Another interesting way to evaluating the integral $int frac{cos(x)}{a+b cos(x)}:dx$












2












$begingroup$


I need help in evaluating the following integral, please:
$$intfrac{cos(x)}{a+bcos(x)}dx$$
so that we get the following result:
$$
int frac{cos x }{a + bcos x}:dx
= frac{a}{bsqrt{a^2-b^2}}
arcsinleft(frac{b+acos x}{a+bcos x}right)
- frac1b
arcsin(cos x)
+ C
$$




Also, I would say I don't have a clue how the substitution can be done. I know another well-known answer which is done using the tangent half-angle substitution, but the result I am asking about exists nowhere online as far as I am concerned. This interesting result is given as a hint to a problem in a physics book and it is completely correct as you can test it yourself. I would appreciate any help. Thanks.



(edit: I came to know that the substitution [u=cos(x)] would lead to the desired formula. So, maybe you can help with that as a hint.)






integration trigonometry indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you know the answer, and wonder how to get there, you can just differentiate and see what happens. Then do it in reverse, and voila!
    $endgroup$
    – Arthur
    Dec 12 '18 at 17:42








  • 1




    $begingroup$
    the answer only seems to be valid for $|b| < |a|$ otherwise the root becomes complex
    $endgroup$
    – gt6989b
    Dec 12 '18 at 17:45










  • $begingroup$
    Use this here mathworld.wolfram.com/WeierstrassSubstitution.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 12 '18 at 17:46










  • $begingroup$
    @Arthur I tried doing so and still can't figure it out.
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:03












  • $begingroup$
    @gt6989b yes, a>b
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:04
















2












$begingroup$


I need help in evaluating the following integral, please:
$$intfrac{cos(x)}{a+bcos(x)}dx$$
so that we get the following result:
$$
int frac{cos x }{a + bcos x}:dx
= frac{a}{bsqrt{a^2-b^2}}
arcsinleft(frac{b+acos x}{a+bcos x}right)
- frac1b
arcsin(cos x)
+ C
$$




Also, I would say I don't have a clue how the substitution can be done. I know another well-known answer which is done using the tangent half-angle substitution, but the result I am asking about exists nowhere online as far as I am concerned. This interesting result is given as a hint to a problem in a physics book and it is completely correct as you can test it yourself. I would appreciate any help. Thanks.



(edit: I came to know that the substitution [u=cos(x)] would lead to the desired formula. So, maybe you can help with that as a hint.)






integration trigonometry indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you know the answer, and wonder how to get there, you can just differentiate and see what happens. Then do it in reverse, and voila!
    $endgroup$
    – Arthur
    Dec 12 '18 at 17:42








  • 1




    $begingroup$
    the answer only seems to be valid for $|b| < |a|$ otherwise the root becomes complex
    $endgroup$
    – gt6989b
    Dec 12 '18 at 17:45










  • $begingroup$
    Use this here mathworld.wolfram.com/WeierstrassSubstitution.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 12 '18 at 17:46










  • $begingroup$
    @Arthur I tried doing so and still can't figure it out.
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:03












  • $begingroup$
    @gt6989b yes, a>b
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:04














2












2








2


1



$begingroup$


I need help in evaluating the following integral, please:
$$intfrac{cos(x)}{a+bcos(x)}dx$$
so that we get the following result:
$$
int frac{cos x }{a + bcos x}:dx
= frac{a}{bsqrt{a^2-b^2}}
arcsinleft(frac{b+acos x}{a+bcos x}right)
- frac1b
arcsin(cos x)
+ C
$$




Also, I would say I don't have a clue how the substitution can be done. I know another well-known answer which is done using the tangent half-angle substitution, but the result I am asking about exists nowhere online as far as I am concerned. This interesting result is given as a hint to a problem in a physics book and it is completely correct as you can test it yourself. I would appreciate any help. Thanks.



(edit: I came to know that the substitution [u=cos(x)] would lead to the desired formula. So, maybe you can help with that as a hint.)






integration trigonometry indefinite-integrals







share|cite|improve this question











$endgroup$




I need help in evaluating the following integral, please:
$$intfrac{cos(x)}{a+bcos(x)}dx$$
so that we get the following result:
$$
int frac{cos x }{a + bcos x}:dx
= frac{a}{bsqrt{a^2-b^2}}
arcsinleft(frac{b+acos x}{a+bcos x}right)
- frac1b
arcsin(cos x)
+ C
$$




Also, I would say I don't have a clue how the substitution can be done. I know another well-known answer which is done using the tangent half-angle substitution, but the result I am asking about exists nowhere online as far as I am concerned. This interesting result is given as a hint to a problem in a physics book and it is completely correct as you can test it yourself. I would appreciate any help. Thanks.



