Menu Close

pi-2-0-tanx-1-3-sinx-cosx-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 193439 by Nimnim111118 last updated on 14/Jun/23
∫^(π/2) _( 0) (((tanx))^(1/3) /((sinx+cosx)^2 ))dx
$$\underset{\:\:\mathrm{0}} {\int}^{\pi/\mathrm{2}} \frac{\sqrt[{\mathrm{3}}]{{tanx}}}{\left({sinx}+{cosx}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by Nimnim111118 last updated on 14/Jun/23
help...
$${help}… \\ $$
Commented by Frix last updated on 14/Jun/23
Use t=((tan x))^(1/3) ⇒ 3∫(t^3 /((t^3 +1)^2 ))dt Now decompose etc I get ((2(√3))/9)π
$$\mathrm{Use}\:{t}=\sqrt[{\mathrm{3}}]{\mathrm{tan}\:{x}} \\ $$$$\Rightarrow \\ $$$$\mathrm{3}\int\frac{{t}^{\mathrm{3}} }{\left({t}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$\mathrm{Now}\:\mathrm{decompose}\:\mathrm{etc} \\ $$$$\mathrm{I}\:\mathrm{get}\:\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\pi \\ $$
Answered by MM42 last updated on 14/Jun/23
I=∫ (((tanx))^(1/3) /((sinx+cosx)^2 ))dx=∫(((tanx))^(1/3) /(1+sin2x))dx= ∫(((tanx))^(1/3) /((1+tanx)^2 ))(1+tan^2 x)dx ; let : tanx=u^3 ⇒(1+tan^2 x)dx=3u^2 du x→0⇒u→0 & x→(π/2)⇒u→∞ ⇒I=∫((3u^3 )/((1+u^3 )^2 )) du ∫(u^3 /((1+u^3 ))) du = ((au^2 +bu+c)/(1+u^3 )) + ∫ ((a^′ u^2 +b′u+c′)/(1+u^3 )) du (ostrohrdsky) =−(2/3)×(u/(u^3 +1))+(2/3)∫ (du/(u^3 +1)) ∫ (du/(u^3 +1)) =∫ ((a/(u+1))+((bu+c)/(u^2 −u+1)))du ( decomposition of fraction) =(1/3)∫ ((1/(u+1))−((u−2)/(u^2 −u+1)))du ⇒I=−(1/3)×(u/(u^3 +1))+(2/9)×∫ ((1/(u+1))−(1/2)×((2u−1−3)/(u^2 −u+1)))du= −(1/3)×(u/(u^3 +1))+(2/9)×∫ ((1/(u+1))−(1/2)×((2u−1)/(u^2 −u+1))+(3/2)×(1/((u−(1/2))^2 +(3/4))))du= ⇒I=−(1/3)×(u/(u^3 +1))+(2/9) ln(u+1)−(1/9)ln(u^2 −u+1)+(2/(3(√3)))tan^(−1) (((2u−1)/( (√3)))) ⇒ans=−(1/3)×(u/(u^3 +1)) +(1/9)ln((((u+1)^2 )/(u^2 −u+1)))+(1/( (√3))) tan^(−1) (((2u−1)/( (√3)))) ∣_0 ^∞ =(π/( 2(√3)))+(π/( 6(√3)))=((2(√(3 ))π)/( 9 )) ✓
$${I}=\int\:\frac{\sqrt[{\mathrm{3}}]{{tanx}}}{\left({sinx}+{cosx}\right)^{\mathrm{2}} }{dx}=\int\frac{\sqrt[{\mathrm{3}}]{{tanx}}}{\mathrm{1}+{sin}\mathrm{2}{x}}{dx}= \\ $$$$\int\frac{\sqrt[{\mathrm{3}}]{{tanx}}}{\left(\mathrm{1}+{tanx}\right)^{\mathrm{2}} }\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx}\:\:\:\:;\:\:{let}\::\:{tanx}={u}^{\mathrm{3}} \Rightarrow\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx}=\mathrm{3}{u}^{\mathrm{2}} {du} \\ $$$${x}\rightarrow\mathrm{0}\Rightarrow{u}\rightarrow\mathrm{0}\:\:\:\&\:\:{x}\rightarrow\frac{\pi}{\mathrm{2}}\Rightarrow{u}\rightarrow\infty \\ $$$$\Rightarrow{I}=\int\frac{\mathrm{3}{u}^{\mathrm{3}} }{\left(\mathrm{1}+{u}^{\mathrm{3}} \right)^{\mathrm{2}} }\:{du} \\ $$$$\int\frac{{u}^{\mathrm{3}} }{\left(\mathrm{1}+{u}^{\mathrm{3}} \right)}\:{du}\:=\:\frac{{au}^{\mathrm{2}} +{bu}+{c}}{\mathrm{1}+{u}^{\mathrm{3}} }\:+\:\int\:\:\frac{{a}^{'} {u}^{\mathrm{2}} +{b}'{u}+{c}'}{\mathrm{1}+{u}^{\mathrm{3}} }\:{du}\:\:\:\left({ostrohrdsky}\right) \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}×\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{3}}\int\:\frac{{du}}{{u}^{\mathrm{3}} +\mathrm{1}}\: \\ $$$$\int\:\frac{{du}}{{u}^{\mathrm{3}} +\mathrm{1}}\:=\int\:\left(\frac{{a}}{{u}+\mathrm{1}}+\frac{{bu}+{c}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}\right){du}\:\:\left(\:{decomposition}\:{of}\:{fraction}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\:\left(\frac{\mathrm{1}}{{u}+\mathrm{1}}−\frac{{u}−\mathrm{2}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}\right){du} \\ $$$$\Rightarrow{I}=−\frac{\mathrm{1}}{\mathrm{3}}×\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{9}}×\int\:\left(\frac{\mathrm{1}}{{u}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{u}−\mathrm{1}−\mathrm{3}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}\right){du}= \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}×\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{9}}×\int\:\left(\frac{\mathrm{1}}{{u}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{u}−\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{1}}{\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}\right){du}= \\ $$$$\Rightarrow{I}=−\frac{\mathrm{1}}{\mathrm{3}}×\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{9}}\:{ln}\left({u}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{9}}{ln}\left({u}^{\mathrm{2}} −{u}+\mathrm{1}\right)+\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{u}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\: \\ $$$$\Rightarrow{ans}=−\frac{\mathrm{1}}{\mathrm{3}}×\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{9}}{ln}\left(\frac{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} −{u}+\mathrm{1}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{u}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\mid_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\pi}{\:\mathrm{2}\sqrt{\mathrm{3}}}+\frac{\pi}{\:\mathrm{6}\sqrt{\mathrm{3}}}=\frac{\mathrm{2}\sqrt{\mathrm{3}\:}\pi}{\:\mathrm{9}\:}\:\checkmark \\ $$$$ \\ $$
Commented by Frix last updated on 15/Jun/23
I=∫((3u^3 )/((u^3 +1)^2 ))du I=−(u/(u^3 +1))+(1/6)ln ((((u+1)^2 )/(u^2 −u+1))) +(1/( (√3)))tan^(−1) ((2u−1)/( (√3))) +C ⇒ Your result is wrong.
$${I}=\int\frac{\mathrm{3}{u}^{\mathrm{3}} }{\left({u}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }{du} \\ $$$${I}=−\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left(\frac{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} −{u}+\mathrm{1}}\right)\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{2}{u}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+{C} \\ $$$$\Rightarrow \\ $$$$\mathrm{Your}\:\mathrm{result}\:\mathrm{is}\:\mathrm{wrong}. \\ $$
Commented by MM42 last updated on 14/Jun/23
why “ I=∫ ((3u^2 )/((u^3 +1)^2 )) du ” ??? in addition : ∫ ((3u^2 )/((u^3 +1)^2 )) du=((−1)/(u^3 +1)) +c ∵
$${why}\:\:“\:{I}=\int\:\frac{\mathrm{3}{u}^{\mathrm{2}} }{\left({u}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }\:{du}\:''\:??? \\ $$$${in}\:{addition}\:: \\ $$$$\int\:\frac{\mathrm{3}{u}^{\mathrm{2}} }{\left({u}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }\:{du}=\frac{−\mathrm{1}}{{u}^{\mathrm{3}} +\mathrm{1}}\:+{c}\:\because \\ $$$$ \\ $$
Commented by MM42 last updated on 14/Jun/23
thank you for your mention.
$${thank}\:{you}\:{for}\:{your}\:{mention}. \\ $$
Commented by Frix last updated on 15/Jun/23
Typo, I corrected it 2→3
$$\mathrm{Typo},\:\mathrm{I}\:\mathrm{corrected}\:\mathrm{it}\:\mathrm{2}\rightarrow\mathrm{3} \\ $$
Commented by Tawa11 last updated on 16/Jun/23
MM42 you are MrW?? I can see mrW hand writing in all MM42 solutions. Sm sorry if am wrong. Just that I miss mrW or you are mjs?
$$\mathrm{MM42}\:\mathrm{you}\:\mathrm{are}\:\mathrm{MrW}?? \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{see}\:\mathrm{mrW}\:\mathrm{hand}\:\mathrm{writing}\:\mathrm{in}\:\mathrm{all}\:\mathrm{MM42}\:\mathrm{solutions}. \\ $$$$\mathrm{Sm}\:\mathrm{sorry}\:\mathrm{if}\:\mathrm{am}\:\mathrm{wrong}. \\ $$$$\mathrm{Just}\:\mathrm{that}\:\mathrm{I}\:\mathrm{miss}\:\mathrm{mrW} \\ $$$$\mathrm{or}\:\mathrm{you}\:\mathrm{are}\:\mathrm{mjs}? \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *