finding integral of $ sqrt {50-50cos(t)} $












0












$begingroup$


I have a cycloide



$ {x(t)=5 (t-sin(t)), y(t)=5(1-cos(t))} $ and



$ {x(t)=5 (t-sin(t)), y(t)=-5(1-cos(t))} $



and the aim is to find the circumference and to find it I have to show and take the differentiate integral:



$$ int sqrt{50-50cos(t)} dt $$



so I need help with it



Is the right anwser $ frac{-20}{sqrt{tan^2( frac{t}{2})+1}} $ ??






integration indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should really add that $mathrm dt$ at the end.
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 5 '18 at 0:47






  • 1




    $begingroup$
    Factor out $sqrt{50}$ and use my least favorite change of variables: $tan(t/2) = u.$
    $endgroup$
    – Sean Roberson
    Dec 5 '18 at 0:49
















0












$begingroup$


I have a cycloide



$ {x(t)=5 (t-sin(t)), y(t)=5(1-cos(t))} $ and



$ {x(t)=5 (t-sin(t)), y(t)=-5(1-cos(t))} $



and the aim is to find the circumference and to find it I have to show and take the differentiate integral:



$$ int sqrt{50-50cos(t)} dt $$



so I need help with it



Is the right anwser $ frac{-20}{sqrt{tan^2( frac{t}{2})+1}} $ ??






integration indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should really add that $mathrm dt$ at the end.
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 5 '18 at 0:47






  • 1




    $begingroup$
    Factor out $sqrt{50}$ and use my least favorite change of variables: $tan(t/2) = u.$
    $endgroup$
    – Sean Roberson
    Dec 5 '18 at 0:49














0












0








0





$begingroup$


I have a cycloide



$ {x(t)=5 (t-sin(t)), y(t)=5(1-cos(t))} $ and



$ {x(t)=5 (t-sin(t)), y(t)=-5(1-cos(t))} $



and the aim is to find the circumference and to find it I have to show and take the differentiate integral:



$$ int sqrt{50-50cos(t)} dt $$



so I need help with it



Is the right anwser $ frac{-20}{sqrt{tan^2( frac{t}{2})+1}} $ ??






integration indefinite-integrals







share|cite|improve this question











$endgroup$




I have a cycloide



$ {x(t)=5 (t-sin(t)), y(t)=5(1-cos(t))} $ and



$ {x(t)=5 (t-sin(t)), y(t)=-5(1-cos(t))} $



and the aim is to find the circumference and to find it I have to show and take the differentiate integral:



$$ int sqrt{50-50cos(t)} dt $$



so I need help with it



Is the right anwser $ frac{-20}{sqrt{tan^2( frac{t}{2})+1}} $ ??






integration indefinite-integrals




integration indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 1:13







Student123

















asked Dec 5 '18 at 0:44









Student123Student123

536




536












  • $begingroup$
    You should really add that $mathrm dt$ at the end.
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 5 '18 at 0:47






  • 1




    $begingroup$
    Factor out $sqrt{50}$ and use my least favorite change of variables: $tan(t/2) = u.$
    $endgroup$
    – Sean Roberson
    Dec 5 '18 at 0:49


















  • $begingroup$
    You should really add that $mathrm dt$ at the end.
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 5 '18 at 0:47






  • 1




    $begingroup$
    Factor out $sqrt{50}$ and use my least favorite change of variables: $tan(t/2) = u.$
    $endgroup$
    – Sean Roberson
    Dec 5 '18 at 0:49
















$begingroup$
You should really add that $mathrm dt$ at the end.
$endgroup$
– Mohammad Zuhair Khan
Dec 5 '18 at 0:47




$begingroup$
You should really add that $mathrm dt$ at the end.
$endgroup$
– Mohammad Zuhair Khan
Dec 5 '18 at 0:47




1




1




$begingroup$
Factor out $sqrt{50}$ and use my least favorite change of variables: $tan(t/2) = u.$
$endgroup$
– Sean Roberson
Dec 5 '18 at 0:49




$begingroup$
Factor out $sqrt{50}$ and use my least favorite change of variables: $tan(t/2) = u.$
$endgroup$
– Sean Roberson
Dec 5 '18 at 0:49










2 Answers
2






active

oldest

votes


















0












$begingroup$

$$sqrt{50-50cos x}=sqrt{50}sqrt{1 - cos x} =sqrt{50}sqrt{2 sin^2 frac{x} {2}}=10left| sinfrac{x} {2}right| $$
Can you finish now using this?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Did I get the right anwser?
    $endgroup$
    – Student123
    Dec 5 '18 at 1:14