(edit: I came to know that the substitution [u=cos(x)] would lead to the desired formula. So, maybe you can help with that as a hint.)






integration trigonometry indefinite-integrals




integration trigonometry indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 22:13







physicist bychoice

















asked Dec 12 '18 at 17:39









physicist bychoicephysicist bychoice

114




114












  • $begingroup$
    If you know the answer, and wonder how to get there, you can just differentiate and see what happens. Then do it in reverse, and voila!
    $endgroup$
    – Arthur
    Dec 12 '18 at 17:42








  • 1




    $begingroup$
    the answer only seems to be valid for $|b| < |a|$ otherwise the root becomes complex
    $endgroup$
    – gt6989b
    Dec 12 '18 at 17:45










  • $begingroup$
    Use this here mathworld.wolfram.com/WeierstrassSubstitution.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 12 '18 at 17:46










  • $begingroup$
    @Arthur I tried doing so and still can't figure it out.
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:03












  • $begingroup$
    @gt6989b yes, a>b
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:04


















  • $begingroup$
    If you know the answer, and wonder how to get there, you can just differentiate and see what happens. Then do it in reverse, and voila!
    $endgroup$
    – Arthur
    Dec 12 '18 at 17:42








  • 1




    $begingroup$
    the answer only seems to be valid for $|b| < |a|$ otherwise the root becomes complex
    $endgroup$
    – gt6989b
    Dec 12 '18 at 17:45










  • $begingroup$
    Use this here mathworld.wolfram.com/WeierstrassSubstitution.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 12 '18 at 17:46










  • $begingroup$
    @Arthur I tried doing so and still can't figure it out.
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:03












  • $begingroup$
    @gt6989b yes, a>b
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:04
















$begingroup$
If you know the answer, and wonder how to get there, you can just differentiate and see what happens. Then do it in reverse, and voila!
$endgroup$
– Arthur
Dec 12 '18 at 17:42






$begingroup$
If you know the answer, and wonder how to get there, you can just differentiate and see what happens. Then do it in reverse, and voila!
$endgroup$
– Arthur
Dec 12 '18 at 17:42






1




1




$begingroup$
the answer only seems to be valid for $|b| < |a|$ otherwise the root becomes complex
$endgroup$
– gt6989b
Dec 12 '18 at 17:45




$begingroup$
the answer only seems to be valid for $|b| < |a|$ otherwise the root becomes complex
$endgroup$
– gt6989b
Dec 12 '18 at 17:45












$begingroup$
Use this here mathworld.wolfram.com/WeierstrassSubstitution.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 12 '18 at 17:46




$begingroup$
Use this here mathworld.wolfram.com/WeierstrassSubstitution.html
$endgroup$
– Dr. Sonnhard Graubner
Dec 12 '18 at 17:46












$begingroup$
@Arthur I tried doing so and still can't figure it out.
$endgroup$
– physicist bychoice
Dec 12 '18 at 19:03






$begingroup$
@Arthur I tried doing so and still can't figure it out.
$endgroup$
– physicist bychoice
Dec 12 '18 at 19:03














$begingroup$
@gt6989b yes, a>b
$endgroup$
– physicist bychoice
Dec 12 '18 at 19:04




$begingroup$
@gt6989b yes, a>b
$endgroup$
– physicist bychoice
Dec 12 '18 at 19:04










2 Answers
2






active

oldest

votes


















2












$begingroup$

Well this probably isn't the way the book does it, but you said you'd appreciate any help. This is how I did it.



