Question Number 179478 by Acem last updated on 29/Oct/22
$${Find}\:\int_{\:\mathrm{0}} ^{\:\frac{\mathrm{1}}{\mathrm{6}}} \:\frac{\mathrm{1}}{{x}^{−\mathrm{5}} \:\left(\mathrm{36}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{dx} \\ $$
Answered by MJS_new last updated on 29/Oct/22
![∫(dx/(x^(−5) (36x^2 +1)^(3/2) ))= [t=(√(36x^2 +1)) → dx=((√(36x^2 +1))/(36x))dt] =(1/(46656))∫(((t^2 −1)^2 )/t^2 )dt=(1/(46656))∫(t^2 −2+(1/t^2 ))dt= =((t^4 −6t^2 −3)/(139968t))=((162x^4 −18x^2 −1)/(17496(√(36x^2 +1))))+C ⇒ answer is ((16−11(√2))/(279936))](https://www.tinkutara.com/question/Q179481.png)
$$\int\frac{{dx}}{{x}^{−\mathrm{5}} \left(\mathrm{36}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{36}{x}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{36}{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{36}{x}}{dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{46656}}\int\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} }{dt}=\frac{\mathrm{1}}{\mathrm{46656}}\int\left({t}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}= \\ $$$$=\frac{{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} −\mathrm{3}}{\mathrm{139968}{t}}=\frac{\mathrm{162}{x}^{\mathrm{4}} −\mathrm{18}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{17496}\sqrt{\mathrm{36}{x}^{\mathrm{2}} +\mathrm{1}}}+{C} \\ $$$$\Rightarrow \\ $$$$\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{16}−\mathrm{11}\sqrt{\mathrm{2}}}{\mathrm{279936}} \\ $$
Commented by Acem last updated on 29/Oct/22
$${Nice}\:{friend}! \\ $$
Answered by ARUNG_Brandon_MBU last updated on 29/Oct/22
![I=∫_0 ^(1/6) (1/(x^(−5) (36x^2 +1)^(3/2) ))dx=∫_0 ^(1/6) (x^5 /((36x^2 +1)^(3/2) ))dx, x=((tanθ)/6) ⇒dx=((sec^2 θ)/6)dθ =∫_0 ^(π/4) (((tan^5 θ)/6^5 )/((tan^2 θ+1)^(3/2) ))(((sec^2 θ)/6)dθ)=(1/6^6 )∫_0 ^(π/4) ((tan^5 θ)/(secθ))dθ =(1/6^6 )∫_0 ^(π/4) ((sin^5 θ)/(cos^4 θ))dθ=(1/6^6 )∫_0 ^(π/4) ((sin^4 θ)/(cos^4 θ))sinθdθ=−(1/6^6 )∫_0 ^(π/4) (((1−cos^2 θ)^2 )/(cos^4 θ))d(cosθ) =−(1/6^6 )∫_0 ^(π/4) ((1−2cos^2 θ+cos^4 θ)/(cos^4 θ))d(cosθ)=−(1/6^6 )[−(1/(3cos^3 θ))+(2/(cosθ))+cosθ]_0 ^(π/4) =(1/6^6 )(((2(√2))/3)−2(√2)−((√2)/2)−(1/3)+2+1)=(1/6^6 )((8/3)−((11(√2))/6))=((16−11(√2))/6^7 )](https://www.tinkutara.com/question/Q179489.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{6}}} \frac{\mathrm{1}}{{x}^{−\mathrm{5}} \left(\mathrm{36}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{6}}} \frac{{x}^{\mathrm{5}} }{\left(\mathrm{36}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx},\:{x}=\frac{\mathrm{tan}\theta}{\mathrm{6}}\:\Rightarrow{dx}=\frac{\mathrm{sec}^{\mathrm{2}} \theta}{\mathrm{6}}{d}\theta \\ $$$$\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\frac{\mathrm{tan}^{\mathrm{5}} \theta}{\mathrm{6}^{\mathrm{5}} }}{\left(\mathrm{tan}^{\mathrm{2}} \theta+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\left(\frac{\mathrm{sec}^{\mathrm{2}} \theta}{\mathrm{6}}{d}\theta\right)=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{tan}^{\mathrm{5}} \theta}{\mathrm{sec}\theta}{d}\theta \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sin}^{\mathrm{5}} \theta}{\mathrm{cos}^{\mathrm{4}} \theta}{d}\theta=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sin}^{\mathrm{4}} \theta}{\mathrm{cos}^{\mathrm{4}} \theta}\mathrm{sin}\theta{d}\theta=−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{4}} \theta}{d}\left(\mathrm{cos}\theta\right) \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}−\mathrm{2cos}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{4}} \theta}{\mathrm{cos}^{\mathrm{4}} \theta}{d}\left(\mathrm{cos}\theta\right)=−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\left[−\frac{\mathrm{1}}{\mathrm{3cos}^{\mathrm{3}} \theta}+\frac{\mathrm{2}}{\mathrm{cos}\theta}+\mathrm{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{2}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\left(\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{11}\sqrt{\mathrm{2}}}{\mathrm{6}}\right)=\frac{\mathrm{16}−\mathrm{11}\sqrt{\mathrm{2}}}{\mathrm{6}^{\mathrm{7}} } \\ $$