  • 2




    $begingroup$
    Yes, your answer is correct. You can also simplify it to:$$frac{-20}{sqrt{tan^2 frac{t}{2}+1}} =frac{-20}{sqrt{sec^2 frac{t}{2}}} =-20left|cos frac{x} {2} right|$$ And of course you need to add $+C$. And if you differentiate it you should see that we find what I left in the answer.
    $endgroup$
    – Zacky
    Dec 5 '18 at 1:20












  • $begingroup$
    how did you simplify that?
    $endgroup$
    – Student123
    Dec 5 '18 at 8:27










  • $begingroup$
    Oh, I just realised that I used $x$ in the last part instead of $t$. To prove it I will denote $frac{t}{2}=z$ then we have: $$frac{1}{sqrt{tan^2 z+1}} =frac{1}{sqrt{frac{sin^2 z}{cos^2 z}+1}}=frac{1}{sqrt{frac{sin^2z+cos^2 z}{cos^2 z}}}=frac{1}{sqrt{frac{1}{cos^2 z}}} =sqrt{cos^2 z} =left|cos zright|$$ $$Rightarrow frac{-20}{sqrt{tan^2 frac{t}{2}+1}}=-20left|cos frac{t}{2} right|$$
    $endgroup$
    – Zacky
    Dec 5 '18 at 9:52












  • $begingroup$
    Something is wrong, because when I replace t from $ 2pi $ to 0, the circumference comes 0
    $endgroup$
    – Student123
    Dec 6 '18 at 0:47





















0












$begingroup$

Hint:



$$sqrt{50-50cos(t)}=sqrt{50}cdot sqrt{1-cos(t)}$$



and



$$sqrt{1-cos(t)}=sqrt{2}cdot|sin(t/2)| $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    did I get the right anwser?
    $endgroup$
    – Student123
    Dec 5 '18 at 1:14











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

$$sqrt{50-50cos x}=sqrt{50}sqrt{1 - cos x} =sqrt{50}sqrt{2 sin^2 frac{x} {2}}=10left| sinfrac{x} {2}right| $$
Can you finish now using this?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Did I get the right anwser?
    $endgroup$
    – Student123
    Dec 5 '18 at 1:14






  • 2




    $begingroup$
    Yes, your answer is correct. You can also simplify it to:$$frac{-20}{sqrt{tan^2 frac{t}{2}+1}} =frac{-20}{sqrt{sec^2 frac{t}{2}}} =-20left|cos frac{x} {2} right|$$ And of course you need to add $+C$. And if you differentiate it you should see that we find what I left in the answer.
    $endgroup$
    – Zacky
    Dec 5 '18 at 1:20












  • $begingroup$
    how did you simplify that?
    $endgroup$
    – Student123
    Dec 5 '18 at 8:27










  • $begingroup$
    Oh, I just realised that I used $x$ in the last part instead of $t$. To prove it I will denote $frac{t}{2}=z$ then we have: $$frac{1}{sqrt{tan^2 z+1}} =frac{1}{sqrt{frac{sin^2 z}{cos^2 z}+1}}=frac{1}{sqrt{frac{sin^2z+cos^2 z}{cos^2 z}}}=frac{1}{sqrt{frac{1}{cos^2 z}}} =sqrt{cos^2 z} =left|cos zright|$$ $$Rightarrow frac{-20}{sqrt{tan^2 frac{t}{2}+1}}=-20left|cos frac{t}{2} right|$$
    $endgroup$
    – Zacky
    Dec 5 '18 at 9:52












  • $begingroup$
    Something is wrong, because when I replace t from $ 2pi $ to 0, the circumference comes 0
    $endgroup$
    – Student123
    Dec 6 '18 at 0:47


















0












$begingroup$

$$sqrt{50-50cos x}=sqrt{50}sqrt{1 - cos x} =sqrt{50}sqrt{2 sin^2 frac{x} {2}}=10left| sinfrac{x} {2}right| $$
Can you finish now using this?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Did I get the right anwser?
    $endgroup$
    – Student123
    Dec 5 '18 at 1:14






  • 2




    $begingroup$
    Yes, your answer is correct. You can also simplify it to:$$frac{-20}{sqrt{tan^2 frac{t}{2}+1}} =frac{-20}{sqrt{sec^2 frac{t}{2}}} =-20left|cos frac{x} {2} right|$$ And of course you need to add $+C$. And if you differentiate it you should see that we find what I left in the answer.
    $endgroup$
    – Zacky
    Dec 5 '18 at 1:20












  • $begingroup$
    how did you simplify that?
    $endgroup$
    – Student123
    Dec 5 '18 at 8:27










