1)find f(x) =∫_0 ^(π/4) ((sint)/(2+x cos(2t)))dt 2) find g(x) =∫_0 ^(π/4) ((sint sin(2t)/((2+x cos(2t))^2 ))dx 3) find the value of ∫_0 ^(π/4) ((sint)/(2+3 cos(2t)))dt and ∫_0 ^(π/4) ((sin(t)sin(2t))/((2+3cos(2t))^2 ))dt


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Question Number 49635 by maxmathsup by imad last updated on 08/Dec/18

$1) f i n d f (x) = \int_{0}^{\frac{π}{4}} \frac{s i n t}{2 + x c o s (2 t)} d t$ $2) f i n d g (x) = \int_{0}^{\frac{π}{4}} \frac{s i n t s i n (2 t}{{(2 + x c o s (2 t))}^{2}} d x$ $3) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{4}} \frac{s i n t}{2 + 3 c o s (2 t)} d t a n d \int_{0}^{\frac{π}{4}} \frac{s i n (t) s i n (2 t)}{{(2 + 3 c o s (2 t))}^{2}} d t$
Commented by Abdo msup. last updated on 15/Dec/18
$1) we have f(x) =∫_0 ^(π/4) ((sint)/(2+x(2cos^2 t−1)))dt =∫_0 ^(π/4) ((sint)/(2xcos^2 t +2−x))dt =_(cost =u) ∫_1 ^(1/(√2)) ((−du)/(2xu^2 +2−x)) =−∫_1 ^(1/(√2)) (du/(2x(u^2 +((2−x)/(2x))))) =(1/(2x)) ∫_((√2)/2) ^1 (du/(u^2 +((2−x)/(2x)))) if ((2−x)/(2x)) >0 we do the changement u=(√((2−x)/(2x)))α f(x)=(1/(2x)) ∫_(((√2)/2)(√((2x)/(2−x)))) ^(√((2x)/(2−x))) (1/(((2−x)/(2x))(1+α^2 ))) (√((2−x)/(2x)))dα =(1/(2−x))(√((2−x)/(2x))) [arctan(α)]_(((√2)/2)(√((2x)/(2−x)))) ^(√((2x)/(2−x))) f(x)= (1/(√((2−x)2x))) { arctan((√((2x)/(2−x))))−arctan(((√2)/2)(√((2x)/(2−x))))} if ((2−x)/(2x))<0 we get f(x)=(1/(2x)) ∫_((√2)/2) ^1 (du/(u^2 −((x−2)/(2x)))) and wedo the changement u=(√((x−2)/(2x)))α and that lead to f(x)=(1/(2x)) ∫_(((√2)/2)(√((x−2)/(2x)))) ^(√((x−2)/(2x))) (1/(((x−2)/(2x))(α^2 −1)))(√((x−2)/(2x)))dα = (1/(√((x−2)2x))) (1/2)∫_(..) ^(..) {(1/(α−1)) −(1/(α+1))}dα =(1/(2(√(2x(x−2)))))[ln∣((α−1)/(α+1))∣]_(((√2)/2)(√((x−2)/(2x)))) ^(√((x−2)/(2x))) =(1/(2(√(2x(x−2))))) {ln∣(((√((x−2)/(2x)))−1)/((√((x−2)/(2x)))+1))∣−ln∣((((√2)/2)(√((x−2)/(2x)))−1)/(((√2)/2)(√((x−2)/(2x)))+1))∣.$
$1) w e h a v e f (x) = \int_{0}^{\frac{π}{4}} \frac{s i n t}{2 + x (2 {c o s}^{2} t - 1)} d t$ $= \int_{0}^{\frac{π}{4}} \frac{s i n t}{2 {x c o s}^{2} t + 2 - x} d t =_{c o s t = u} \int_{1}^{\frac{1}{\sqrt{2}}} \frac{- d u}{2 {x u}^{2} + 2 - x}$ $= - \int_{1}^{\frac{1}{\sqrt{2}}} \frac{d u}{2 x (u^{2} + \frac{2 - x}{2 x})} = \frac{1}{2 x} \int_{\frac{\sqrt{2}}{2}}^{1} \frac{d