1-calculate-f-a-dx-x-2-ax-1-with-a-lt-2-2-calculate-g-a-x-x-2-ax-1-2-3-find-values-of-integrals-dx-x-2-2-x-1-and-

Question Number 54376 by Abdo msup. last updated on 02/Feb/19

1) c a l c u l a t e f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + a x + 1}

w i t h ∣ a ∣< 2

2) c a l c u l a t e g (a) = \int_{- \infty}^{+ \infty} \frac{x}{{(x^{2} + a x + 1)}^{2}}

3) f i n d v a l u e s o f i n t e g r a l s \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + \sqrt{2} x + 1}

a n d \int_{- \infty}^{+ \infty} \frac{x}{{(x^{2} + \sqrt{2} x + 1)}^{2}}

4) c a l c u l a t e A (θ) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 2 c o s θ + 1}

θ i s a g i v e n r e a l .

Commented by maxmathsup by imad last updated on 03/Feb/19

1) w e h a v e f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + a x + 1} \Rightarrow f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 2 x \frac{a}{2} + \frac{a^{2}}{4} + 1 - \frac{a^{2}}{4}}

= \int_{- \infty}^{+ \infty} \frac{d x}{{(x + \frac{a}{2})}^{2} + \frac{4 - a^{2}}{4}} =_{x + \frac{a}{2} = \frac{1}{2} \sqrt{4 - a^{2}} t} \int_{- \infty}^{+ \infty} \frac{1}{\frac{4 - a^{2}}{4} (1 + t^{2})} \frac{\sqrt{4 - a^{2}} d t}{2}

= 2 \frac{1}{\sqrt{4 - a^{2}}} \int_{- \infty}^{+ \infty} \frac{d t}{1 + t^{2}} = \frac{2 π}{\sqrt{4 - a^{2}}} \Rightarrow f (a) = \frac{2 π}{\sqrt{4 - a^{2}}} .

2) w e h a v e f^{^{'}} (a) = - \int_{- \infty}^{+ \infty} \frac{x d x}{{(x^{2} + a x + 1)}^{2}} = - g (a) \Rightarrow

g (a) = - f^{^{'}} (a) = - 2 π {({(4 - a^{2})}^{- \frac{1}{2}})}^{^{'}} = - 2 π . (- \frac{1}{2}) (- 2 a) {(4 - a^{2})}^{- \frac{3}{2}}

= - 2 π a \frac{1}{(4 - a^{2}) \sqrt{4 - a^{2}}} = \frac{- 2 π a}{(4 - a^{2}) \sqrt{4 - a^{2}}}

Commented by maxmathsup by imad last updated on 03/Feb/19

3) \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + \sqrt{2} x + 1} = f (\sqrt{2}) = \frac{2 π}{\sqrt{4 - 2}} = \frac{2 π}{\sqrt{2}} = π \sqrt{2} .

\int_{- \infty}^{+ \infty} \frac{x d x}{{(x^{2} + \sqrt{2} x + 1)}^{2}} = g (\sqrt{2}) = \frac{- 2 π \sqrt{2}}{2 \sqrt{2}} = - π .

Commented by maxmathsup by imad last updated on 03/Feb/19

4) \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 2 c o s θ x + 1} = f (2 c o s θ) = \frac{2 π}{\sqrt{4 - 4 {c o s}^{2} θ}} = \frac{2 π}{2 \sqrt{1 - {c o s}^{2} θ}} = \frac{π}{∣ s i n θ ∣}

(w e s u p p o s e h e r e θ \neq k π k f r o m Z) .