Commented by Acem last updated on 29/Oct/22
![Great friend! Important! for (√((1+ tan^2 θ)^3 )) = ∣sec θ∣^3 be aware, it must compting values of θ and see if it ∈ ]−(π/2), (π/2)[ or not if it is then you can deem that ∣sec θ∣^3 = sec^3 θ otherwise ∣sec θ∣^3 = − sec^3 θ](https://www.tinkutara.com/question/Q179493.png)
$${Great}\:{friend}!\: \\ $$$$ \\ $$$$\boldsymbol{{Important}}!\:{for}\:\sqrt{\left(\mathrm{1}+\:\mathrm{tan}^{\mathrm{2}} \:\theta\right)^{\mathrm{3}} }\:=\:\mid\mathrm{sec}\:\theta\mid^{\mathrm{3}} \\ $$$$\:{be}\:{aware},\:{it}\:{must}\:{compting}\:{values}\:{of}\:\theta\:{and} \\ $$$$\left.\:{see}\:{if}\:{it}\:\in\:\right]−\frac{\pi}{\mathrm{2}},\:\frac{\pi}{\mathrm{2}}\left[\:{or}\:{not}\right. \\ $$$$\:{if}\:{it}\:{is}\:{then}\:{you}\:{can}\:{deem}\:{that}\:\mid\mathrm{sec}\:\theta\mid^{\mathrm{3}} =\:\mathrm{sec}^{\mathrm{3}} \:\theta \\ $$$$\:{otherwise}\:\mid\mathrm{sec}\:\theta\mid^{\mathrm{3}} =\:−\:\mathrm{sec}^{\mathrm{3}} \:\theta \\ $$$$ \\ $$
Answered by Acem last updated on 29/Oct/22
![x= (1/6) tan θ , dx= (1/6) sec^2 θ dθ (36x^2 +1)^(3/2) = (√((tan^2 θ+1)^3 ))= ∣sec θ∣^3 x=0 ⇒ θ= 0 , x= (1/6) ⇒ θ= (π/4) as θ ∈ ]−(π/2), (π/2)[ ⇒ ∣sec θ∣^3 = sec^3 θ a= (1/6^6 ) ∫_0 ^( (π/4)) ((tan^5 θ sec^2 θ)/(sec^3 θ)) dθ= (1/6^6 ) ∫_0 ^( (π/4)) ((sin^4 θ)/(cos^4 θ)) sin θ dθ u= cos θ , du= −sin θ dθ θ_1 = 0 ⇒ u_1 = 1 , θ_2 = (π/4) ⇒ u_2 = (1/( (√2))) a= ((−1)/6^6 ) ∫_1 ^( (1/( (√2)))) (((1−u^2 )^2 )/u^4 ) du = ((−1)/6^6 ) (u + (2/u) − (1/(3u^3 ))) ∣_( 1) ^(1/( (√2))) a= (1/6^7 ) (16− 11 (√2))](https://www.tinkutara.com/question/Q179492.png)
$$\:{x}=\:\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{tan}\:\theta\:,\:{dx}=\:\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{sec}^{\mathrm{2}} \:\theta\:{d}\theta \\ $$$$\:\left(\mathrm{36}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\:\sqrt{\left(\mathrm{tan}^{\mathrm{2}} \:\theta+\mathrm{1}\right)^{\mathrm{3}} }=\:\mid\mathrm{sec}\:\theta\mid^{\mathrm{3}} \\ $$$$\:{x}=\mathrm{0}\:\Rightarrow\:\theta=\:\mathrm{0}\:\:\:\:\:,\:\:\:\:{x}=\:\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow\:\theta=\:\frac{\pi}{\mathrm{4}} \\ $$$$\left.\:{as}\:\theta\:\in\:\right]−\frac{\pi}{\mathrm{2}},\:\frac{\pi}{\mathrm{2}}\left[\:\:\Rightarrow\:\mid\mathrm{sec}\:\theta\mid^{\mathrm{3}} \:=\:\mathrm{sec}^{\mathrm{3}} \:\theta\right. \\ $$$$\:{a}=\:\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{tan}^{\mathrm{5}} \:\theta\:\mathrm{sec}^{\mathrm{2}} \:\theta}{\mathrm{sec}^{\mathrm{3}} \:\theta}\:{d}\theta=\:\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{sin}^{\mathrm{4}} \:\theta}{\mathrm{cos}^{\mathrm{4}} \:\theta}\:\mathrm{sin}\:\theta\:{d}\theta \\ $$$$\:{u}=\:\mathrm{cos}\:\theta\:,\:{du}=\:−\mathrm{sin}\:\theta\:{d}\theta \\ $$$$\:\theta_{\mathrm{1}} =\:\mathrm{0}\:\Rightarrow\:{u}_{\mathrm{1}} =\:\mathrm{1}\:\:\:\:,\:\:\:\theta_{\mathrm{2}} =\:\frac{\pi}{\mathrm{4}}\:\Rightarrow\:{u}_{\mathrm{2}} =\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:{a}=\:\frac{−\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\:\int_{\mathrm{1}} ^{\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\frac{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{{u}^{\mathrm{4}} }\:{du} \\ $$$$\:\:\:\:=\:\frac{−\mathrm{1}}{\mathrm{6}^{\mathrm{6}} }\:\left({u}\:+\:\frac{\mathrm{2}}{{u}}\:−\:\frac{\mathrm{1}}{\mathrm{3}{u}^{\mathrm{3}} }\right)\:\:\mid_{\:\:\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \\ $$$$\:{a}=\:\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{7}} }\:\left(\mathrm{16}−\:\mathrm{11}\:\sqrt{\mathrm{2}}\right) \\ $$