$$I=intfrac{cos x}{a+bcos x}dx$$
$$Ib=intfrac{bcos x}{a+bcos x}dx$$
$$Ib=intfrac{a+bcos x}{a+bcos x}dx-aintfrac{dx}{a+bcos x}$$
$$Ib=x-aintfrac{dx}{a+bcos x}$$
Then we focus on
$$J=intfrac{dx}{a+bcos x}$$
We may write the integral as
$$J=-intfrac{sec^2(frac x2)}{(b-a)tan^2(frac x2)-b-a}dx$$
$$J=frac1{a+b}intfrac{sec^2(frac x2)dx}{frac{a-b}{a+b}tan^2(frac x2)+1}$$
Then we let
$$tan(x/2)=sqrt{frac{a+b}{a-b}}u Rightarrow sec^2(x/2)dx=2sqrt{frac{a+b}{a-b}}du$$
Which gives
$$J=frac1{a+b}intfrac{2sqrt{frac{a+b}{a-b}}du}{frac{a-b}{a+b}big(sqrt{frac{a+b}{a-b}}ubig)^2+1}$$
$$J=frac2{sqrt{a^2-b^2}}intfrac{du}{u^2+1}$$
$$J=frac2{sqrt{a^2-b^2}}arctan u$$
$$J=frac2{sqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]$$
Hence we have
$$Ib=x-frac{2a}{sqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]$$
Which means
$$I=frac{x}b-frac{2a}{bsqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]+C$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    You are right it is not the desired result, but you did it quite beautifully. Thanks and I do appreciate it. I still can't upvote because I am still a new member and it requires reputation, which I, sadly, don't have enough of. :')
    $endgroup$
    – physicist bychoice
    Dec 13 '18 at 15:49



















0












$begingroup$

Hint: Use $$u=tan(frac{x}{2})$$ so $$du=frac{1}{2}sec^2(frac{x}{2})dx$$ and $$sin(x)=frac{2u}{u^2-1},cos(x)=frac{1-u^2}{1+u^2}$$ and $$dx=frac{2du}{1+u^2}$$ and we get
$$intfrac{2(1-u^2)}{(u^2+1)^2left(a+frac{b(1-u^2)}{u^2+1}right)}du$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This way of substitution, I suppose, will lead to a different result and that tangent half-angle substitution is so common but as I have seen will not result to the the answer given above. I can't see how this substitution can lead to the desired result. If you give another hint, it will be easier for me. And Thanks.
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:12











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Well this probably isn't the way the book does it, but you said you'd appreciate any help. This is how I did it.



$$I=intfrac{cos x}{a+bcos x}dx$$
$$Ib=intfrac{bcos x}{a+bcos x}dx$$
$$Ib=intfrac{a+bcos x}{a+bcos x}dx-aintfrac{dx}{a+bcos x}$$
$$Ib=x-aintfrac{dx}{a+bcos x}$$
Then we focus on
$$J=intfrac{dx}{a+bcos x}$$
We may write the integral as
$$J=-intfrac{sec^2(frac x2)}{(b-a)tan^2(frac x2)-b-a}dx$$
$$J=frac1{a+b}intfrac{sec^2(frac x2)dx}{frac{a-b}{a+b}tan^2(frac x2)+1}$$
Then we let
$$tan(x/2)=sqrt{frac{a+b}{a-b}}u Rightarrow sec^2(x/2)dx=2sqrt{frac{a+b}{a-b}}du$$
Which gives
$$J=frac1{a+b}intfrac{2sqrt{frac{a+b}{a-b}}du}{frac{a-b}{a+b}big(sqrt{frac{a+b}{a-b}}ubig)^2+1}$$
$$J=frac2{sqrt{a^2-b^2}}intfrac{du}{u^2+1}$$
$$J=frac2{sqrt{a^2-b^2}}arctan u$$
$$J=frac2{sqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]$$
Hence we have
$$Ib=x-frac{2a}{sqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]$$
Which means
$$I=frac{x}b-frac{2a}{bsqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]+C$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    You are right it is not the desired result, but you did it quite beautifully. Thanks and I do appreciate it. I still can't upvote because I am still a new member and it requires reputation, which I, sadly, don't have enough of. :')
    $endgroup$
    – physicist bychoice
    Dec 13 '18 at 15:49
















2












$begingroup$

Well this probably isn't the way the book does it, but you said you'd appreciate any help. This is how I did it.



$$I=intfrac{cos x}{a+bcos x}dx$$
$$Ib=intfrac{bcos x}{a+bcos x}dx$$
$$Ib=intfrac{a+bcos x}{a+bcos x}dx-aintfrac{dx}{a+bcos x}$$
$$Ib=x-aintfrac{dx}{a+bcos x}$$
Then we focus on
$$J=intfrac{dx}{a+bcos x}$$
We may write the integral as
$$J=-intfrac{sec^2(frac x2)}{(b-a)tan^2(frac x2)-b-a}dx$$
$$J=frac1{a+b}intfrac{sec^2(frac x2)dx}{frac{a-b}{a+b}tan^2(frac x2)+1}$$
Then we let
$$tan(x/2)=sqrt{frac{a+b}{a-b}}u Rightarrow sec^2(x/2)dx=2sqrt{frac{a+b}{a-b}}du$$
Which gives
$$J=frac1{a+b}intfrac{2sqrt{frac{a+b}{a-b}}du}{frac{a-b}{a+b}big(sqrt{frac{a+b}{a-b}}ubig)^2+1}$$
$$J=frac2{sqrt{a^2-b^2}}intfrac{du}{u^2+1}$$
$$J=frac2{sqrt{a^2-b^2}}arctan u$$
$$J=frac2{sqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]$$
Hence we have
$$Ib=x-frac{2a}{sqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]$$
Which means
$$I=frac{x}b-frac{2a}{bsqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]+C$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    You are right it is not the desired result, but you did it quite beautifully. Thanks and I do appreciate it. I still can't upvote because I am still a new member and it requires reputation, which I, sadly, don't have enough of. :')
    $endgroup$
    – physicist bychoice
    Dec 13 '18 at 15:49














