Integral$int_{pi/6}^{pi/2} frac{sin^{1/3} (x)}{sin^{1/3} (x)+cos^{1/3} (x)}$












1












$begingroup$


$$int_{pi/6}^{pi/2} frac{sin^{1/3} (x)}{sin^{1/3} (x)+cos^{1/3} (x)}$$



I tried using the substitution $t^3=tan x$. Which gives me



$$intfrac{3t^3}{(t+1)(t^6+1)}$$



How should I proceed?






calculus definite-integrals







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  • $begingroup$
    Partial fractions? By the way, you seem to have lost your limits.
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 5:42










  • $begingroup$
    Typing on phone, so skipped it. Is a better way to type mathjax on phone
    $endgroup$
    – Piyush Divyanakar
    Dec 19 '18 at 5:43










  • $begingroup$
    @Piyush Divyanakar, you should try with lower limit $frac {π}{4}$
    $endgroup$
    – Awe Kumar Jha
    Dec 19 '18 at 12:39


















1












$begingroup$


$$int_{pi/6}^{pi/2} frac{sin^{1/3} (x)}{sin^{1/3} (x)+cos^{1/3} (x)}$$



I tried using the substitution $t^3=tan x$. Which gives me



$$intfrac{3t^3}{(t+1)(t^6+1)}$$



How should I proceed?






calculus definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Partial fractions? By the way, you seem to have lost your limits.
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 5:42










  • $begingroup$
    Typing on phone, so skipped it. Is a better way to type mathjax on phone
    $endgroup$
    – Piyush Divyanakar
    Dec 19 '18 at 5:43










  • $begingroup$
    @Piyush Divyanakar, you should try with lower limit $frac {π}{4}$
    $endgroup$
    – Awe Kumar Jha
    Dec 19 '18 at 12:39
















1












1








1





$begingroup$


$$int_{pi/6}^{pi/2} frac{sin^{1/3} (x)}{sin^{1/3} (x)+cos^{1/3} (x)}$$



I tried using the substitution $t^3=tan x$. Which gives me



$$intfrac{3t^3}{(t+1)(t^6+1)}$$



How should I proceed?






calculus definite-integrals







share|cite|improve this question











$endgroup$




$$int_{pi/6}^{pi/2} frac{sin^{1/3} (x)}{sin^{1/3} (x)+cos^{1/3} (x)}$$



I tried using the substitution $t^3=tan x$. Which gives me



$$intfrac{3t^3}{(t+1)(t^6+1)}$$



How should I proceed?






calculus definite-integrals




calculus definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 9:50









Arjang

5,63062363




5,63062363










asked Dec 19 '18 at 5:40









Piyush DivyanakarPiyush Divyanakar

3,348327




3,348327












  • $begingroup$
    Partial fractions? By the way, you seem to have lost your limits.
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 5:42










  • $begingroup$
    Typing on phone, so skipped it. Is a better way to type mathjax on phone
    $endgroup$
    – Piyush Divyanakar
    Dec 19 '18 at 5:43










  • $begingroup$
    @Piyush Divyanakar, you should try with lower limit $frac {π}{4}$
    $endgroup$
    – Awe Kumar Jha
    Dec 19 '18 at 12:39




















  • $begingroup$
    Partial fractions? By the way, you seem to have lost your limits.
    $endgroup$
    – Lord Shark the Unknown
    Dec 19 '18 at 5:42










  • $begingroup$
    Typing on phone, so skipped it. Is a better way to type mathjax on phone
    $endgroup$
    – Piyush Divyanakar
    Dec 19 '18 at 5:43










  • $begingroup$
    @Piyush Divyanakar, you should try with lower limit $frac {π}{4}$
    $endgroup$
    – Awe Kumar Jha
    Dec 19 '18 at 12:39


















$begingroup$
Partial fractions? By the way, you seem to have lost your limits.
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 5:42




