 2 x - 1}

Δ^{^{'}} = 1 + 2 = 3 \Rightarrow x_{1} = \frac{- 1 + \sqrt{3}}{2} a n d x_{2} = \frac{- 1 - \sqrt{3}}{2} \Rightarrow

F (x) = \frac{1}{2 (x - x_{1}) (x - x_{2})} = \frac{a}{x - x_{1}} + \frac{b}{x - x_{2}}

a = \frac{1}{2 (x_{1} - x_{2})} = \frac{1}{2 . \sqrt{3}}

b = \frac{1}{2 (x_{2} - x_{1})} = \frac{1}{2 (- \sqrt{3})} = - \frac{1}{2 \sqrt{3}} \Rightarrow

F (x) = \frac{1}{2 \sqrt{3}} {\frac{1}{x - \frac{- 1 + \sqrt{3}}{2}} - \frac{1}{x - \frac{- 1 - \sqrt{3}}{2}}}

= \frac{1}{2 \sqrt{3}} {\frac{2}{2 x + 1 - \sqrt{3}} - \frac{2}{2 x + 1 + \sqrt{3}}} \Rightarrow

I = \frac{1}{\sqrt{3}} \int_{0}^{\frac{π}{4}} \frac{d x}{2 c o s x + 1 - \sqrt{3}} - \frac{1}{\sqrt{3}} \int_{0}^{\frac{π}{4}} \frac{d x}{2 c o s x + 1 + \sqrt{3}}

= H - K

c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

H = \frac{1}{\sqrt{3}} \int_{0}^{\sqrt{2} - 1} \frac{1}{2 \frac{1 - t^{2}}{1 + t^{2}} + 1 - \sqrt{3}} \frac{2 d t}{1 + t^{2}}

= \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{2} - 1} \frac{d t}{2 - 2 t^{2} + 1 - \sqrt{3} + (1 - \sqrt{3}) t^{2}}

= \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{2} - 1} \frac{d t}{3 - \sqrt{3} - (1 + \sqrt{3}) t^{2}}

= \frac{2}{\sqrt{3} (3 - \sqrt{3})} \int_{0}^{\sqrt{2} - 1} \frac{d t}{1 - {(\sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}} t)}^{2}}

=_{\sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}} t = u} \frac{2}{\sqrt{3} (3 - \sqrt{3})} \int_{0}^{(\sqrt{2} - 1) \sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}}} \frac{1}{1 - u^{2}} \sqrt{\frac{3 - \sqrt{3}}{1 + \sqrt{3}}} d u

= \frac{2}{\sqrt{3} \sqrt{(3 - \sqrt{3) (1 + \sqrt{3)}}}} \int_{0}^{(\sqrt{2 - 1}) \sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}}} \frac{d u}{1 - u^{2}}

= \frac{1}{\sqrt{3} \sqrt{(3 - \sqrt{3}) (1 + \sqrt{3})}} {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{(\sqrt{2} - 1) \sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}}}

= \frac{\sqrt{3}}{3 \sqrt{(3 - \sqrt{3}) (1 + \sqrt{3})}} l n (\frac{1 + λ}{1 - λ}) w i t h

λ = (\sqrt{2} - 1) \sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}}

w e f o l l o w t h e s a m e ? m e t h o d t o c a l c u l a t e K \dots .

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jun/18

\int_{0}^{\frac{Π}{4}} \frac{d x}{2 c o s x + 2 {c o s}^{2} x - 1}

= \frac{1}{2} \int_{0}^{\frac{Π}{4}} \frac{d x}{{c o s}^{2} x + c o s x - \frac{1}{2}}

a^{2} + a - \frac{1}{2}

a^{2} + 2 . a . \frac{1}{2} + \frac{1}{4} - \frac{1}{4} - \frac{1}{2}

{(a + \frac{1}{2})}^{2} - \frac{3}{4}

(a + \frac{1}{2} + \frac{\sqrt{3}}{2}) (a + \frac{1}{2} - \frac{\sqrt{3}}{2})

(a + \frac{1 + \sqrt{3}}{2}) (a + \frac{1 - \sqrt{3}}{2})

(a + \frac{1 + \sqrt{3}}{2}) - (a + \frac{1 - \sqrt{3}}{2})

\sqrt{3}

= \frac{1}{2 \sqrt{3}} \int_{0}^{\frac{Π}{4}} \frac{\sqrt{3}}{(c o s x + \frac{1 + \sqrt{3}}{2}) (c o s x + \frac{1 - \sqrt{3}}{2})} d x

= \frac{1}{2 \sqrt{3}} \int_{0}^{\frac{Π}{4}} \frac{d x}{c o s x + \frac{1 - \sqrt{3}}{2}} - \frac{1}{2 \sqrt{3}} \int_{0}^{\frac{Π}{4}} \frac{d x}{c o s x + \frac{1 - \sqrt{3}}{2}}

