let f(x) =∫_0 ^π (dt/(x+sint)) with xreal 1) find a explicit form of f(x) 2) find also g(x) =∫_0 ^π (dt/((x+sint)^2 )) 3) give f^((n)) (x) at form of integral 4) calculate ∫_0 ^π (dt/(3+sint)) and ∫


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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
$1)f(x) =∫_0 ^π (dt/(x+sint)) changement tan((t/2))=u give f(x) =∫_0 ^∞ (1/(x+((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) =∫_0 ^∞ ((2du)/(x+xu^2 +2u)) =∫_0 ^∞ ((2du)/(xu^2 +2u +x)) xu^2 +2u +x =0→Δ^′ =1−x^2 case 1 1−x^2 >0 ⇒∣x∣<1 ⇒u_1 =((−1+(√(1−x^2 )))/x) and u_2 =((−1−(√(1−x^2 )))/x) ( x≠0) ⇒f(x) =∫_0 ^∞ ((2du)/(x(u−u_1 )(u−u_2 ))) =(2/x)(1/((2(√(1−x^2 )))/x))∫_0 ^∞ ((1/(u−u_1 ))−(1/(u−u_2 )))du =(1/(√(1−x^2 )))[ln∣((u−u_1 )/(u−u_2 ))∣]_0 ^(+∞) =(1/(√(1−x^2 ))){−ln(∣(u_1 /u_2 )∣)} =(1/(√(1−x^2 )))ln∣((1+(√(1−x^2 )))/(1−(√(1−x^2 ))))∣ case 2 1−x^2 <0 ⇒∣x∣>1 ⇒xu^2 +2u +x =x(u^2 +((2u)/x) +1) =x{ u^2 +2(u/x) +(1/x^2 ) +1−(1/x^2 )} =x{ (u+(1/x))^2 +((x^2 −1)/x^2 )} we use the changement u+(1/x) =((√(x^2 −1))/(∣x∣)) u ⇒ f(x) = ∫_0 ^∞ ((2du)/(x{(u+(1/x))^(2 ) +((x^2 −1)/x^2 )})) =(2/x)∫_0 ^∞ (1/(((x^2 −1)/x^2 )(1+u^2 )))((√(x^2 −1))/(∣x∣)) du =((2x^2 )/(x∣x∣)) ((√(x^2 −1))/(x^2 −1)) ∫_0 ^∞ (du/(1+u^2 )) =2ξ(x)(1/(√(x^2 −1))) (π/2) =((πξ(x))/(√(x^2 −1))) with ξ(x) =1 if x>0 and ξ(x)=−1 if x<0 x =0 ⇒f(x)=∫_0 ^π (dt/(sint)) =_(tan((t/2))=u) ∫_0 ^∞ (1/((2u)/(1+u^2 ))) ((2du)/(1+u^2 )) =∫_0 ^∞ (du/u) f(x)dont exist (the integral diverges!)$
$1) f (x) = \int_{0}^{π} \frac{d t}{x + s i n t} c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e$ $f (x) = \int_{0}^{\infty} \frac{1}{x + \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int_{0}^{\infty} \frac{2 d u}{x + {x u}^{2} + 2 u} = \int_{0}^{\infty} \frac{2 d u}{{x u}^{2} + 2 u + x}$ ${x u}^{2} + 2 u + x = 0 \to Δ^{^{'}} = 1 - x^{2}$ $c a s e 1 1 - x^{2} > 0 \Rightarrow∣ x ∣< 1 \Rightarrow u_{1} = \frac{- 1 + \sqrt{1 - x^{2}}}{x} a n d u_{2} = \frac{- 1 - \sqrt{1 - x^{2}}}{x}$ $(x \neq 0) \Rightarrow f (x) = \int_{0}^{\infty} \frac{2 d u}{x (u - u_{1}) (u - u_{2})}$ $= \frac{2}{x} \frac{1}{\frac{2 \sqrt{1 - x^{2}}}{x}} \int_{0}^{\infty} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) d u = \frac{1}{\sqrt{1 - x^{2}}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{0}^{+ \infty}$ $= \frac{1}{\sqrt{1 - x^{2}}} {- l n (∣ \frac{u_{1}}{u_{2}} ∣)} = \frac{1}{\sqrt{1 - x^{2}}} l n ∣ \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} ∣$ $c a s e 2 1 - x^{2} < 0 \Rightarrow∣ x ∣> 1 \Rightarrow {x u}^{2} + 2 u + x = x (u^{2} + \frac{2 u}{x} + 1)$ $= x {u^{2} + 2 \frac{u}{x} + \frac{1}{x^{2}} + 1 - \frac{1}{x^{2}}} = x {{(u + \frac{1}{x})}^{2} + \frac{x^{2} - 1}{x^{2}}} w e u s e t h e$ $c h a n g e m e n t u + \frac{1}{x} = \frac{\sqrt{x^{2} - 1}}{∣ x ∣} u \Rightarrow$ $f (x) = \int_{0}^{\infty} \frac{2 d u}{x {{(u + \frac{1}{x})}^{2} + \frac{x^{2} - 1}{x^{2}}}} = \frac{2}{x} \int_{0}^{\infty} \frac{1}{\frac{x^{2} - 1}{x^{2}} (1 + u^{2})} \frac{\sqrt{x^{2} - 1}}{∣ x ∣} d u$ $= \frac{2 x^{2}}{x ∣ x ∣} \frac{\sqrt{x^{2} - 1}}{x^{2} - 1} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} = 2 ξ (x) \frac{1}{\sqrt{x^{2} - 1}} \frac{π}{2} = \frac{π ξ (x)}{\sqrt{x^{2} - 1}} w i t h$ $ξ (x) = 1 i f x > 0 a n d ξ (x) = - 1 i f x < 0$ $x = 0 \Rightarrow f (x) = \int_{0}^{π} \frac{d t}{s i n t} =_{t a n (\frac{t}{2}) = u} \int_{0}^{\infty} \frac{1}{\frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int_{0}^{\infty} \frac{d u}{u}$ $f (x) d o n t e x i s t (t h e i n t e g r a l d i v e r g e s!)$
