L=∫_0 ^1 (√((4−3x)/(4+5x)))dx


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Question Number 208871 by Shrodinger last updated on 26/Jun/24

$L = \int_{0}^{1} \sqrt{\frac{4 - 3 x}{4 + 5 x}} d x$
	Answered by Sutrisno last updated on 26/Jun/24

	$m i s a l$ $\sqrt{\frac{4 - 3 x}{4 + 5 x}} = p \to x = \frac{- 4 p^{2} + 4}{5 p^{2} + 3} \to d x = \frac{- 64 p}{{(5 p^{2} + 3)}^{2}} d p$ $L = \int_{1}^{\frac{1}{3}} p . \frac{- 64 p}{{(5 p^{2} + 3)}^{2}} d p$ $L = \int_{1}^{\frac{1}{3}} \frac{- 64 p^{2}}{{(5 p^{2} + 3)}^{2}} d p$ $m i s a l \sqrt{5} p = \sqrt{3} t a n θ \to d p = \frac{\sqrt{3} {s e c}^{2} θ}{\sqrt{5}} d θ$ $∙ s i n θ = \frac{\sqrt{5} p}{\sqrt{5 p^{2} + 3}} ∙ c o s θ = \frac{\sqrt{3} p}{\sqrt{5 p^{2} + 3}}$ $L = - 64 \int \frac{{(\frac{\sqrt{3}}{\sqrt{5}} t a n θ)}^{2}}{{(3 {t a n}^{2} θ + 3)}^{2}} . \frac{\sqrt{3} {s e c}^{2} θ}{\sqrt{5}} d θ$ $L = - \frac{64 \sqrt{3}}{15 \sqrt{5}} \int \frac{{t a n}^{2} θ {s e c}^{2} θ}{{({s e c}^{2} θ)}^{2}} d θ$ $L = - \frac{64 \sqrt{3}}{15 \sqrt{5}} \int \frac{{t a n}^{2} θ}{{s e c}^{2} θ} d θ$ $L = - \frac{64 \sqrt{3}}{15 \sqrt{5}} \int {s i n}^{2} θ d θ$ $L = - \frac{64 \sqrt{3}}{15 \sqrt{5}} \int \frac{1 - c o s 2 θ}{2} d θ$ $L = - \frac{32 \sqrt{3}}{15 \sqrt{5}} \int 1 - c o s 2 θ d θ$ $L = - \frac{32 \sqrt{3}}{15 \sqrt{5}} (θ - \frac{1}{2} s i n 2 θ)$ $L = - \frac{32 \sqrt{3}}{15 \sqrt{5}} (θ - s i n θ c o s θ)$ $L = - \frac{32 \sqrt{3}}{15 \sqrt{5}} ({t a n}^{- 1} (\frac{\sqrt{5} u}{\sqrt{3}}) - \frac{\sqrt{5} u}{\sqrt{5 u^{2} + 3}} . \frac{\sqrt{3}}{\sqrt{5 u^{2} + 3}})$ $L = - \frac{32 \sqrt{3}}{15 \sqrt{5}} {({t a n}^{- 1} (\frac{\sqrt{5} u}{\sqrt{3}}) - \frac{\sqrt{15} u}{5 u^{2} + 3})}_{1}^{\frac{1}{3}}$ $L = - \frac{32 \sqrt{3}}{15 \sqrt{5}} ({t a n}^{- 1} (\frac{\sqrt{5}}{3 \sqrt{3}}) - {t a n}^{- 1} (\frac{\sqrt{5}}{\sqrt{3}}) + \frac{1}{8} \sqrt{15})$
	Answered by mathzup last updated on 26/Jun/24
	$let (√((4−3x)/(4+5x)))=t ⇒((4−3x)/(4+5x))=t^2 ⇒ 4−3x=4t^2 +5t^2 x ⇔4−4t^2 =(5t^2 +3)x ⇔ x=((4−4t^2 )/(3+5t^2 )) and (dx/dt)=((−8t(3+5t^2 )−(4−4t^2 )10t)/((5t^2 +3)^2 )) =((−24t−40t^3 −40t+40t^3 )/((5t^2 +3)^2 ))=((−64t)/((5t^2 +3)^2 )) ⇒I=∫_1 ^(1/3) t(((−64t)/((5t^2 +3)^2 )))dt =((64)/5)∫_(1/3) ^1 ((5t^2 +3−3)/((5t^2 +3)^2 ))dt =((64)/5)∫_(1/3) ^1 (dt/(5t^2 +3)) −((3.64)/5)∫_(1/3) ^1 (dt/((5t^2 +3)^2 )) but ∫_(1/3) ^1 (dt/(5t^2 +3))=(1/5)∫_(1/3) ^1 (dt/(t^2 +((√(3/5)))^2 ))dt (t=(√(3/5))u) =(1/5)(√(3/5))∫_((1/3)(√(3/5))) ^(√(3/5)) (du/((3/5)(1+u^2 ))) =(1/5)(√(5/3)){arctan((√(3/5))−(1/3)(√(3/5))} ∫_(1/3) ^1 (dt/((5t^2 +3)^2 ))=(1/(25))∫_(1/3) ^1 (dt/((t^2 +(3/5))^2 )) (t=(√(3/5))tanθ) =(1/(25))×(√(3/5))∫_(arctan((1/3)(√(5/3)))) ^(arctan((√(5/3)))) ((1+tan^2 θ)/(((3/5))^2 (1+tan^2 θ))) =(1/(25))×(√(3/5))×((5/3))^2 ∫ (dθ/(1+tan^2 θ)) now its easy....$
