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Question Number 51997 by maxmathsup by imad last updated on 01/Jan/19

l e t f (x) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x s i n t} w i t h x > - 1

1) c a l c u l a t e f (o), f (1) a n d f (2)

2) g i v e f a t f o r m o f f u n c t i o n

Commented by maxmathsup by imad last updated on 02/Jan/19

1) w e h a v e f (0) = \int_{0}^{\frac{π}{2}} d t = \frac{π}{2}

f (1) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + s i n t} =_{t a n (\frac{t}{2}) = u} \int_{0}^{1} \frac{1}{1 + \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = 2 \int_{0}^{1} \frac{d u}{1 + u^{2} + 2 u}

= 2 \int_{0}^{1} \frac{d u}{{(u + 1)}^{2}} = {[- \frac{2}{u + 1}]}_{0}^{1} = - 2 (\frac{1}{2} - 1) = - 1 + 2 = 1 \Rightarrow f (1) = 1

f (2) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + 2 s i n t} =_{t a n (\frac{t}{2}) = u} \int_{0}^{1} \frac{1}{1 + 2 \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= 2 \int_{0}^{1} \frac{d u}{1 + u^{2} + 4 u} = 2 \int_{0}^{1} \frac{d u}{u^{2} + 4 u + 1} . r o o t s o f u^{2} + 4 u + 1

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

f (2) = 2 \int_{0}^{1} \frac{d u}{(u - u_{1}) (u - u_{2})} = \frac{2}{u_{1} - u_{2}} \int_{0}^{1} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) d u

= \frac{2}{2 \sqrt{3}} \int_{0}^{1} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}} d u = \frac{1}{\sqrt{3}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{0}^{1} = \frac{1}{\sqrt{3}} {l n ∣ \frac{1 - u_{1}}{1 - u_{2}} ∣ - l n ∣ \frac{u_{1}}{u_{2}} ∣}

= \frac{1}{\sqrt{3}} {l n ∣ \frac{3 - \sqrt{3}}{3 + \sqrt{3}} ∣ - l n ∣ \frac{2 - \sqrt{3}}{2 + \sqrt{3}} ∣ .

Commented by maxmathsup by imad last updated on 02/Jan/19

2) c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e f (x) = \int_{0}^{1} \frac{1}{1 + x \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= 2 \int_{0}^{1} \frac{d u}{1 + u^{2} + 2 x u} = 2 \int_{0}^{1} \frac{d u}{u^{2} + 2 x u + 1} l e t p (u) = u^{2} + 2 x u + 1

Δ^{^{'}} = x^{2} - 1

c a s e 1 Δ^{^{'}} > 0 \Leftrightarrow∣ x ∣> 1 \Rightarrow u_{1} = - x + \sqrt{x^{2} - 1} a n d u_{2} = - x - \sqrt{x^{2} - 1}

\Rightarrow f (x) = 2 \int_{0}^{1} \frac{d u}{(u - u_{1}) (u - u_{2})} = \frac{2}{u_{1} - u_{2}} \int_{0}^{1} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}} d u

= \frac{2}{2 \sqrt{1 - x^{2}}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{0}^{1} = \frac{1}{\sqrt{1 - x^{2}}} {l n ∣ \frac{1 - u_{1}}{1 - u_{2}} ∣ - l n ∣ \frac{u_{1}}{u_{2}} ∣}

= \frac{1}{\sqrt{1 - x^{2}}} {l n ∣ \frac{1 + x - \sqrt{x^{2} - 1}}{1 + x + \sqrt{x^{2} - 1}} ∣ - l n ∣ \frac{x - \sqrt{x^{2} - 1}}{x + \sqrt{x^{2} - 1}} ∣}

c a s e 2 Δ^{^{'}} < 0 \Leftrightarrow∣ x ∣< 1 \Rightarrow p (u) = u^{2} + 2 x u + x^{2} + 1 - x^{2} = {(u + x)}^{2} + 1 - x^{2}

w e d o t h e c h a n g e m e n t u + x = \sqrt{1 - x^{2}} α \Rightarrow

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

= \frac{2}{\sqrt{1 - x^{2}}} {[a r c t a n (α)]}_{\frac{x}{\sqrt{1 - x^{2}}}}^{\frac{1 + x}{\sqrt{1 - x^{2}}}} = \frac{2}{\sqrt{1 - x^{2}}} {a r c t a n (\sqrt{\frac{1 + x}{1 - x}}) - a r c t a n (\frac{x}{\sqrt{1 - x^{2}}})} .

Answered by Smail last updated on 02/Jan/19

l e t u = t a n (t / 2) \Rightarrow d t = \frac{2 d u}{1 + u^{2}}

s i n t = \frac{2 u}{1 + u^{2}}

f (x) = 2 \int_{0}^{1} \frac{d u}{(1 + u^{2}) (1 + \frac{2 x u}{1 + u^{2}})}

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

i f - 1 < x ⩽ 1

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

θ = \frac{u + x}{\sqrt{1 - x^{2}}} \Rightarrow d θ = \frac{d u}{\sqrt{1 - x^{2}}}

f (x) = \frac{2}{\sqrt{1 - x^{2}}} \int_{x / \sqrt{1 - x^{2}}}^{(1 + x) / \sqrt{1 - x^{2}}} \frac{d θ}{θ^{2} + 1}

= \frac{2}{\sqrt{1 - x^{2}}} {[{t a n}^{- 1} (θ)]}_{x / \sqrt{1 - x^{2}}}^{(1 + x) / \sqrt{1 - x^{2}}}

= \frac{2}{\sqrt{1 - x^{2}}} ({t a n}^{- 1} (\frac{1 + x}{\sqrt{1 - x^{2}}}) - {t a n}^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}))

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