u}{u^{2} + \frac{2 - x}{2 x}}$ $i f \frac{2 - x}{2 x} > 0 w e d o t h e c h a n g e m e n t u = \sqrt{\frac{2 - x}{2 x}} α$ $f (x) = \frac{1}{2 x} \int_{\frac{\sqrt{2}}{2} \sqrt{\frac{2 x}{2 - x}}}^{\sqrt{\frac{2 x}{2 - x}}} \frac{1}{\frac{2 - x}{2 x} (1 + α^{2})} \sqrt{\frac{2 - x}{2 x}} d α$ $= \frac{1}{2 - x} \sqrt{\frac{2 - x}{2 x}} {[a r c t a n (α)]}_{\frac{\sqrt{2}}{2} \sqrt{\frac{2 x}{2 - x}}}^{\sqrt{\frac{2 x}{2 - x}}}$ $f (x) = \frac{1}{\sqrt{(2 - x) 2 x}} {a r c t a n (\sqrt{\frac{2 x}{2 - x}}) - a r c t a n (\frac{\sqrt{2}}{2} \sqrt{\frac{2 x}{2 - x}})}$ $i f \frac{2 - x}{2 x} < 0 w e g e t f (x) = \frac{1}{2 x} \int_{\frac{\sqrt{2}}{2}}^{1} \frac{d u}{u^{2} - \frac{x - 2}{2 x}} a n d w e d o$ $t h e c h a n g e m e n t u = \sqrt{\frac{x - 2}{2 x}} α a n d t h a t l e a d t o$ $f (x) = \frac{1}{2 x} \int_{\frac{\sqrt{2}}{2} \sqrt{\frac{x - 2}{2 x}}}^{\sqrt{\frac{x - 2}{2 x}}} \frac{1}{\frac{x - 2}{2 x} (α^{2} - 1)} \sqrt{\frac{x - 2}{2 x}} d α$ $= \frac{1}{\sqrt{(x - 2) 2 x}} \frac{1}{2} \int_{. .}^{. .} {\frac{1}{α - 1} - \frac{1}{α + 1}} d α$ $= \frac{1}{2 \sqrt{2 x (x - 2)}} {[l n ∣ \frac{α - 1}{α + 1} ∣]}_{\frac{\sqrt{2}}{2} \sqrt{\frac{x - 2}{2 x}}}^{\sqrt{\frac{x - 2}{2 x}}}$ $= \frac{1}{2 \sqrt{2 x (x - 2)}} {l n ∣ \frac{\sqrt{\frac{x - 2}{2 x}} - 1}{\sqrt{\frac{x - 2}{2 x}} + 1} ∣ - l n ∣ \frac{\frac{\sqrt{2}}{2} \sqrt{\frac{x - 2}{2 x}} - 1}{\frac{\sqrt{2}}{2} \sqrt{\frac{x - 2}{2 x}} + 1} ∣ .$
	Answered by Smail last updated on 09/Dec/18

	$f (x) = \int_{0}^{π / 4} \frac{s i n t}{2 + x c o s (2 t)} d t$ $= \int_{0}^{π / 4} \frac{s i n t}{2 + x (2 {c o s}^{2} (t) - 1)} d t$ $u = c o s t \Rightarrow d u = - s i n t d t$ $f (x) = - \int_{1}^{\sqrt{2} / 2} \frac{d u}{2 + 2 {x u}^{2} - x}$ $= - \int_{1}^{1 / \sqrt{2}} \frac{d u}{(2 - x) (\frac{2 {x u}^{2}}{2 - x} + 1)}$ $y = \sqrt{\frac{2 x}{2 - x}} u \Rightarrow d u = \sqrt{\frac{2 - x}{2 x}} d y$ $f (x) = - \frac{1}{2 - x} \times \sqrt{\frac{2 - x}{2 x}} \int_{\sqrt{2 x / (2 - x)}}^{\sqrt{x / (2 - x)}} \frac{d y}{y^{2} + 1}$ $= - \frac{1}{\sqrt{2 x (2 - x)}} [{t a n}^{- 1} (\sqrt{\frac{x}{2 - x}}) - {t a n}^{- 1} (\sqrt{\frac{2 x}{2 - x}})]$
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