Integration of $frac1x^2-a^2$ by trigonometric substitution?

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2
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$$int frac1x^2-a^2dx$$




Now, I know this can be done by splitting the function into two integrable functions,



$displaystyledfrac12aint bigg(dfrac1x-a - dfrac1x+abigg)dx$



And then doing the usual stuff.



My question is, how can we do this by using trigonometric substitution?



The only thing that gets in my mind is $x=asectheta$, but then got stuck on proceeding further.



Any help would be appreciated.




integration trigonometry indefinite-integrals




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  • 2




    Try $x=acos2theta$.
    – Antonio DJC
    Jul 30 at 7:30










  • The integrand is discontinuous at $x = pm a$, so the antiderivative ought to be split into three intervals, each with a different added constant.
    – mr_e_man
    Jul 30 at 7:57














up vote
2
down vote

favorite













$$int frac1x^2-a^2dx$$




Now, I know this can be done by splitting the function into two integrable functions,



$displaystyledfrac12aint bigg(dfrac1x-a - dfrac1x+abigg)dx$



And then doing the usual stuff.



My question is, how can we do this by using trigonometric substitution?



The only thing that gets in my mind is $x=asectheta$, but then got stuck on proceeding further.



Any help would be appreciated.




integration trigonometry indefinite-integrals




share|cite|improve this question

















  • 2




    Try $x=acos2theta$.
    – Antonio DJC
    Jul 30 at 7:30










  • The integrand is discontinuous at $x = pm a$, so the antiderivative ought to be split into three intervals, each with a different added constant.
    – mr_e_man
    Jul 30 at 7:57












up vote
2
down vote

favorite









up vote
2
down vote

favorite












$$int frac1x^2-a^2dx$$




Now, I know this can be done by splitting the function into two integrable functions,



$displaystyledfrac12aint bigg(dfrac1x-a - dfrac1x+abigg)dx$



And then doing the usual stuff.



My question is, how can we do this by using trigonometric substitution?



The only thing that gets in my mind is $x=asectheta$, but then got stuck on proceeding further.



Any help would be appreciated.




integration trigonometry indefinite-integrals




share|cite|improve this question














$$int frac1x^2-a^2dx$$




Now, I know this can be done by splitting the function into two integrable functions,



$displaystyledfrac12aint bigg(dfrac1x-a - dfrac1x+abigg)dx$



And then doing the usual stuff.



My question is, how can we do this by using trigonometric substitution?



The only thing that gets in my mind is $x=asectheta$, but then got stuck on proceeding further.



Any help would be appreciated.





integration trigonometry indefinite-integrals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 30 at 8:24









user 108128

19k41544




19k41544









asked Jul 30 at 7:25









Piano Land

35812




35812







  • 2




    Try $x=acos2theta$.
    – Antonio DJC
    Jul 30 at 7:30










  • The integrand is discontinuous at $x = pm a$, so the antiderivative ought to be split into three intervals, each with a different added constant.
    – mr_e_man
    Jul 30 at 7:57












  • 2




    Try $x=acos2theta$.
    – Antonio DJC
    Jul 30 at 7:30










  • The integrand is discontinuous at $x = pm a$, so the antiderivative ought to be split into three intervals, each with a different added constant.
    – mr_e_man
    Jul 30 at 7:57







2




2




Try $x=acos2theta$.
– Antonio DJC
Jul 30 at 7:30




Try $x=acos2theta$.
– Antonio DJC
Jul 30 at 7:30












The integrand is discontinuous at $x = pm a$, so the antiderivative ought to be split into three intervals, each with a different added constant.
– mr_e_man
Jul 30 at 7:57




The integrand is discontinuous at $x = pm a$, so the antiderivative ought to be split into three intervals, each with a different added constant.
– mr_e_man
Jul 30 at 7:57










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










We have
$$I=int frac1x^2-a^2dx$$



Let $x=acostheta$, then $dx=-asintheta dtheta$.



beginalign
I
&= int frac1x^2-a^2dx \
&= int frac-asinthetaa^2(cos^2theta-1)dtheta \
&=int frac-asintheta-a^2sin^2thetadtheta \
&= frac1aint csctheta dtheta \
&= -frac1alnleftlvertcsctheta+cotthetarightrvert +C\
&= -frac1alnleftlvertfrac1sintheta+fraccosthetasinthetarightrvert +C\
&= -frac1alnleftlvertfrac1+x/apmsqrt1-x^2/a^2rightrvert +C\
&= -frac1alnleftlvertfracsqrt1+x/asqrt1+x/asqrt1-x/asqrt1+x/arightrvert +C\
&= -frac12alnleftlvertfrac1+x/a1-x/arightrvert +C\
&= -frac12alnleftlvertfraca+xa-xrightrvert +C\
endalign






share|cite|improve this answer





















  • Simpler than the logarithm is the inverse hyperbolic tangent: $-frac1a textartanh(costheta) = -frac1atextartanh(frac xa)$. The derivative of $textartanh(x)$ is $1/(1-x^2)$.
    – mr_e_man
    Jul 30 at 7:50


















up vote
3
down vote













In addition to other answers: for $x > a$ a substitution $$x = a cosh(x)$$ seems appropriate. The rest of the solution should work just the same.






