let f(x) =∫_(π/4) ^(π/3) (dt/(2+xsint)) 1) find a explicit form of f(x) 2)determine also g(x)=∫_(π/4) ^(π/3) ((sint)/((2+xsint)^2 ))dt 3) find the value of ∫_(π/4) ^(π/3) (dt/(2+3sint)) and ∫


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Question Number 58487 by Mr X pcx last updated on 23/Apr/19

$l e t f (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{d t}{2 + x s i n t}$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) d e t e r m i n e a l s o g (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{s i n t}{{(2 + x s i n t)}^{2}} d t$ $3) f i n d t h e v a l u e o f \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{d t}{2 + 3 s i n t}$ $a n d \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{s i n t}{{(2 + 3 s i n t)}^{2}} d t$
Commented by maxmathsup by imad last updated on 24/Apr/19
$changement tan((t/2)) =u give f(x) =∫_((√2)−1) ^(1/(√3)) ((2du)/((1+u^2 )(2+x ((2u)/(1+u^2 ))))) = ∫_((√2)−1) ^(1/(√3)) ((2du)/(2(1+u^2 ) +2xu)) = ∫_((√2)−1) ^(1/(√3)) (du/(u^2 +2xu +1)) =∫_((√2)−1) ^(1/(√3)) (du/((u+x)^2 +1−x^2 )) case 1 if 1−x^2 >0 ⇒∣x∣<1 we do the changement u+x =(√(1−x^2 ))α ⇒f(x) = ∫_(((√2)−1 +x)/(√(1−x^2 ))) ^(((1/(√3))+x)/(√(1−x^2 ))) (((√(1−x^2 ))dα)/((1−x^2 )(1+α^2 ))) =(1/(√(1−x^2 ))) [ arctanα]_(((√2)−1+x)/(√(1−x^2 ))) ^(((1/(√3))+x)/(√(1−x^2 ))) ⇒f(x) =(1/(√(1−x^2 ))){ arctan(((1+x(√3))/((√3)(√(1−x^2 )))))−arctan((((√2)−1+x)/(√(1−x^2 ))))} case 2 if 1−x^2 <0 ⇒∣x∣>1 ⇒f(x)=∫_((√2)−1) ^(1/(√3)) (du/((u+x)^2 −((√(x^2 −1)))^2 )) =∫_((√2)−1) ^(1/(√3)) (du/((u+x+(√(x^2 −1)))(u+x−(√(x^2 −1))))) =(1/(2(√(x^2 −1))))∫_((√2)−1) ^(1/(√3)) {(1/(u+x −(√(x^2 −1)))) −(1/(u+x +(√(x^2 −1))))}du =(1/(2(√(x^2 −1)))) [ln∣((u+x −(√(x^2 −1)))/(u+x +(√(x^2 −1)))) ∣]_(u=(√2)−1) ^(u=(1/(√3))) =(1/(2(√(x^2 −1)))){ln∣(((1/((√3) ))+x−(√(x^2 −1)))/((1/(√3))+x+(√(x^2 −1))))∣−ln∣(((√2)−1+x −(√(x^2 −1)))/((√2)−1+x +(√(x^2 −1))))∣ } .$
$c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e f (x) = \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} \frac{2 d u}{(1 + u^{2}) (2 + x \frac{2 u}{1 + u^{2}})}$ $= \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} \frac{2 d u}{2 (1 + u^{2}) + 2 x u} = \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} \frac{d u}{u^{2} + 2 x u + 1} = \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} \frac{d u}{{(u + x)}^{2} + 1 - x^{2}}$ $c a s e 1 i f 1 - x^{2} > 0 \Rightarrow∣ x ∣< 1 w e d o t h e c h a n g e m e n t u + x = \sqrt{1 - x^{2}} α$ $\Rightarrow f (x) = \int_{\frac{\sqrt{2} - 1 + x}{\sqrt{1 - x^{2}}}}^{\frac{\frac{1}{\sqrt{3}} + x}{\sqrt{1 - x^{2}}}} \frac{\sqrt{1 - x^{2}} d α}{(1 - x^{2}) (1 + α^{2})}$ $= \frac{1}{\sqrt{1 - x^{2}}} {[a r c t a n α]}_{\frac{\sqrt{2} - 1 + x}{\sqrt{1 - x^{2}}}}^{\frac{\frac{1}{\sqrt{3}} + x}{\sqrt{1 - x^{2}}}} \Rightarrow f (x) = \frac{1}{\sqrt{1 - x^{2}}} {a r c t a n (\frac{1 + x \sqrt{3}}{\sqrt{3} \sqrt{1 - x^{2}}}) - a r c t a n (\frac{\sqrt{2} - 1 + x}{\sqrt{1 - x^{2}}})}$ $c a s e 2 i f 1 - x^{2} < 0 \Rightarrow∣ x ∣> 1 \Rightarrow f (x) = \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} \frac{d u}{{(u + x)}^{2} - {(\sqrt{x^{2} - 1})}^{2}}$ $= \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} \frac{d u}{(u + x + \sqrt{x^{2} - 1}) (u + x - \sqrt{x^{2} - 1})}$ $= \frac{1}{2 \sqrt{x^{2} - 1}} \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} {\frac{1}{u + x - \sqrt{x^{2} - 1}} - \frac{1}{u + x + \sqrt{x^{2} - 1}}} d u$ $= \frac{1}{2 \sqrt{x^{2} - 1}} {[l n ∣ \frac{u + x - \sqrt{x^{2} - 1}}{u + x + \sqrt{x^{2} - 1}} ∣]}_{u = \sqrt{2} - 1}^{u = \frac{1}{\sqrt{3}}}$ $= \frac{1}{2 \sqrt{x^{2} - 1}} {l n ∣ \frac{\frac{1}{\sqrt{3}} + x - \sqrt{x^{2} - 1}}{\frac{1}{\sqrt{3}} + x + \sqrt{x^{2} - 1}} ∣ - l n ∣ \frac{\sqrt{2} - 1 + x - \sqrt{x^{2} - 1}}{\sqrt{2} - 1 + x + \sqrt{x^{2} - 1}} ∣} .$
Commented by maxmathsup by imad last updated on 25/Apr/19