  • $begingroup$
    Oh, I just realised that I used $x$ in the last part instead of $t$. To prove it I will denote $frac{t}{2}=z$ then we have: $$frac{1}{sqrt{tan^2 z+1}} =frac{1}{sqrt{frac{sin^2 z}{cos^2 z}+1}}=frac{1}{sqrt{frac{sin^2z+cos^2 z}{cos^2 z}}}=frac{1}{sqrt{frac{1}{cos^2 z}}} =sqrt{cos^2 z} =left|cos zright|$$ $$Rightarrow frac{-20}{sqrt{tan^2 frac{t}{2}+1}}=-20left|cos frac{t}{2} right|$$
    $endgroup$
    – Zacky
    Dec 5 '18 at 9:52












  • $begingroup$
    Something is wrong, because when I replace t from $ 2pi $ to 0, the circumference comes 0
    $endgroup$
    – Student123
    Dec 6 '18 at 0:47
















0












0








0





$begingroup$

$$sqrt{50-50cos x}=sqrt{50}sqrt{1 - cos x} =sqrt{50}sqrt{2 sin^2 frac{x} {2}}=10left| sinfrac{x} {2}right| $$
Can you finish now using this?






share|cite|improve this answer









$endgroup$



$$sqrt{50-50cos x}=sqrt{50}sqrt{1 - cos x} =sqrt{50}sqrt{2 sin^2 frac{x} {2}}=10left| sinfrac{x} {2}right| $$
Can you finish now using this?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 0:50









ZackyZacky

5,4981856




5,4981856












  • $begingroup$
    Did I get the right anwser?
    $endgroup$
    – Student123
    Dec 5 '18 at 1:14






  • 2




    $begingroup$
    Yes, your answer is correct. You can also simplify it to:$$frac{-20}{sqrt{tan^2 frac{t}{2}+1}} =frac{-20}{sqrt{sec^2 frac{t}{2}}} =-20left|cos frac{x} {2} right|$$ And of course you need to add $+C$. And if you differentiate it you should see that we find what I left in the answer.
    $endgroup$
    – Zacky
    Dec 5 '18 at 1:20












  • $begingroup$
    how did you simplify that?
    $endgroup$
    – Student123
    Dec 5 '18 at 8:27










  • $begingroup$
    Oh, I just realised that I used $x$ in the last part instead of $t$. To prove it I will denote $frac{t}{2}=z$ then we have: $$frac{1}{sqrt{tan^2 z+1}} =frac{1}{sqrt{frac{sin^2 z}{cos^2 z}+1}}=frac{1}{sqrt{frac{sin^2z+cos^2 z}{cos^2 z}}}=frac{1}{sqrt{frac{1}{cos^2 z}}} =sqrt{cos^2 z} =left|cos zright|$$ $$Rightarrow frac{-20}{sqrt{tan^2 frac{t}{2}+1}}=-20left|cos frac{t}{2} right|$$
    $endgroup$
    – Zacky
    Dec 5 '18 at 9:52












  • $begingroup$
    Something is wrong, because when I replace t from $ 2pi $ to 0, the circumference comes 0
    $endgroup$
    – Student123
    Dec 6 '18 at 0:47




















  • $begingroup$
    Did I get the right anwser?
    $endgroup$
    – Student123
    Dec 5 '18 at 1:14






  • 2




    $begingroup$
    Yes, your answer is correct. You can also simplify it to:$$frac{-20}{sqrt{tan^2 frac{t}{2}+1}} =frac{-20}{sqrt{sec^2 frac{t}{2}}} =-20left|cos frac{x} {2} right|$$ And of course you need to add $+C$. And if you differentiate it you should see that we find what I left in the answer.
    $endgroup$
    – Zacky
    Dec 5 '18 at 1:20












  • $begingroup$
    how did you simplify that?
    $endgroup$
    – Student123
    Dec 5 '18 at 8:27










  • $begingroup$
    Oh, I just realised that I used $x$ in the last part instead of $t$. To prove it I will denote $frac{t}{2}=z$ then we have: $$frac{1}{sqrt{tan^2 z+1}} =frac{1}{sqrt{frac{sin^2 z}{cos^2 z}+1}}=frac{1}{sqrt{frac{sin^2z+cos^2 z}{cos^2 z}}}=frac{1}{sqrt{frac{1}{cos^2 z}}} =sqrt{cos^2 z} =left|cos zright|$$ $$Rightarrow frac{-20}{sqrt{tan^2 frac{t}{2}+1}}=-20left|cos frac{t}{2} right|$$
    $endgroup$
    – Zacky
    Dec 5 '18 at 9:52












  • $begingroup$
    Something is wrong, because when I replace t from $ 2pi $ to 0, the circumference comes 0
    $endgroup$
    – Student123
    Dec 6 '18 at 0:47


















$begingroup$
Did I get the right anwser?
$endgroup$
– Student123
Dec 5 '18 at 1:14