2












2








2





$begingroup$

Well this probably isn't the way the book does it, but you said you'd appreciate any help. This is how I did it.



$$I=intfrac{cos x}{a+bcos x}dx$$
$$Ib=intfrac{bcos x}{a+bcos x}dx$$
$$Ib=intfrac{a+bcos x}{a+bcos x}dx-aintfrac{dx}{a+bcos x}$$
$$Ib=x-aintfrac{dx}{a+bcos x}$$
Then we focus on
$$J=intfrac{dx}{a+bcos x}$$
We may write the integral as
$$J=-intfrac{sec^2(frac x2)}{(b-a)tan^2(frac x2)-b-a}dx$$
$$J=frac1{a+b}intfrac{sec^2(frac x2)dx}{frac{a-b}{a+b}tan^2(frac x2)+1}$$
Then we let
$$tan(x/2)=sqrt{frac{a+b}{a-b}}u Rightarrow sec^2(x/2)dx=2sqrt{frac{a+b}{a-b}}du$$
Which gives
$$J=frac1{a+b}intfrac{2sqrt{frac{a+b}{a-b}}du}{frac{a-b}{a+b}big(sqrt{frac{a+b}{a-b}}ubig)^2+1}$$
$$J=frac2{sqrt{a^2-b^2}}intfrac{du}{u^2+1}$$
$$J=frac2{sqrt{a^2-b^2}}arctan u$$
$$J=frac2{sqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]$$
Hence we have
$$Ib=x-frac{2a}{sqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]$$
Which means
$$I=frac{x}b-frac{2a}{bsqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]+C$$






share|cite|improve this answer









$endgroup$



Well this probably isn't the way the book does it, but you said you'd appreciate any help. This is how I did it.



$$I=intfrac{cos x}{a+bcos x}dx$$
$$Ib=intfrac{bcos x}{a+bcos x}dx$$
$$Ib=intfrac{a+bcos x}{a+bcos x}dx-aintfrac{dx}{a+bcos x}$$
$$Ib=x-aintfrac{dx}{a+bcos x}$$
Then we focus on
$$J=intfrac{dx}{a+bcos x}$$
We may write the integral as
$$J=-intfrac{sec^2(frac x2)}{(b-a)tan^2(frac x2)-b-a}dx$$
$$J=frac1{a+b}intfrac{sec^2(frac x2)dx}{frac{a-b}{a+b}tan^2(frac x2)+1}$$
Then we let
$$tan(x/2)=sqrt{frac{a+b}{a-b}}u Rightarrow sec^2(x/2)dx=2sqrt{frac{a+b}{a-b}}du$$
Which gives
$$J=frac1{a+b}intfrac{2sqrt{frac{a+b}{a-b}}du}{frac{a-b}{a+b}big(sqrt{frac{a+b}{a-b}}ubig)^2+1}$$
$$J=frac2{sqrt{a^2-b^2}}intfrac{du}{u^2+1}$$
$$J=frac2{sqrt{a^2-b^2}}arctan u$$
$$J=frac2{sqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]$$
Hence we have
$$Ib=x-frac{2a}{sqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]$$
Which means
$$I=frac{x}b-frac{2a}{bsqrt{a^2-b^2}}arctanbigg[sqrt{frac{a-b}{a+b}}tanbigg(frac x2bigg)bigg]+C$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 21:47









clathratusclathratus

4,640337




4,640337








  • 1




    $begingroup$
    You are right it is not the desired result, but you did it quite beautifully. Thanks and I do appreciate it. I still can't upvote because I am still a new member and it requires reputation, which I, sadly, don't have enough of. :')
    $endgroup$
    – physicist bychoice
    Dec 13 '18 at 15:49