$2) w e h a v e f^{^{'}} (x) = \int_{\frac{π}{4}}^{\frac{π}{3}} - \frac{s i n t}{{(2 + x s i n t)}^{2}} d t = - g (x) \Rightarrow g (x) = - f^{^{'}} (x) r e s t t o c a l c u l a t e$ $f^{^{'}} (x)$
Commented by maxmathsup by imad last updated on 25/Apr/19
$3) ∫_(π/4) ^(π/3) (dt/(2+3cost)) =f(3) =(1/(2(√(3^2 −1)))){ln∣(((1/(√3))+3−(√(3^2 −1)))/((1/(√3)) +3 +(√(3^2 −1))))∣−ln∣(((√2)−1 +3−(√(3^2 −1)))/((√2)−1 +3+(√(3^2 −1))))∣ =(1/(4(√2))){ln∣(((1/(√3))+3−2(√2))/((1/((√3) )) +3+2(√2)))∣ −ln∣((2−(√2))/(2+3(√2)))∣ .$
$3) \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{d t}{2 + 3 c o s t} = f (3) = \frac{1}{2 \sqrt{3^{2} - 1}} {l n ∣ \frac{\frac{1}{\sqrt{3}} + 3 - \sqrt{3^{2} - 1}}{\frac{1}{\sqrt{3}} + 3 + \sqrt{3^{2} - 1}} ∣ - l n ∣ \frac{\sqrt{2} - 1 + 3 - \sqrt{3^{2} - 1}}{\sqrt{2} - 1 + 3 + \sqrt{3^{2} - 1}} ∣$ $= \frac{1}{4 \sqrt{2}} {l n ∣ \frac{\frac{1}{\sqrt{3}} + 3 - 2 \sqrt{2}}{\frac{1}{\sqrt{3}} + 3 + 2 \sqrt{2}} ∣ - l n ∣ \frac{2 - \sqrt{2}}{2 + 3 \sqrt{2}} ∣ .$
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