$begingroup$
Did I get the right anwser?
$endgroup$
– Student123
Dec 5 '18 at 1:14




2




2




$begingroup$
Yes, your answer is correct. You can also simplify it to:$$frac{-20}{sqrt{tan^2 frac{t}{2}+1}} =frac{-20}{sqrt{sec^2 frac{t}{2}}} =-20left|cos frac{x} {2} right|$$ And of course you need to add $+C$. And if you differentiate it you should see that we find what I left in the answer.
$endgroup$
– Zacky
Dec 5 '18 at 1:20






$begingroup$
Yes, your answer is correct. You can also simplify it to:$$frac{-20}{sqrt{tan^2 frac{t}{2}+1}} =frac{-20}{sqrt{sec^2 frac{t}{2}}} =-20left|cos frac{x} {2} right|$$ And of course you need to add $+C$. And if you differentiate it you should see that we find what I left in the answer.
$endgroup$
– Zacky
Dec 5 '18 at 1:20














$begingroup$
how did you simplify that?
$endgroup$
– Student123
Dec 5 '18 at 8:27




$begingroup$
how did you simplify that?
$endgroup$
– Student123
Dec 5 '18 at 8:27












$begingroup$
Oh, I just realised that I used $x$ in the last part instead of $t$. To prove it I will denote $frac{t}{2}=z$ then we have: $$frac{1}{sqrt{tan^2 z+1}} =frac{1}{sqrt{frac{sin^2 z}{cos^2 z}+1}}=frac{1}{sqrt{frac{sin^2z+cos^2 z}{cos^2 z}}}=frac{1}{sqrt{frac{1}{cos^2 z}}} =sqrt{cos^2 z} =left|cos zright|$$ $$Rightarrow frac{-20}{sqrt{tan^2 frac{t}{2}+1}}=-20left|cos frac{t}{2} right|$$
$endgroup$
– Zacky
Dec 5 '18 at 9:52






$begingroup$
Oh, I just realised that I used $x$ in the last part instead of $t$. To prove it I will denote $frac{t}{2}=z$ then we have: $$frac{1}{sqrt{tan^2 z+1}} =frac{1}{sqrt{frac{sin^2 z}{cos^2 z}+1}}=frac{1}{sqrt{frac{sin^2z+cos^2 z}{cos^2 z}}}=frac{1}{sqrt{frac{1}{cos^2 z}}} =sqrt{cos^2 z} =left|cos zright|$$ $$Rightarrow frac{-20}{sqrt{tan^2 frac{t}{2}+1}}=-20left|cos frac{t}{2} right|$$
$endgroup$
– Zacky
Dec 5 '18 at 9:52














$begingroup$
Something is wrong, because when I replace t from $ 2pi $ to 0, the circumference comes 0
$endgroup$
– Student123
Dec 6 '18 at 0:47






$begingroup$
Something is wrong, because when I replace t from $ 2pi $ to 0, the circumference comes 0
$endgroup$
– Student123
Dec 6 '18 at 0:47













0












$begingroup$

Hint:



$$sqrt{50-50cos(t)}=sqrt{50}cdot sqrt{1-cos(t)}$$



and



$$sqrt{1-cos(t)}=sqrt{2}cdot|sin(t/2)| $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    did I get the right anwser?
    $endgroup$
    – Student123
    Dec 5 '18 at 1:14
















0












$begingroup$

Hint:



$$sqrt{50-50cos(t)}=sqrt{50}cdot sqrt{1-cos(t)}$$



and



$$sqrt{1-cos(t)}=sqrt{2}cdot|sin(t/2)| $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    did I get the right anwser?
    $endgroup$
    – Student123
    Dec 5 '18 at 1:14














0












0








0





$begingroup$

Hint:



$$sqrt{50-50cos(t)}=sqrt{50}cdot sqrt{1-cos(t)}$$



and



$$sqrt{1-cos(t)}=sqrt{2}cdot|sin(t/2)| $$






share|cite|improve this answer









$endgroup$



Hint:



$$sqrt{50-50cos(t)}=sqrt{50}cdot sqrt{1-cos(t)}$$



and



$$sqrt{1-cos(t)}=sqrt{2}cdot|sin(t/2)| $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 0:52









Thomas ShelbyThomas Shelby

2,480221




2,480221












  • $begingroup$
    did I get the right anwser?
    $endgroup$
    – Student123
    Dec 5 '18 at 1:14


















  • $begingroup$
    did I get the right anwser?
    $endgroup$
    – Student123
    Dec 5 '18 at 1:14
















$begingroup$
did I get the right anwser?
$endgroup$
– Student123
Dec 5 '18 at 1:14




$begingroup$
did I get the right anwser?
$endgroup$
– Student123
Dec 5 '18 at 1:14


















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