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Question Number 64873 by mathmax by abdo last updated on 22/Jul/19

l e t f (x) = \int_{0}^{π} \frac{d t}{x + s i n t} w i t h x r e a l

1) f i n d a e x p l i c i t f o r m o f f (x)

2) f i n d a l s o g (x) = \int_{0}^{π} \frac{d t}{{(x + s i n t)}^{2}}

3) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l

4) c a l c u l a t e \int_{0}^{π} \frac{d t}{3 + s i n t} a n d \int_{0}^{π} \frac{d t}{{(3 + s i n t)}^{2}}

Commented by mathmax by abdo last updated on 23/Jul/19

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

{x u}^{2} + 2 u + x = 0 \to Δ^{^{'}} = 1 - x^{2}

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

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

= \frac{2}{x} \frac{1}{\frac{2 \sqrt{1 - x^{2}}}{x}} \int_{0}^{\infty} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) d u = \frac{1}{\sqrt{1 - x^{2}}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{0}^{+ \infty}

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

c a s e 2 1 - x^{2} < 0 \Rightarrow∣ x ∣> 1 \Rightarrow {x u}^{2} + 2 u + x = x (u^{2} + \frac{2 u}{x} + 1)

= x {u^{2} + 2 \frac{u}{x} + \frac{1}{x^{2}} + 1 - \frac{1}{x^{2}}} = x {{(u + \frac{1}{x})}^{2} + \frac{x^{2} - 1}{x^{2}}} w e u s e t h e

c h a n g e m e n t u + \frac{1}{x} = \frac{\sqrt{x^{2} - 1}}{∣ x ∣} u \Rightarrow

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

= \frac{2 x^{2}}{x ∣ x ∣} \frac{\sqrt{x^{2} - 1}}{x^{2} - 1} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} = 2 ξ (x) \frac{1}{\sqrt{x^{2} - 1}} \frac{π}{2} = \frac{π ξ (x)}{\sqrt{x^{2} - 1}} w i t h

ξ (x) = 1 i f x > 0 a n d ξ (x) = - 1 i f x < 0

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

f (x) d o n t e x i s t (t h e i n t e g r a l d i v e r g e s!)

Commented by mathmax by abdo last updated on 23/Jul/19

2) w e h a v e f^{^{'}} (x) = - \int_{0}^{π} \frac{d t}{{(x + s i n t)}^{2}} = - g (x) \Rightarrow g (x) = - f^{^{'}} (x)

i f ∣ x ∣> 1 f (x) = \frac{π ξ (x)}{\sqrt{x^{2} - 1}} \Rightarrow f^{^{'}} (x) = π ξ (x) {{(x^{2} - 1)}^{- \frac{1}{2}}}^{(1)}

= π ξ (x) (- \frac{1}{2}) (2 x) {(x^{2} - 1)}^{- \frac{3}{2}} = - \frac{π x ξ (x)}{(x^{2} - 1) \sqrt{x^{2} - 1}} \Rightarrow

g (x) = \frac{π x ξ (x)}{(x^{2} - 1) \sqrt{x^{2} - 1}}

i f ∣ x ∣< 1 (x \neq 0) f (x) = \frac{1}{\sqrt{1 - x^{2}}} l n ∣ \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} ∣ \Rightarrow

f^{^{'}} (x) = {{(1 - x^{2})}^{- \frac{1}{2}}}^{(1)} l n ∣ \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} ∣ + \frac{1}{\sqrt{1 - x^{2}}} \frac{w^{^{'}} (x)}{w (x)}

w i t h w (x) = \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} \Rightarrow w^{^{'}} (x) = \frac{\frac{- 2 x}{2 \sqrt{1 - x^{2}}} (1 - \sqrt{1 - x^{2}}) - (\frac{x}{\sqrt{1 - x^{2}}}) (1 + \sqrt{1 - x^{2}})}{{(1 - \sqrt{1 - x^{2}})}^{2}}

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

f^{^{'}} (x) = \frac{x}{(1 - x^{2}) \sqrt{1 - x^{2}}} l n ∣ \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} ∣ + \frac{1}{\sqrt{1 - x^{2}}} \frac{- 2 x}{\sqrt{1 - x^{2}} (1 - \sqrt{1 - x^{2}})} \times \frac{1 - \sqrt{1 - x^{2}}}{1 + \sqrt{1 - x^{2}}}

= \dots .

Commented by mathmax by abdo last updated on 23/Jul/19

3) w e h a v e f (x) = \int_{0}^{π} \frac{d t}{(x + s i n t)} \Rightarrow f^{(n)} (x) = \int_{0}^{π} \frac{{(- 1)}^{n} n!}{{(x + s i n t)}^{n + 1}} d t

= {(- 1)}^{n} n! \int_{0}^{π} \frac{d t}{{(x + s i n t)}^{n + 1}}

Commented by mathmax by abdo last updated on 23/Jul/19

4) \int_{0}^{π} \frac{d t}{3 + s i n t} = f (3) = \frac{π \times 1}{\sqrt{3^{2} - 1}} = \frac{π}{\sqrt{8}} = \frac{π}{2 \sqrt{2}}

\int_{0}^{π} \frac{d t}{{(3 + s i n t)}^{2}} = g (3) = \frac{3 π \times 1}{(3^{2} - 1) \sqrt{3^{2} - 1}} = \frac{3 π}{8 \sqrt{8}} = \frac{3 π}{8 (2 \sqrt{2})} = \frac{3 π}{16 \sqrt{2}} .