  • 1




    $begingroup$
    You are right it is not the desired result, but you did it quite beautifully. Thanks and I do appreciate it. I still can't upvote because I am still a new member and it requires reputation, which I, sadly, don't have enough of. :')
    $endgroup$
    – physicist bychoice
    Dec 13 '18 at 15:49








1




1




$begingroup$
You are right it is not the desired result, but you did it quite beautifully. Thanks and I do appreciate it. I still can't upvote because I am still a new member and it requires reputation, which I, sadly, don't have enough of. :')
$endgroup$
– physicist bychoice
Dec 13 '18 at 15:49




$begingroup$
You are right it is not the desired result, but you did it quite beautifully. Thanks and I do appreciate it. I still can't upvote because I am still a new member and it requires reputation, which I, sadly, don't have enough of. :')
$endgroup$
– physicist bychoice
Dec 13 '18 at 15:49











0












$begingroup$

Hint: Use $$u=tan(frac{x}{2})$$ so $$du=frac{1}{2}sec^2(frac{x}{2})dx$$ and $$sin(x)=frac{2u}{u^2-1},cos(x)=frac{1-u^2}{1+u^2}$$ and $$dx=frac{2du}{1+u^2}$$ and we get
$$intfrac{2(1-u^2)}{(u^2+1)^2left(a+frac{b(1-u^2)}{u^2+1}right)}du$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This way of substitution, I suppose, will lead to a different result and that tangent half-angle substitution is so common but as I have seen will not result to the the answer given above. I can't see how this substitution can lead to the desired result. If you give another hint, it will be easier for me. And Thanks.
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:12
















0












$begingroup$

Hint: Use $$u=tan(frac{x}{2})$$ so $$du=frac{1}{2}sec^2(frac{x}{2})dx$$ and $$sin(x)=frac{2u}{u^2-1},cos(x)=frac{1-u^2}{1+u^2}$$ and $$dx=frac{2du}{1+u^2}$$ and we get
$$intfrac{2(1-u^2)}{(u^2+1)^2left(a+frac{b(1-u^2)}{u^2+1}right)}du$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This way of substitution, I suppose, will lead to a different result and that tangent half-angle substitution is so common but as I have seen will not result to the the answer given above. I can't see how this substitution can lead to the desired result. If you give another hint, it will be easier for me. And Thanks.
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:12














0












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Hint: Use $$u=tan(frac{x}{2})$$ so $$du=frac{1}{2}sec^2(frac{x}{2})dx$$ and $$sin(x)=frac{2u}{u^2-1},cos(x)=frac{1-u^2}{1+u^2}$$ and $$dx=frac{2du}{1+u^2}$$ and we get
$$intfrac{2(1-u^2)}{(u^2+1)^2left(a+frac{b(1-u^2)}{u^2+1}right)}du$$






share|cite|improve this answer









$endgroup$



Hint: Use $$u=tan(frac{x}{2})$$ so $$du=frac{1}{2}sec^2(frac{x}{2})dx$$ and $$sin(x)=frac{2u}{u^2-1},cos(x)=frac{1-u^2}{1+u^2}$$ and $$dx=frac{2du}{1+u^2}$$ and we get
$$intfrac{2(1-u^2)}{(u^2+1)^2left(a+frac{b(1-u^2)}{u^2+1}right)}du$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 18:13









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.2k42866




76.2k42866












  • $begingroup$
    This way of substitution, I suppose, will lead to a different result and that tangent half-angle substitution is so common but as I have seen will not result to the the answer given above. I can't see how this substitution can lead to the desired result. If you give another hint, it will be easier for me. And Thanks.
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:12


















  • $begingroup$
    This way of substitution, I suppose, will lead to a different result and that tangent half-angle substitution is so common but as I have seen will not result to the the answer given above. I can't see how this substitution can lead to the desired result. If you give another hint, it will be easier for me. And Thanks.
    $endgroup$
    – physicist bychoice
    Dec 12 '18 at 19:12
















$begingroup$
This way of substitution, I suppose, will lead to a different result and that tangent half-angle substitution is so common but as I have seen will not result to the the answer given above. I can't see how this substitution can lead to the desired result. If you give another hint, it will be easier for me. And Thanks.
$endgroup$
– physicist bychoice
Dec 12 '18 at 19:12




$begingroup$
This way of substitution, I suppose, will lead to a different result and that tangent half-angle substitution is so common but as I have seen will not result to the the answer given above. I can't see how this substitution can lead to the desired result. If you give another hint, it will be easier for me. And Thanks.
$endgroup$
– physicist bychoice
Dec 12 '18 at 19:12


















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