$begingroup$
Partial fractions? By the way, you seem to have lost your limits.
$endgroup$
– Lord Shark the Unknown
Dec 19 '18 at 5:42












$begingroup$
Typing on phone, so skipped it. Is a better way to type mathjax on phone
$endgroup$
– Piyush Divyanakar
Dec 19 '18 at 5:43




$begingroup$
Typing on phone, so skipped it. Is a better way to type mathjax on phone
$endgroup$
– Piyush Divyanakar
Dec 19 '18 at 5:43












$begingroup$
@Piyush Divyanakar, you should try with lower limit $frac {π}{4}$
$endgroup$
– Awe Kumar Jha
Dec 19 '18 at 12:39






$begingroup$
@Piyush Divyanakar, you should try with lower limit $frac {π}{4}$
$endgroup$
– Awe Kumar Jha
Dec 19 '18 at 12:39












2 Answers
2






active

oldest

votes


















3












$begingroup$

I will put you on the path:



begin{align}
frac{3x^3}{left(x + 1right)left(x^6 + 1right)} &= frac{3x^3}{left(x + 1right)left(x^2 + 1right)left(x^2 + sqrt{3}x + 1right)left(x^2 - sqrt{3}x + 1right)} \
&= -frac{3}{2}frac{1}{x + 1} - frac{1}{2}left[frac{1}{x^2 + 1} +frac{x}{x^2 + 1}right] + frac{1}{2left(2 + sqrt{3}right)}frac{x + 1}{x^2 - sqrt{3}x + 1} \
&qquad + frac{2 +sqrt{3}}{2}frac{x + 1}{x^2 + sqrt{3}x + 1}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry I didn't finish it, I was falling asleep as I was typing. Are you ok from here? or would you like some further assistance to move forward? Will type up if so :-)
    $endgroup$
    – DavidG
    Dec 20 '18 at 0:36



















2












$begingroup$

As Lord Shark the Unknown commented, the first thing to do is partial fraction decomposition



Since $t^6+1=(t^2+1)(t^4-t^2+1)$, then
$$frac{3t^3}{(t+1)(t^6+1)}=-frac{3}{2 (t+1)}-frac{t+1}{2 left(t^2+1right)}+frac{2 t^3-t^2-t+2}{t^4-t^2+1}$$I suppose that the last term will impose to work with the roots of unity.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I agree with you, Maxima gives the indefinite integral in terms of roots of $t^4-t^2+1=0$
    $endgroup$
    – Awe Kumar Jha
    Dec 19 '18 at 12:37










  • $begingroup$
    Is there a simpler way to calculate the definite integral. I tried using this method and it is too long from an exam point of view.
    $endgroup$
    – Piyush Divyanakar
    Dec 19 '18 at 15:41











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

I will put you on the path:



begin{align}
frac{3x^3}{left(x + 1right)left(x^6 + 1right)} &= frac{3x^3}{left(x + 1right)left(x^2 + 1right)left(x^2 + sqrt{3}x + 1right)left(x^2 - sqrt{3}x + 1right)} \
&= -frac{3}{2}frac{1}{x + 1} - frac{1}{2}left[frac{1}{x^2 + 1} +frac{x}{x^2 + 1}right] + frac{1}{2left(2 + sqrt{3}right)}frac{x + 1}{x^2 - sqrt{3}x + 1} \
&qquad + frac{2 +sqrt{3}}{2}frac{x + 1}{x^2 + sqrt{3}x + 1}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry I didn't finish it, I was falling asleep as I was typing. Are you ok from here? or would you like some further assistance to move forward? Will type up if so :-)
    $endgroup$
    – DavidG
    Dec 20 '18 at 0:36
