= \frac{1}{2 \sqrt{3}} {\int_{0}^{\frac{Π}{4}} \frac{d x}{a + c o s x} - \int_{0}^{\frac{Π}{4}} \frac{d x}{b + c o s x}}

n o w i t c a n b e s o l v e d b u t l e n t h y \dots m e t h o d \dots \int

\int \frac{d x}{a + c o s x}

\int \frac{d x}{a + \frac{1 - {t a n}^{2} \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}}}

\int \frac{{s e c}^{2} \frac{x}{2}}{a + {a t a n}^{2} \frac{x}{2} + 1 - {t a n}^{2} \frac{x}{2}} d x

\int \frac{{s e c}^{2} \frac{x}{2}}{(a + 1) + {t a n}^{2} \frac{x}{2} (a - 1)} d x

\frac{1}{a - 1} \int \frac{{s e c}^{2} \frac{x}{2}}{\frac{a + 1}{a - 1} + {t a n}^{2} \frac{x}{2}} d x

t = t a n \frac{x}{2} d t = {s e c}^{2} \frac{x}{2} \times \frac{1}{2} d x

\frac{2}{a - 1} \int \frac{d t}{{\sqrt{\frac{a + 1}{a - 1}}}^{2} + t^{2}}

\frac{2}{a - 1} \times \frac{\sqrt{a - 1}}{\sqrt{a + 1}} \times {t a n}^{- 1} {\frac{t}{\frac{\sqrt{a - 1}}{\sqrt{a + 1}}}} \dots l i k e t h i s w a y

Answered by MJS last updated on 17/Jun/18

Weierstrass

\int \frac{d x}{2 \cos x + \cos 2 x} =

[t = \tan \frac{x}{2} \to d x = \frac{2 d t}{1 + t^{2}}]

= - 2 \int \frac{t^{2} + 1}{t^{4} + 6 t^{2} - 3} d t =

= - 2 \int \frac{t^{2} + 1}{(t - \sqrt{- 3 + 2 \sqrt{3}}) (t + \sqrt{- 3 + 2 \sqrt{3}}) (t^{2} + 3 + 2 \sqrt{3})} d t =

= - 2 (\int \frac{A}{t - \sqrt{- 3 + 2 \sqrt{3}}} d t + \int \frac{B}{t + \sqrt{- 3 + 2 \sqrt{3}}} d t + \int \frac{C}{t^{2} + 3 + 2 \sqrt{3}} d t) =

= - \frac{\sqrt{2 \sqrt{3}}}{6} \int \frac{d t}{t - \sqrt{- 3 + 2 \sqrt{3}}} + \frac{\sqrt{2 \sqrt{3}}}{6} \int \frac{d t}{t + \sqrt{- 3 + 2 \sqrt{3}}} - \frac{3 + \sqrt{3}}{3} \int \frac{d t}{t^{2} + 3 + 2 \sqrt{3}}) =

[\int \frac{d x}{x \pm a} = \ln ∣ x \pm a ∣; \int \frac{d x}{x^{2} + a} = \frac{1}{\sqrt{a}} \arctan \frac{x}{\sqrt{a}}]

= \frac{\sqrt{2 \sqrt{3}}}{6} (\ln ∣ t + \sqrt{- 3 + 2 \sqrt{3}} ∣ - \ln ∣ t - \sqrt{- 3 + 2 \sqrt{3}} ∣) - (1 + \frac{\sqrt{3}}{3}) \frac{1}{\sqrt{3 + 2 \sqrt{3}}} \arctan \frac{t}{\sqrt{3 + 2 \sqrt{3}}} =

= \frac{\sqrt{2 \sqrt{3}}}{6} \ln ∣ \frac{t + \sqrt{- 3 + 2 \sqrt{3}}}{t - \sqrt{- 3 + 2 \sqrt{3}}} ∣ - \frac{\sqrt{2 \sqrt{3}}}{3} \arctan \frac{\sqrt{2 \sqrt{3}} (3 - \sqrt{3}) t}{6} =

= \frac{\sqrt{2 \sqrt{3}}}{6} (\ln ∣ \frac{t + \sqrt{- 3 + 2 \sqrt{3}}}{t - \sqrt{- 3 + 2 \sqrt{3}}} ∣ - 2 \arctan \frac{\sqrt{2 \sqrt{3}} (3 - \sqrt{3}) t}{6}) =

= \frac{\sqrt{2 \sqrt{3}}}{6} (\ln ∣ \frac{\tan \frac{x}{2} + \sqrt{- 3 + 2 \sqrt{3}}}{\tan \frac{x}{2} - \sqrt{- 3 + 2 \sqrt{3}}} ∣ - 2 \arctan \frac{\sqrt{2 \sqrt{3}} (3 - \sqrt{3}) \tan \frac{x}{2}}{6}) + C

Commented by math khazana by abdo last updated on 18/Jun/18

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