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Question Number 106948 by abdomathmax last updated on 08/Aug/20

let f (a) = \int_{0}^{π} \frac{dx}{a + \cos^{2} x} with a > 0

1) explicite f (a)

2) explicite g (a) = \int_{0}^{π} \frac{dx}{{(a + \cos^{2} x)}^{2}}

3) find tbe valued of intevrsls

\int_{0}^{π} \frac{dx}{1 + \cos^{2} x} and \int_{0}^{π} \frac{dx}{{(1 + \cos^{2} x)}^{2}}

Answered by Ar Brandon last updated on 08/Aug/20

f (a) = \int_{0}^{π} \frac{dx}{a + \cos^{2} x} = \int_{0}^{π} \frac{\sec^{2} x}{{asec}^{2} x + 1} dx = \int_{0}^{π} \frac{d (tanx)}{a (1 + \tan^{2} x) + 1}

= \int_{0}^{π} \frac{d (tanx)}{(a + 1) + {atan}^{2} x} = \frac{1}{a} \int_{0}^{π} \frac{d (tanx)}{(\frac{a + 1}{a}) + \tan^{2} x}

= \frac{1}{a} \cdot \sqrt{\frac{a}{a + 1}} {[Arctan (\sqrt{\frac{a}{a + 1}} \cdot tanx)]}_{0}^{π}

= \frac{π}{a} \cdot \sqrt{\frac{a}{a + 1}} = \frac{π}{\sqrt{a^{2} + a}}

Answered by Ar Brandon last updated on 08/Aug/20

g (a) = \int_{0}^{π} \frac{dx}{{(a + \cos^{2} x)}^{2}} = \int_{0}^{π} \frac{\sec^{4} x}{{({asec}^{2} x + 1)}^{2}} dx

= \int_{0}^{π} \frac{\sec^{2} x \cdot d (tanx)}{{(a + {atan}^{2} x + 1)}^{2}} = \int_{0}^{π} \frac{1 + \tan^{2} x}{{(a + 1 + {atan}^{2} x)}^{2}} d (tanx)

= \int_{0}^{- 0} \frac{1 + t^{2}}{{(a + 1 + {at}^{2})}^{2}} dt = \int_{0}^{- 0} \frac{1 + t^{2}}{{(a + 1)}^{2} + 2 a (a + 1) + a^{2} t^{4}} dt

= \frac{1}{2 a} \int_{0}^{- 0} \frac{(a + 1 + {at}^{2}) - (a + 1 - {at}^{2})}{{(a + 1)}^{2} + 2 a (a + 1) + a^{2} t^{4}} dt + \int_{0}^{- 0} \frac{1}{{(a + 1)}^{2} + 2 a (a + 1) + a^{2} t^{4}} dt

= \frac{1}{2 a} {\int_{0}^{- 0} \frac{(a + 1) + {at}^{2}}{{(a + 1)}^{2} + 2 a (a + 1) + a^{2} t^{4}} dt

- \int_{0}^{- 0} \frac{(a + 1) - {at}^{2}}{{(a + 1)}^{2} + 2 a (a + 1) + a^{2} t^{4}} dt \begin{matrix}  \end{matrix}} + I

= \frac{1}{2 a} {\int_{0}^{- 0} \frac{1 + \frac{a + 1}{{at}^{2}}}{{(\frac{a + 1}{at})}^{2} + \frac{2 (a + 1)}{t^{2}} + {at}^{2}} + \int_{0}^{- 0} \frac{1 - \frac{a + 1}{{at}^{2}}}{{(\frac{a + 1}{at})}^{2} + \frac{2 (a + 1)}{t^{2}} + {at}^{2}} dt} + I

\dots

Answered by Ar Brandon last updated on 08/Aug/20

f (1) = \int_{0}^{π} \frac{dx}{1 + \cos^{2} x} = \frac{π}{\sqrt{1^{2} + 1}} = \frac{π}{\sqrt{2}}

g (1) = \int_{0}^{π} \frac{dx}{{(1 + \cos^{2} x)}^{2}} = \int_{0}^{π} \frac{\sec^{4} x}{{(\sec^{2} x + 1)}^{2}} dx

= \int_{0}^{π} \frac{1 + \tan^{2} x}{{(2 + \tan^{2} x)}^{2}} d (tanx) = \int_{0}^{- 0} \frac{1 + t^{2}}{{(2 + t^{2})}^{2}} dt

= \int_{0}^{- 0} \frac{2 + t^{2}}{{(2 + t^{2})}^{2}} dt - \int_{0}^{- 0} \frac{1}{{(2 + t^{2})}^{2}} dt