share|cite|improve this answer





















  • (+1) It leads to the shortest solution.
    – Bernard
    Jul 30 at 8:52

















up vote
2
down vote













You might use $x=asectheta$: $dx=asecthetatantheta,dtheta$ and the integral becomes
$$
intfrac1a^2(sec^2theta-1)asecthetatantheta,dtheta
=
frac1aintfraccos^2thetasin^2thetafrac1costhetafracsinthetacostheta,dtheta
=
frac1aintfrac1sintheta,dtheta
$$
and this is a known fellow: set $theta=2u$, so the integral becomes (leaving aside the factor $1/a$):
$$
intfrac1sin ucos u,du=
intfrac1cos^2u+sin^2usin ucos u,du=
intfraccos usin u,du+intfracsin ucos u,du
$$
You thus get
$$
loglvertsin urvert-loglvertcos urvert+c=
logleftlverttanfractheta2rightrvert+c
$$
Now back substitute.



Oh, well, but if we do $x=acos t$, instead?
$$
intfrac1a^2(cos^2t-1)(-asin t),dt=
frac1aintfrac1sin t,dt
$$
Speedier, isn't it? Anyway, not easier than observing that
$$
frac1x^2-a^2=frac12aleft(frac1x-a-frac1x+aright)
$$






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    We have
    $$I=int frac1x^2-a^2dx$$



    Let $x=acostheta$, then $dx=-asintheta dtheta$.



    beginalign
    I
    &= int frac1x^2-a^2dx \
    &= int frac-asinthetaa^2(cos^2theta-1)dtheta \
    &=int frac-asintheta-a^2sin^2thetadtheta \
    &= frac1aint csctheta dtheta \
    &= -frac1alnleftlvertcsctheta+cotthetarightrvert +C\
    &= -frac1alnleftlvertfrac1sintheta+fraccosthetasinthetarightrvert +C\
    &= -frac1alnleftlvertfrac1+x/apmsqrt1-x^2/a^2rightrvert +C\
    &= -frac1alnleftlvertfracsqrt1+x/asqrt1+x/asqrt1-x/asqrt1+x/arightrvert +C\
    &= -frac12alnleftlvertfrac1+x/a1-x/arightrvert +C\
    &= -frac12alnleftlvertfraca+xa-xrightrvert +C\
    endalign






    share|cite|improve this answer





















    • Simpler than the logarithm is the inverse hyperbolic tangent: $-frac1a textartanh(costheta) = -frac1atextartanh(frac xa)$. The derivative of $textartanh(x)$ is $1/(1-x^2)$.
      – mr_e_man
      Jul 30 at 7:50















    up vote
    2
    down vote



    accepted










    We have
    $$I=int frac1x^2-a^2dx$$



    Let $x=acostheta$, then $dx=-asintheta dtheta$.



    beginalign
    I
    &= int frac1x^2-a^2dx \
    &= int frac-asinthetaa^2(cos^2theta-1)dtheta \
    &=int frac-asintheta-a^2sin^2thetadtheta \
    &= frac1aint csctheta dtheta \
    &= -frac1alnleftlvertcsctheta+cotthetarightrvert +C\
    &= -frac1alnleftlvertfrac1sintheta+fraccosthetasinthetarightrvert +C\
    &= -frac1alnleftlvertfrac1+x/apmsqrt1-x^2/a^2rightrvert +C\
    &= -frac1alnleftlvertfracsqrt1+x/asqrt1+x/asqrt1-x/asqrt1+x/arightrvert +C\
    &= -frac12alnleftlvertfrac1+x/a1-x/arightrvert +C\
    &= -frac12alnleftlvertfraca+xa-xrightrvert +C\
    endalign






    share|cite|improve this answer





















    • Simpler than the logarithm is the inverse hyperbolic tangent: $-frac1a textartanh(costheta) = -frac1atextartanh(frac xa)$. The derivative of $textartanh(x)$ is $1/(1-x^2)$.
      – mr_e_man
      Jul 30 at 7:50













    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    We have
    $$I=int frac1x^2-a^2dx$$



    Let $x=acostheta$, then $dx=-asintheta dtheta$.



    beginalign
    I
    &= int frac1x^2-a^2dx \
    &= int frac-asinthetaa^2(cos^2theta-1)dtheta \
    &=int frac-asintheta-a^2sin^2thetadtheta \
    &= frac1aint csctheta dtheta \
    &= -frac1alnleftlvertcsctheta+cotthetarightrvert +C\
    &= -frac1alnleftlvertfrac1sintheta+fraccosthetasinthetarightrvert +C\
    &= -frac1alnleftlvertfrac1+x/apmsqrt1-x^2/a^2rightrvert +C\
    &= -frac1alnleftlvertfracsqrt1+x/asqrt1+x/asqrt1-x/asqrt1+x/arightrvert +C\
    &= -frac12alnleftlvertfrac1+x/a1-x/arightrvert +C\
    &= -frac12alnleftlvertfraca+xa-xrightrvert +C\
    endalign






    share|cite|improve this answer













    We have
    $$I=int frac1x^2-a^2dx$$



    Let $x=acostheta$, then $dx=-asintheta dtheta$.



    beginalign
    I
    &= int frac1x^2-a^2dx \
    &= int frac-asinthetaa^2(cos^2theta-1)dtheta \
    &=int frac-asintheta-a^2sin^2thetadtheta \
    &= frac1aint csctheta dtheta \
    &= -frac1alnleftlvertcsctheta+cotthetarightrvert +C\
    &= -frac1alnleftlvertfrac1sintheta+fraccosthetasinthetarightrvert +C\
    &= -frac1alnleftlvertfrac1+x/apmsqrt1-x^2/a^2rightrvert +C\
    &= -frac1alnleftlvertfracsqrt1+x/asqrt1+x/asqrt1-x/asqrt1+x/arightrvert +C\
    &= -frac12alnleftlvertfrac1+x/a1-x/arightrvert +C\
    &= -frac12alnleftlvertfraca+xa-xrightrvert +C\
    endalign







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Jul 30 at 7:47









    Karn Watcharasupat

    3,7992426




    3,7992426











    • Simpler than the logarithm is the inverse hyperbolic tangent: $-frac1a textartanh(costheta) = -frac1atextartanh(frac xa)$. The derivative of $textartanh(x)$ is $1/(1-x^2)$.
      – mr_e_man
      Jul 30 at 7:50

















    • Simpler than the logarithm is the inverse hyperbolic tangent: $-frac1a textartanh(costheta) = -frac1atextartanh(frac xa)$. The derivative of $textartanh(x)$ is $1/(1-x^2)$.
      – mr_e_man
      Jul 30 at 7:50
















    Simpler than the logarithm is the inverse hyperbolic tangent: $-frac1a textartanh(costheta) = -frac1atextartanh(frac xa)$. The derivative of $textartanh(x)$ is $1/(1-x^2)$.
    – mr_e_man
    Jul 30 at 7:50





    Simpler than the logarithm is the inverse hyperbolic tangent: $-frac1a textartanh(costheta) = -frac1atextartanh(frac xa)$. The derivative of $textartanh(x)$ is $1/(1-x^2)$.
    – mr_e_man
    Jul 30 at 7:50











    up vote
    3
    down vote













    In addition to other answers: for $x > a$ a substitution $$x = a cosh(x)$$ seems appropriate. The rest of the solution should work just the same.






    share|cite|improve this answer





















    • (+1) It leads to the shortest solution.
      – Bernard
      Jul 30 at 8:52














    up vote
    3
    down vote













    In addition to other answers: for $x > a$ a substitution $$x = a cosh(x)$$ seems appropriate. The rest of the solution should work just the same.






    share|cite|improve this answer





















    • (+1) It leads to the shortest solution.
      – Bernard
      Jul 30 at 8:52












    up vote
    3
    down vote










    up vote
    3
    down vote









    In addition to other answers: for $x > a$ a substitution $$x = a cosh(x)$$ seems appropriate. The rest of the solution should work just the same.






    share|cite|improve this answer













    In addition to other answers: for $x > a$ a substitution $$x = a cosh(x)$$ seems appropriate. The rest of the solution should work just the same.







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Jul 30 at 7:47









    denklo

    913




    913











    • (+1) It leads to the shortest solution.
      – Bernard
      Jul 30 at 8:52
















    • (+1) It leads to the shortest solution.
      – Bernard
      Jul 30 at 8:52















    (+1) It leads to the shortest solution.
    – Bernard
    Jul 30 at 8:52




    (+1) It leads to the shortest solution.
    – Bernard
    Jul 30 at 8:52










    up vote
    2
    down vote













    You might use $x=asectheta$: $dx=asecthetatantheta,dtheta$ and the integral becomes
    $$
    intfrac1a^2(sec^2theta-1)asecthetatantheta,dtheta
    =
    frac1aintfraccos^2thetasin^2thetafrac1costhetafracsinthetacostheta,dtheta
    =
    frac1aintfrac1sintheta,dtheta
    $$
    and this is a known fellow: set $theta=2u$, so the integral becomes (leaving aside the factor $1/a$):
    $$
    intfrac1sin ucos u,du=
    intfrac1cos^2u+sin^2usin ucos u,du=
    intfraccos usin u,du+intfracsin ucos u,du
    $$
    You thus get
    $$
    loglvertsin urvert-loglvertcos urvert+c=
    logleftlverttanfractheta2rightrvert+c
    $$
    Now back substitute.