3












$begingroup$

I will put you on the path:



begin{align}
frac{3x^3}{left(x + 1right)left(x^6 + 1right)} &= frac{3x^3}{left(x + 1right)left(x^2 + 1right)left(x^2 + sqrt{3}x + 1right)left(x^2 - sqrt{3}x + 1right)} \
&= -frac{3}{2}frac{1}{x + 1} - frac{1}{2}left[frac{1}{x^2 + 1} +frac{x}{x^2 + 1}right] + frac{1}{2left(2 + sqrt{3}right)}frac{x + 1}{x^2 - sqrt{3}x + 1} \
&qquad + frac{2 +sqrt{3}}{2}frac{x + 1}{x^2 + sqrt{3}x + 1}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry I didn't finish it, I was falling asleep as I was typing. Are you ok from here? or would you like some further assistance to move forward? Will type up if so :-)
    $endgroup$
    – DavidG
    Dec 20 '18 at 0:36














3












3








3





$begingroup$

I will put you on the path:



begin{align}
frac{3x^3}{left(x + 1right)left(x^6 + 1right)} &= frac{3x^3}{left(x + 1right)left(x^2 + 1right)left(x^2 + sqrt{3}x + 1right)left(x^2 - sqrt{3}x + 1right)} \
&= -frac{3}{2}frac{1}{x + 1} - frac{1}{2}left[frac{1}{x^2 + 1} +frac{x}{x^2 + 1}right] + frac{1}{2left(2 + sqrt{3}right)}frac{x + 1}{x^2 - sqrt{3}x + 1} \
&qquad + frac{2 +sqrt{3}}{2}frac{x + 1}{x^2 + sqrt{3}x + 1}
end{align}






share|cite|improve this answer









$endgroup$



I will put you on the path:



begin{align}
frac{3x^3}{left(x + 1right)left(x^6 + 1right)} &= frac{3x^3}{left(x + 1right)left(x^2 + 1right)left(x^2 + sqrt{3}x + 1right)left(x^2 - sqrt{3}x + 1right)} \
&= -frac{3}{2}frac{1}{x + 1} - frac{1}{2}left[frac{1}{x^2 + 1} +frac{x}{x^2 + 1}right] + frac{1}{2left(2 + sqrt{3}right)}frac{x + 1}{x^2 - sqrt{3}x + 1} \
&qquad + frac{2 +sqrt{3}}{2}frac{x + 1}{x^2 + sqrt{3}x + 1}
end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 13:43









DavidGDavidG

2,3691724




2,3691724












  • $begingroup$
    Sorry I didn't finish it, I was falling asleep as I was typing. Are you ok from here? or would you like some further assistance to move forward? Will type up if so :-)
    $endgroup$
    – DavidG
    Dec 20 '18 at 0:36


















  • $begingroup$
    Sorry I didn't finish it, I was falling asleep as I was typing. Are you ok from here? or would you like some further assistance to move forward? Will type up if so :-)
    $endgroup$
    – DavidG
    Dec 20 '18 at 0:36
















$begingroup$
Sorry I didn't finish it, I was falling asleep as I was typing. Are you ok from here? or would you like some further assistance to move forward? Will type up if so :-)
$endgroup$
– DavidG
Dec 20 '18 at 0:36




$begingroup$
Sorry I didn't finish it, I was falling asleep as I was typing. Are you ok from here? or would you like some further assistance to move forward? Will type up if so :-)
$endgroup$
– DavidG
Dec 20 '18 at 0:36











2












$begingroup$

As Lord Shark the Unknown commented, the first thing to do is partial fraction decomposition



Since $t^6+1=(t^2+1)(t^4-t^2+1)$, then
$$frac{3t^3}{(t+1)(t^6+1)}=-frac{3}{2 (t+1)}-frac{t+1}{2 left(t^2+1right)}+frac{2 t^3-t^2-t+2}{t^4-t^2+1}$$I suppose that the last term will impose to work with the roots of unity.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I agree with you, Maxima gives the indefinite integral in terms of roots of $t^4-t^2+1=0$
    $endgroup$
    – Awe Kumar Jha
    Dec 19 '18 at 12:37