\underset{t = \sqrt{2} \tan θ}{=} \int_{0}^{- 0} \frac{dt}{2 + t^{2}} - \frac{1}{4} \int_{0}^{π} \frac{\sqrt{2} \sec^{2} θ d θ}{{(1 + \tan^{2} θ)}^{2}}

= {[\frac{1}{\sqrt{2}} Arctan (\frac{t}{\sqrt{2}})]}_{0}^{- 0} - \frac{\sqrt{2}}{4} \int_{0}^{π} \frac{d θ}{\sec^{2} θ}

= \frac{π}{\sqrt{2}} - \frac{\sqrt{2}}{4} \cdot \frac{1}{2} \int_{0}^{π} (1 + \cos 2 θ) d θ = \frac{π}{\sqrt{2}} - \frac{1}{4 \sqrt{2}} {[θ + \frac{\sin 2 θ}{2}]}_{0}^{π}

= \frac{π}{\sqrt{2}} - \frac{π}{4 \sqrt{2}} = \frac{3 π}{4 \sqrt{2}}

Commented by Ar Brandon last updated on 08/Aug/20

You're welcome, senior.��

Commented by abdomathmax last updated on 08/Aug/20

thank you sir brandon

Answered by mathmax by abdo last updated on 08/Aug/20

1) residus method f (a) = \int_{0}^{π} \frac{dx}{a + \cos^{2} x} \Rightarrow f (a) = \int_{0}^{π} \frac{dx}{a + \frac{1 + \cos (2 x)}{2}}

= 2 \int_{0}^{π} \frac{dx}{2 a + 1 + \cos (2 x)} =_{2 x = t} 2 \int_{0}^{2 π} \frac{dt}{2 (2 a + 1 + cost)} = \int_{0}^{2 π} \frac{dt}{2 a + 1 + cost}

=_{e^{it} = z} \int_{∣ z ∣= 1} \frac{dz}{iz (2 a + 1 + \frac{z + z^{- 1}}{2})} = \int \frac{- 2 idz}{z (4 a + 2 + z + z^{- 1})}

= \int \frac{- 2 idz}{(4 a + 2) z + z^{2} + 1} let φ (z) = \frac{- 2 i}{z^{2} + (4 a + 2) z + 1} poles of φ ?

Δ^{^{'}} = {(2 a + 1)}^{2} - 1 = {4 a}^{2} + 4 a + 1 - 1 = {4 a}^{2} + 4 a \Rightarrow

z_{1} = - (2 a + 1) + 2 \sqrt{a^{2} + a} and z_{2} = - (2 a + 1) - 2 \sqrt{a^{2} + a}

∣ z_{1} ∣ - 1 =∣ 2 \sqrt{a^{2} + a} - (2 a + 1) ∣ - 1 = 2 a + 1 - 2 \sqrt{a^{2} + a} - 1

= 2 a - 2 \sqrt{a^{2} + a} = 2 (a - \sqrt{a^{2} + a}) < 0 \Rightarrow∣ z_{1} ∣< 1

∣ z_{2} ∣ - 1 = 2 a + 1 + 2 \sqrt{a^{2} + a} - 1 > 0 \Rightarrow∣ z_{2} ∣> 1 (to eliminate from residus)

\Rightarrow \int_{∣ z ∣= 1} φ (2) dz = 2 i π Res (φ, z_{1}) = 2 i π \times \frac{- 2 i}{z_{1} - z_{2}}

= \frac{4 π}{4 \sqrt{a^{2} + a}} = \frac{π}{\sqrt{a^{2} + a}} \Rightarrow f (a) = \frac{π}{\sqrt{a^{2} + a}} (a > 0)

2) we have f^{^{'}} (a) = - \int_{0}^{π} \frac{dx}{{(a + \cos^{2} x)}^{2}} = - g (a) \Rightarrow

g (a) = - f^{^{'}} (a) but f^{^{'}} (a) = - π \times \frac{{(\sqrt{a^{2} + a})}^{^{'}}}{a^{2} + a \Rightarrow} = - π \times \frac{2 a + 1}{2 (a^{2} + a) \sqrt{a^{2} + a}}

\Rightarrow g (a) = \frac{(2 a + 1) π}{2 (a^{2} + a) \sqrt{a^{2} + a}}

3) \int_{0}^{π} \frac{dx}{1 + \cos^{2} x} = f (1) = \frac{π}{\sqrt{2}}

\int_{0}^{π} \frac{dx}{{(1 + \cos^{2} x)}^{2}} = g (1) = \frac{3 π}{4 \sqrt{2}}