    Oh, well, but if we do $x=acos t$, instead?
    $$
    intfrac1a^2(cos^2t-1)(-asin t),dt=
    frac1aintfrac1sin t,dt
    $$
    Speedier, isn't it? Anyway, not easier than observing that
    $$
    frac1x^2-a^2=frac12aleft(frac1x-a-frac1x+aright)
    $$






    share|cite|improve this answer

























      up vote
      2
      down vote













      You might use $x=asectheta$: $dx=asecthetatantheta,dtheta$ and the integral becomes
      $$
      intfrac1a^2(sec^2theta-1)asecthetatantheta,dtheta
      =
      frac1aintfraccos^2thetasin^2thetafrac1costhetafracsinthetacostheta,dtheta
      =
      frac1aintfrac1sintheta,dtheta
      $$
      and this is a known fellow: set $theta=2u$, so the integral becomes (leaving aside the factor $1/a$):
      $$
      intfrac1sin ucos u,du=
      intfrac1cos^2u+sin^2usin ucos u,du=
      intfraccos usin u,du+intfracsin ucos u,du
      $$
      You thus get
      $$
      loglvertsin urvert-loglvertcos urvert+c=
      logleftlverttanfractheta2rightrvert+c
      $$
      Now back substitute.



      Oh, well, but if we do $x=acos t$, instead?
      $$
      intfrac1a^2(cos^2t-1)(-asin t),dt=
      frac1aintfrac1sin t,dt
      $$
      Speedier, isn't it? Anyway, not easier than observing that
      $$
      frac1x^2-a^2=frac12aleft(frac1x-a-frac1x+aright)
      $$






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        You might use $x=asectheta$: $dx=asecthetatantheta,dtheta$ and the integral becomes
        $$
        intfrac1a^2(sec^2theta-1)asecthetatantheta,dtheta
        =
        frac1aintfraccos^2thetasin^2thetafrac1costhetafracsinthetacostheta,dtheta
        =
        frac1aintfrac1sintheta,dtheta
        $$
        and this is a known fellow: set $theta=2u$, so the integral becomes (leaving aside the factor $1/a$):
        $$
        intfrac1sin ucos u,du=
        intfrac1cos^2u+sin^2usin ucos u,du=
        intfraccos usin u,du+intfracsin ucos u,du
        $$
        You thus get
        $$
        loglvertsin urvert-loglvertcos urvert+c=
        logleftlverttanfractheta2rightrvert+c
        $$
        Now back substitute.



        Oh, well, but if we do $x=acos t$, instead?
        $$
        intfrac1a^2(cos^2t-1)(-asin t),dt=
        frac1aintfrac1sin t,dt
        $$
        Speedier, isn't it? Anyway, not easier than observing that
        $$
        frac1x^2-a^2=frac12aleft(frac1x-a-frac1x+aright)
        $$






        share|cite|improve this answer













        You might use $x=asectheta$: $dx=asecthetatantheta,dtheta$ and the integral becomes
        $$
        intfrac1a^2(sec^2theta-1)asecthetatantheta,dtheta
        =
        frac1aintfraccos^2thetasin^2thetafrac1costhetafracsinthetacostheta,dtheta
        =
        frac1aintfrac1sintheta,dtheta
        $$
        and this is a known fellow: set $theta=2u$, so the integral becomes (leaving aside the factor $1/a$):
        $$
        intfrac1sin ucos u,du=
        intfrac1cos^2u+sin^2usin ucos u,du=
        intfraccos usin u,du+intfracsin ucos u,du
        $$
        You thus get
        $$
        loglvertsin urvert-loglvertcos urvert+c=
        logleftlverttanfractheta2rightrvert+c
        $$
        Now back substitute.



        Oh, well, but if we do $x=acos t$, instead?
        $$
        intfrac1a^2(cos^2t-1)(-asin t),dt=
        frac1aintfrac1sin t,dt
        $$
        Speedier, isn't it? Anyway, not easier than observing that
        $$
        frac1x^2-a^2=frac12aleft(frac1x-a-frac1x+aright)
        $$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 30 at 7:56









        egreg

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