	$l e t \sqrt{\frac{4 - 3 x}{4 + 5 x}} = t \Rightarrow \frac{4 - 3 x}{4 + 5 x} = t^{2} \Rightarrow$ $4 - 3 x = 4 t^{2} + 5 t^{2} x \Leftrightarrow 4 - 4 t^{2} = (5 t^{2} + 3) x \Leftrightarrow$ $x = \frac{4 - 4 t^{2}}{3 + 5 t^{2}} a n d \frac{d x}{d t} = \frac{- 8 t (3 + 5 t^{2}) - (4 - 4 t^{2}) 10 t}{{(5 t^{2} + 3)}^{2}}$ $= \frac{- 24 t - 40 t^{3} - 40 t + 40 t^{3}}{{(5 t^{2} + 3)}^{2}} = \frac{- 64 t}{{(5 t^{2} + 3)}^{2}}$ $\Rightarrow I = \int_{1}^{\frac{1}{3}} t (\frac{- 64 t}{{(5 t^{2} + 3)}^{2}}) d t$ $= \frac{64}{5} \int_{\frac{1}{3}}^{1} \frac{5 t^{2} + 3 - 3}{{(5 t^{2} + 3)}^{2}} d t$ $= \frac{64}{5} \int_{\frac{1}{3}}^{1} \frac{d t}{5 t^{2} + 3} - \frac{3.64}{5} \int_{\frac{1}{3}}^{1} \frac{d t}{{(5 t^{2} + 3)}^{2}}$ $b u t \int_{\frac{1}{3}}^{1} \frac{d t}{5 t^{2} + 3} = \frac{1}{5} \int_{\frac{1}{3}}^{1} \frac{d t}{t^{2} + {(\sqrt{\frac{3}{5}})}^{2}} d t (t = \sqrt{\frac{3}{5}} u)$ $= \frac{1}{5} \sqrt{\frac{3}{5}} \int_{\frac{1}{3} \sqrt{\frac{3}{5}}}^{\sqrt{\frac{3}{5}}} \frac{d u}{\frac{3}{5} (1 + u^{2})}$ $= \frac{1}{5} \sqrt{\frac{5}{3}} {a r c t a n (\sqrt{\frac{3}{5}} - \frac{1}{3} \sqrt{\frac{3}{5}}}$ $\int_{\frac{1}{3}}^{1} \frac{d t}{{(5 t^{2} + 3)}^{2}} = \frac{1}{25} \int_{\frac{1}{3}}^{1} \frac{d t}{{(t^{2} + \frac{3}{5})}^{2}} (t = \sqrt{\frac{3}{5}} t a n θ)$ $= \frac{1}{25} \times \sqrt{\frac{3}{5}} \int_{a r c t a n (\frac{1}{3} \sqrt{\frac{5}{3}})}^{a r c t a n (\sqrt{\frac{5}{3}})} \frac{1 + {t a n}^{2} θ}{{(\frac{3}{5})}^{2} (1 + {t a n}^{2} θ)}$ $= \frac{1}{25} \times \sqrt{\frac{3}{5}} \times {(\frac{5}{3})}^{2} \int \frac{d θ}{1 + {t a n}^{2} θ}$ $n o w i t s e a s y . . . .$
	Answered by Frix last updated on 26/Jun/24

	$\int \sqrt{\frac{4 - 3 x}{4 + 5 x}} d x \overset{t = \sqrt{\frac{3 (4 + 5 x)}{5 (4 - 3 x)}}}{=} \frac{64 \sqrt{15}}{75} \int \frac{d t}{{(t^{2} + 1)}^{2}} =$ $= \frac{32 \sqrt{15}}{75} (\frac{t}{t^{2} + 1} + \tan^{- 1} t) =$ $= \frac{\sqrt{(4 - 3 x) (4 + 5 x)}}{5} + \frac{32 \sqrt{15}}{75} \tan^{- 1} \sqrt{\frac{3 (4 + 5 x)}{5 (4 - 3 x)}} + C$ $\Rightarrow$ $L = - \frac{1}{5} + \frac{32 \sqrt{15}}{75} \tan^{- 1} \frac{\sqrt{15}}{7}$
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