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

2) \frac{1}{2} \int \frac{2 x + a - a}{{(x^{2} + a x + 1)}^{2}} d x

\frac{1}{2} \int \frac{d (x^{2} + a x + 1)}{{(x^{2} + a x + 1)}^{2}} d x - \frac{a}{2} \int \frac{d x}{{[{(x + \frac{a}{2})}^{2} + {(\sqrt{1 - \frac{a^{2}}{4}})}^{2}]}^{2}}

\frac{1}{2} \times \frac{- 1}{(x^{2} + a x + 1)} - \frac{a}{2} I_{2}

I_{2} l e t (x + \frac{a}{2}) = \sqrt{1 - \frac{a^{2}}{4}} t a n θ = k t a n θ

d x = k {s e c}^{2} θ d θ

s o I_{2} = \int \frac{{k s e c}^{2} θ d θ}{{[k^{2} {t a n}^{2} θ + k^{2}]}^{2}}

\int \frac{{k s e c}^{2} θ}{k^{4} {s e c}^{4} θ} d θ

\frac{1}{k^{3}} \int (\frac{1 + c o s 2 θ}{2}) d θ

\frac{1}{2 k^{3}} θ + \frac{1}{4 k^{3}} s i n 2 θ + c

\frac{1}{2 k^{3}} {t a n}^{- 1} (\frac{x + \frac{a}{2}}{k}) + \frac{1}{4 k^{3}} \times \frac{2 (\frac{x + \frac{a}{2}}{k})}{1 + {(\frac{x + \frac{a}{2}}{k})}^{2}} + c

s o

\frac{- 1}{2 (x^{2} + a x + 1)} - \frac{a}{2} [\frac{1}{2 k^{3}} {t a n}^{- 1} (\frac{x + \frac{a}{2}}{k}) + \frac{1}{2 k^{3}} \times \frac{(\frac{x + \frac{a}{2}}{k})}{1 + {(\frac{x + \frac{a}{2}}{k})}^{2}}]

∣ \frac{- 1}{2 (x^{2} + a x + 1)} - \frac{a}{4 k^{3}} [{t a n}^{- 1} (\frac{x + \frac{a}{2}}{k}) + \frac{\frac{x + \frac{a}{2}}{k}}{1 + {(\frac{x + \frac{a}{2}}{k})}^{2}}] ∣_{- \infty}^{\infty}

= 0 - \frac{a}{4 k^{3}} [{{t a n}^{- 1} (\infty) - {t a n}^{- 1} (- \infty)} + 0]

= - \frac{a}{4 k^{3}} \times \frac{π}{1} = \frac{- a π}{4 {(1 - \frac{a^{2}}{4})}^{\frac{3}{2}}}

3) s e c o n d p a r t = \frac{- \sqrt{2} π}{4 {(1 - \frac{1}{2})}^{\frac{3}{2}}} = \frac{- \sqrt{2} π}{4} \times \sqrt{2} \times \sqrt{2} \times \sqrt{2} = - π a n s w e r

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

1) \int \frac{d x}{x^{2} + 2 x \frac{a}{2} + \frac{a^{2}}{4} + 1 - \frac{a^{2}}{4}} d x

\int \frac{d x}{{(\sqrt{1 - \frac{a^{2}}{4}})}^{2} + {(x + \frac{a}{2})}^{2}}

s o a n s w e r i s

∣ \frac{1}{\sqrt{1 - \frac{a^{2}}{4}}} {t a n}^{- 1} (\frac{x + \frac{a}{2}}{\sqrt{1 - \frac{a^{2}}{4}}}) ∣_{- \infty}^{\infty} [\frac{a^{2}}{4} < 1]

= \frac{1}{\sqrt{1 - \frac{a^{2}}{4}}} \times {\frac{π}{2} - (- \frac{π}{2})} = \frac{π}{\sqrt{1 - \frac{a^{2}}{4}}}

3) \frac{π}{\sqrt{1 - \frac{2}{4}}} = π \sqrt{2}

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

4) \int_{- \infty}^{\infty} \frac{d x}{x^{2} + b^{2}} [[[b^{2} = 1 + 2 c o s θ]

\frac{1}{b} ∣ {t a n}^{- 1} (\frac{x}{b}) ∣_{- \infty}^{\infty}

\frac{1}{b} [{t a n}^{- 1} (\infty) - {t a n}^{- 1} (- \infty)]

= \frac{1}{\sqrt{1 + 2 c o s θ}} \times {\frac{π}{2} - \frac{- π}{2}}

= \frac{π}{\sqrt{1 + 2 c o s θ}}