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L-0-1-4-3x-4-5x-dx-

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Question Number 208871 by Shrodinger last updated on 26/Jun/24
L=∫_0 ^1 (√((4−3x)/(4+5x)))dx
$${L}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{4}−\mathrm{3}{x}}{\mathrm{4}+\mathrm{5}{x}}}{dx} \\ $$
Answered by Sutrisno last updated on 26/Jun/24
misal (√((4−3x)/(4+5x)))=p→x=((−4p^2 +4)/(5p^2 +3))→dx=((−64p)/((5p^2 +3)^2 ))dp L=∫_1 ^(1/3) p.((−64p)/((5p^2 +3)^2 ))dp L=∫_1 ^(1/3) ((−64p^2 )/((5p^2 +3)^2 ))dp misal (√5)p=(√3)tanθ→dp=(((√3)sec^2 θ)/( (√5)))dθ •sinθ=(((√5)p)/( (√(5p^2 +3)))) •cosθ=(((√3)p)/( (√(5p^2 +3)))) L=−64∫(((((√3)/( (√5)))tanθ)^2 )/((3tan^2 θ+3)^2 )).(((√3)sec^2 θ)/( (√5)))dθ L=−((64(√3))/(15(√5)))∫((tan^2 θsec^2 θ)/((sec^2 θ)^2 ))dθ L=−((64(√3))/(15(√5)))∫((tan^2 θ)/(sec^2 θ))dθ L=−((64(√3))/(15(√5)))∫sin^2 θdθ L=−((64(√3))/(15(√5)))∫((1−cos2θ)/2)dθ L=−((32(√3))/(15(√5)))∫1−cos2θdθ L=−((32(√3))/(15(√5)))(θ−(1/2)sin2θ) L=−((32(√3))/(15(√5)))(θ−sinθcosθ) L=−((32(√3))/(15(√5)))(tan^(−1) ((((√5)u)/( (√3))))−(((√5)u)/( (√(5u^2 +3)))).((√3)/( (√(5u^2 +3))))) L=−((32(√3))/(15(√5)))(tan^(−1) ((((√5)u)/( (√3))))−(((√(15))u)/( 5u^2 +3)))_1 ^(1/3) L=−((32(√3))/(15(√5)))(tan^(−1) (((√5)/( 3(√3))))−tan^(−1) (((√5)/( (√3))))+(1/8)(√(15)))
$${misal} \\ $$$$\sqrt{\frac{\mathrm{4}−\mathrm{3}{x}}{\mathrm{4}+\mathrm{5}{x}}}={p}\rightarrow{x}=\frac{−\mathrm{4}{p}^{\mathrm{2}} +\mathrm{4}}{\mathrm{5}{p}^{\mathrm{2}} +\mathrm{3}}\rightarrow{dx}=\frac{−\mathrm{64}{p}}{\left(\mathrm{5}{p}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dp} \\ $$$${L}=\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {p}.\frac{−\mathrm{64}{p}}{\left(\mathrm{5}{p}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dp} \\ $$$${L}=\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \frac{−\mathrm{64}{p}^{\mathrm{2}} }{\left(\mathrm{5}{p}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dp} \\ $$$${misal}\:\sqrt{\mathrm{5}}{p}=\sqrt{\mathrm{3}}{tan}\theta\rightarrow{dp}=\frac{\sqrt{\mathrm{3}}{sec}^{\mathrm{2}} \theta}{\:\sqrt{\mathrm{5}}}{d}\theta \\ $$$$\bullet{sin}\theta=\frac{\sqrt{\mathrm{5}}{p}}{\:\sqrt{\mathrm{5}{p}^{\mathrm{2}} +\mathrm{3}}}\:\:\bullet{cos}\theta=\frac{\sqrt{\mathrm{3}}{p}}{\:\sqrt{\mathrm{5}{p}^{\mathrm{2}} +\mathrm{3}}} \\ $$$${L}=−\mathrm{64}\int\frac{\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}}}{tan}\theta\right)^{\mathrm{2}} }{\left(\mathrm{3}{tan}^{\mathrm{2}} \theta+\mathrm{3}\right)^{\mathrm{2}} }.