Commented by mathmax by abdo last updated on 23/Jul/19
$2) we have f^′ (x) =−∫_0 ^π (dt/((x +sint)^2 )) =−g(x) ⇒g(x)=−f^′ (x) if ∣x∣>1 f(x) =((πξ(x))/(√(x^2 −1))) ⇒f^′ (x) =πξ(x){(x^2 −1)^(−(1/2)) }^((1)) =πξ(x)(−(1/2))(2x)(x^2 −1)^(−(3/2)) =−((πxξ(x))/((x^2 −1)(√(x^2 −1)))) ⇒ g(x) =((πxξ(x))/((x^2 −1)(√(x^2 −1)))) if ∣x∣<1 (x≠0) f(x)=(1/(√(1−x^2 )))ln∣((1+(√(1−x^2 )))/(1−(√(1−x^2 ))))∣ ⇒ f^′ (x)={(1−x^2 )^(−(1/2)) }^((1)) ln∣((1+(√(1−x^2 )))/(1−(√(1−x^2 ))))∣+(1/(√(1−x^2 ))) ((w^′ (x))/(w(x))) with w(x) =((1+(√(1−x^2 )))/(1−(√(1−x^2 )))) ⇒w^′ (x) =((((−2x)/(2(√(1−x^2 ))))(1−(√(1−x^2 )))−((x/(√(1−x^2 ))))(1+(√(1−x^2 ))))/((1−(√(1−x^2 )))^2 )) =((−(x/(√(1−x^2 ))) +x −(x/(√(1−x^2 )))−x)/((1−(√(1−x^2 )))^2 )) =((−2x)/((√(1−x^2 ))(1−(√(1−x^2 ))))) ⇒ f^′ (x) =(x/((1−x^2 )(√(1−x^2 ))))ln∣((1+(√(1−x^2 )))/(1−(√(1−x^2 ))))∣+(1/(√(1−x^2 ))) ((−2x)/((√(1−x^2 ))(1−(√(1−x^2 )))))×((1−(√(1−x^2 )))/(1+(√(1−x^2 )))) =....$
$2) w e h a v e f^{^{'}} (x) = - \int_{0}^{π} \frac{d t}{{(x + s i n t)}^{2}} = - g (x) \Rightarrow g (x) = - f^{^{'}} (x)$ $i f ∣ x ∣> 1 f (x) = \frac{π ξ (x)}{\sqrt{x^{2} - 1}} \Rightarrow f^{^{'}} (x) = π ξ (x) {{(x^{2} - 1)}^{- \frac{1}{2}}}^{(1)}$ $= π ξ (x) (- \frac{1}{2}) (2 x) {(x^{2} - 1)}^{- \frac{3}{2}} = - \frac{π x ξ (x)}{(x^{2} - 1) \sqrt{x^{2} - 1}} \Rightarrow$ $g (x) = \frac{π x ξ (x)}{(x^{2} - 1) \sqrt{x^{2} - 1}}$ $i f ∣ x ∣< 1 (x \neq 0) f (x) = \frac{1}{\sqrt{1 - x^{2}}} l n ∣ \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} ∣ \Rightarrow$ $f^{^{'}} (x) = {{(1 - x^{2})}^{- \frac{1}{2}}}^{(1)} l n ∣ \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} ∣ + \frac{1}{\sqrt{1 - x^{2}}} \frac{w^{^{'}} (x)}{w (x)}$ $w i t h w (x) = \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} \Rightarrow w^{^{'}} (x) = \frac{\frac{- 2 x}{2 \sqrt{1 - x^{2}}} (1 - \sqrt{1 - x^{2}}) - (\frac{x}{\sqrt{1 - x^{2}}}) (1 + \sqrt{1 - x^{2}})}{{(1 - \sqrt{1 - x^{2}})}^{2}}$ $= \frac{- \frac{x}{\sqrt{1 - x^{2}}} + x - \frac{x}{\sqrt{1 - x^{2}}} - x}{{(1 - \sqrt{1 - x^{2}})}^{2}} = \frac{- 2 x}{\sqrt{1 - x^{2}} (1 - \sqrt{1 - x^{2}})} \Rightarrow$ $f^{^{'}} (x) = \frac{x}{(1 - x^{2}) \sqrt{1 - x^{2}}} l n ∣ \frac{1 + \sqrt{1 - x^{2}}}{1 - \sqrt{1 - x^{2}}} ∣ + \frac{1}{\sqrt{1 - x^{2}}} \frac{- 2 x}{\sqrt{1 - x^{2}} (1 - \sqrt{1 - x^{2}})} \times \frac{1 - \sqrt{1 - x^{2}}}{1 + \sqrt{1 - x^{2}}}$ $= . . . .$
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

$4) \int_{0}^{π} \frac{d t}{3 + s i n t} = f (3) = \frac{π \times 1}{\sqrt{3^{2} - 1}} = \frac{π}{\sqrt{8}} = \frac{π}{2 \sqrt{2}}$ $\int_{0}^{π} \frac{d t}{{(3 + s i n t)}^{2}} = g (3) = \frac{3 π \times 1}{(3^{2} - 1) \sqrt{3^{2} - 1}} = \frac{3 π}{8 \sqrt{8}} = \frac{3 π}{8 (2 \sqrt{2})} = \frac{3 π}{16 \sqrt{2}} .$
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