  • $begingroup$
    Is there a simpler way to calculate the definite integral. I tried using this method and it is too long from an exam point of view.
    $endgroup$
    – Piyush Divyanakar
    Dec 19 '18 at 15:41
















2












$begingroup$

As Lord Shark the Unknown commented, the first thing to do is partial fraction decomposition



Since $t^6+1=(t^2+1)(t^4-t^2+1)$, then
$$frac{3t^3}{(t+1)(t^6+1)}=-frac{3}{2 (t+1)}-frac{t+1}{2 left(t^2+1right)}+frac{2 t^3-t^2-t+2}{t^4-t^2+1}$$I suppose that the last term will impose to work with the roots of unity.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I agree with you, Maxima gives the indefinite integral in terms of roots of $t^4-t^2+1=0$
    $endgroup$
    – Awe Kumar Jha
    Dec 19 '18 at 12:37










  • $begingroup$
    Is there a simpler way to calculate the definite integral. I tried using this method and it is too long from an exam point of view.
    $endgroup$
    – Piyush Divyanakar
    Dec 19 '18 at 15:41














2












2








2





$begingroup$

As Lord Shark the Unknown commented, the first thing to do is partial fraction decomposition



Since $t^6+1=(t^2+1)(t^4-t^2+1)$, then
$$frac{3t^3}{(t+1)(t^6+1)}=-frac{3}{2 (t+1)}-frac{t+1}{2 left(t^2+1right)}+frac{2 t^3-t^2-t+2}{t^4-t^2+1}$$I suppose that the last term will impose to work with the roots of unity.






share|cite|improve this answer









$endgroup$



As Lord Shark the Unknown commented, the first thing to do is partial fraction decomposition



Since $t^6+1=(t^2+1)(t^4-t^2+1)$, then
$$frac{3t^3}{(t+1)(t^6+1)}=-frac{3}{2 (t+1)}-frac{t+1}{2 left(t^2+1right)}+frac{2 t^3-t^2-t+2}{t^4-t^2+1}$$I suppose that the last term will impose to work with the roots of unity.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 6:42









Claude LeiboviciClaude Leibovici

123k1157134




123k1157134












  • $begingroup$
    I agree with you, Maxima gives the indefinite integral in terms of roots of $t^4-t^2+1=0$
    $endgroup$
    – Awe Kumar Jha
    Dec 19 '18 at 12:37










  • $begingroup$
    Is there a simpler way to calculate the definite integral. I tried using this method and it is too long from an exam point of view.
    $endgroup$
    – Piyush Divyanakar
    Dec 19 '18 at 15:41


















  • $begingroup$
    I agree with you, Maxima gives the indefinite integral in terms of roots of $t^4-t^2+1=0$
    $endgroup$
    – Awe Kumar Jha
    Dec 19 '18 at 12:37










  • $begingroup$
    Is there a simpler way to calculate the definite integral. I tried using this method and it is too long from an exam point of view.
    $endgroup$
    – Piyush Divyanakar
    Dec 19 '18 at 15:41
















$begingroup$
I agree with you, Maxima gives the indefinite integral in terms of roots of $t^4-t^2+1=0$
$endgroup$
– Awe Kumar Jha
Dec 19 '18 at 12:37




$begingroup$
I agree with you, Maxima gives the indefinite integral in terms of roots of $t^4-t^2+1=0$
$endgroup$
– Awe Kumar Jha
Dec 19 '18 at 12:37












$begingroup$
Is there a simpler way to calculate the definite integral. I tried using this method and it is too long from an exam point of view.
$endgroup$
– Piyush Divyanakar
Dec 19 '18 at 15:41




$begingroup$
Is there a simpler way to calculate the definite integral. I tried using this method and it is too long from an exam point of view.
$endgroup$
– Piyush Divyanakar
Dec 19 '18 at 15:41


















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