\frac{\sqrt{\mathrm{3}}{sec}^{\mathrm{2}} \theta}{\:\sqrt{\mathrm{5}}}{d}\theta \\ $$$${L}=−\frac{\mathrm{64}\sqrt{\mathrm{3}}}{\mathrm{15}\sqrt{\mathrm{5}}}\int\frac{{tan}^{\mathrm{2}} \theta{sec}^{\mathrm{2}} \theta}{\left({sec}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }{d}\theta \\ $$$${L}=−\frac{\mathrm{64}\sqrt{\mathrm{3}}}{\mathrm{15}\sqrt{\mathrm{5}}}\int\frac{{tan}^{\mathrm{2}} \theta}{{sec}^{\mathrm{2}} \theta}{d}\theta \\ $$$${L}=−\frac{\mathrm{64}\sqrt{\mathrm{3}}}{\mathrm{15}\sqrt{\mathrm{5}}}\int{sin}^{\mathrm{2}} \theta{d}\theta \\ $$$${L}=−\frac{\mathrm{64}\sqrt{\mathrm{3}}}{\mathrm{15}\sqrt{\mathrm{5}}}\int\frac{\mathrm{1}−{cos}\mathrm{2}\theta}{\mathrm{2}}{d}\theta \\ $$$${L}=−\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{15}\sqrt{\mathrm{5}}}\int\mathrm{1}−{cos}\mathrm{2}\theta{d}\theta \\ $$$${L}=−\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{15}\sqrt{\mathrm{5}}}\left(\theta−\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}\theta\right) \\ $$$${L}=−\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{15}\sqrt{\mathrm{5}}}\left(\theta−{sin}\theta{cos}\theta\right) \\ $$$${L}=−\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{15}\sqrt{\mathrm{5}}}\left({tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{5}}{u}}{\:\sqrt{\mathrm{3}}}\right)−\frac{\sqrt{\mathrm{5}}{u}}{\:\sqrt{\mathrm{5}{u}^{\mathrm{2}} +\mathrm{3}}}.\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}{u}^{\mathrm{2}} +\mathrm{3}}}\right) \\ $$$${L}=−\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{15}\sqrt{\mathrm{5}}}\left({tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{5}}{u}}{\:\sqrt{\mathrm{3}}}\right)−\frac{\sqrt{\mathrm{15}}{u}}{\:\mathrm{5}{u}^{\mathrm{2}} +\mathrm{3}}\right)_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${L}=−\frac{\mathrm{32}\sqrt{\mathrm{3}}}{\mathrm{15}\sqrt{\mathrm{5}}}\left({tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{5}}}{\:\mathrm{3}\sqrt{\mathrm{3}}}\right)−{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{3}}}\right)+\frac{\mathrm{1}}{\mathrm{8}}\sqrt{\mathrm{15}}\right) \\ $$$$ \\ $$
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....
