find-f-a-0-pi-4-dx-1-acos-2-x-a-from-R-

Question Number 44304 by abdo.msup.com last updated on 26/Sep/18

f i n d f (a) = \int_{0}^{\frac{π}{4}} \frac{d x}{1 + {a c o s}^{2} x}

a f r o m R .

Commented by maxmathsup by imad last updated on 27/Sep/18

f (a) = \int_{0}^{\frac{π}{4}} \frac{d x}{1 + {a c o s}^{2} x} = \int_{0}^{\frac{π}{4}} \frac{d x}{1 + a \frac{1 + c o s (2 x)}{2}}

= \int_{0}^{\frac{π}{4}} \frac{2 d x}{2 + a + a c o s (2 x)} =_{2 x = t} \int_{0}^{\frac{π}{2}} \frac{d t}{2 + a + a c o s t} c h a n g e m e n t t a n (\frac{t}{2}) = u

g i v e f (a) = \int_{0}^{1} \frac{1}{2 + a + a \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int_{0}^{1} \frac{2 d u}{2 + a + (2 + a) u^{2} + a - {a u}^{2}}

= \int_{0}^{1} \frac{2 d u}{2 u^{2} + 2 + 2 a} = \int_{0}^{1} \frac{d u}{u^{2} + 1 + a}

c a s e 1 1 + a > 0 \Rightarrow f (a) =_{u = \sqrt{1 + a} α} \int_{0}^{\sqrt{1 + a}} \frac{\sqrt{1 + a} d α}{(1 + a) (1 + α^{2})}

= \frac{1}{\sqrt{1 + a}} {[a r c t a n (α)]}_{0}^{\sqrt{1 + a}} = \frac{a r c t a n (\sqrt{1 + a})}{\sqrt{1 + a}}

c a s e 2 1 + a < 0 \Rightarrow f (a) = \int_{0}^{1} \frac{d u}{u^{2} - {(\sqrt{- 1 - a})}^{2}}

= \frac{1}{2 \sqrt{- 1 - a}} \int_{0}^{1} {\frac{1}{u - \sqrt{- 1 - a}} - \frac{1}{u + \sqrt{- 1 - a}}} d u

= \frac{1}{2 \sqrt{- 1 - a}} {[l n ∣ \frac{u - \sqrt{- 1 - a}}{u + \sqrt{- 1 - a}} ∣]}_{0}^{1} = \frac{1}{2 \sqrt{- 1 - a}} l n ∣ \frac{1 - \sqrt{- 1 - a}}{1 + \sqrt{- 1 - a}} ∣ .

Commented by maxmathsup by imad last updated on 27/Sep/18

i f a = - 1 f (a) i s d i v e r g e n t .

Answered by Smail last updated on 27/Sep/18

f (a) = \int_{0}^{π / 4} \frac{d x}{1 + {a c o s}^{2} x}

t = t a n x \Rightarrow d x = \frac{d t}{1 + t^{2}}

f (a) = \int_{0}^{1} \frac{d t}{(1 + t^{2}) (1 + \frac{a}{1 + t^{2}})}

= \int_{0}^{1} \frac{d t}{1 + t^{2} + a} = \int_{0}^{1} \frac{d t}{(1 + a) (\frac{t^{2}}{a + 1} + 1)}

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

u = \frac{t}{\sqrt{a + 1}} \Rightarrow d t = \sqrt{a + 1} d u

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

= \frac{\sqrt{a + 1}}{a + 1} {[{t a n}^{- 1} (u)]}_{0}^{1 / \sqrt{a + 1}}

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

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