$${let}\:\sqrt{\frac{\mathrm{4}−\mathrm{3}{x}}{\mathrm{4}+\mathrm{5}{x}}}={t}\:\Rightarrow\frac{\mathrm{4}−\mathrm{3}{x}}{\mathrm{4}+\mathrm{5}{x}}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{4}−\mathrm{3}{x}=\mathrm{4}{t}^{\mathrm{2}} +\mathrm{5}{t}^{\mathrm{2}} {x}\:\Leftrightarrow\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} =\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}\right){x}\:\Leftrightarrow \\ $$$${x}=\frac{\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} }{\mathrm{3}+\mathrm{5}{t}^{\mathrm{2}} }\:{and}\:\frac{{dx}}{{dt}}=\frac{−\mathrm{8}{t}\left(\mathrm{3}+\mathrm{5}{t}^{\mathrm{2}} \right)−\left(\mathrm{4}−\mathrm{4}{t}^{\mathrm{2}} \right)\mathrm{10}{t}}{\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\frac{−\mathrm{24}{t}−\mathrm{40}{t}^{\mathrm{3}} −\mathrm{40}{t}+\mathrm{40}{t}^{\mathrm{3}} }{\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }=\frac{−\mathrm{64}{t}}{\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{I}=\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {t}\left(\frac{−\mathrm{64}{t}}{\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }\right){dt} \\ $$$$=\frac{\mathrm{64}}{\mathrm{5}}\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{1}} \:\:\frac{\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}−\mathrm{3}}{\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{64}}{\mathrm{5}}\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}}\:−\frac{\mathrm{3}.\mathrm{64}}{\mathrm{5}}\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{1}} \frac{{dt}}{\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${but}\:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{5}}\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{1}} \frac{{dt}}{{t}^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{3}}{\mathrm{5}}}\right)^{\mathrm{2}} }{dt}\:\left({t}=\sqrt{\frac{\mathrm{3}}{\mathrm{5}}}{u}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\sqrt{\frac{\mathrm{3}}{\mathrm{5}}}\int_{\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\mathrm{3}}{\mathrm{5}}}} ^{\sqrt{\frac{\mathrm{3}}{\mathrm{5}}}} \:\:\:\frac{{du}}{\frac{\mathrm{3}}{\mathrm{5}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\left\{{arctan}\left(\sqrt{\frac{\mathrm{3}}{\mathrm{5}}}−\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\mathrm{3}}{\mathrm{5}}}\right\}\right. \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{1}} \frac{{dt}}{\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{25}}\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{1}} \frac{{dt}}{\left({t}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} }\:\left({t}=\sqrt{\frac{\mathrm{3}}{\mathrm{5}}}{tan}\theta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}×\sqrt{\frac{\mathrm{3}}{\mathrm{5}}}\int_{{arctan}\left(\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)} ^{{arctan}\left(\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\right)} \frac{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} \left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}×\sqrt{\frac{\mathrm{3}}{\mathrm{5}}}×\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} \int\:\:\frac{{d}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta} \\ $$$${now}\:{its}\:{easy}…. \\ $$
Answered by Frix last updated on 26/Jun/24
∫(√((4−3x)/(4+5x)))dx =^(t=(√((3(4+5x))/(5(4−3x))))) ((64(√(15)))/(75))∫(dt/((t^2 +1)^2 ))= =((32(√(15)))/(75))((t/(t^2 +1))+tan^(−1) t)= =((√((4−3x)(4+5x)))/5)+((32(√(15)))/(75))tan^(−1) (√((3(4+5x))/(5(4−3x)))) +C ⇒ L=−(1/5)+((32(√(15)))/(75))tan^(−1) ((√(15))/7)
$$\int\sqrt{\frac{\mathrm{4}−\mathrm{3}{x}}{\mathrm{4}+\mathrm{5}{x}}}{dx}\:\overset{{t}=\sqrt{\frac{\mathrm{3}\left(\mathrm{4}+\mathrm{5}{x}\right)}{\mathrm{5}\left(\mathrm{4}−\mathrm{3}{x}\right)}}} {=}\:\frac{\mathrm{64}\sqrt{\mathrm{15}}}{\mathrm{75}}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{32}\sqrt{\mathrm{15}}}{\mathrm{75}}\left(\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+\mathrm{tan}^{−\mathrm{1}} \:{t}\right)= \\ $$$$=\frac{\sqrt{\left(\mathrm{4}−\mathrm{3}{x}\right)\left(\mathrm{4}+\mathrm{5}{x}\right)}}{\mathrm{5}}+\frac{\mathrm{32}\sqrt{\mathrm{15}}}{\mathrm{75}}\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\frac{\mathrm{3}\left(\mathrm{4}+\mathrm{5}{x}\right)}{\mathrm{5}\left(\mathrm{4}−\mathrm{3}{x}\right)}}\:+{C} \\ $$$$\Rightarrow \\ $$$${L}=−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{32}\sqrt{\mathrm{15}}}{\mathrm{75}}\mathrm{tan}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{15}}}{\mathrm{7}} \\ $$

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