Mixing
Integration of $frac1x^2-a^2$ by trigonometric substitution?
Clash Royale CLAN TAG#URR8PPP
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
edited Jul 30 at 8:24
user 108128
19k41544
asked Jul 30 at 7:25
Piano Land
35812
add a comment |Â
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
edited Jul 30 at 8:24
user 108128
19k41544
asked Jul 30 at 7:25
Piano Land
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
add a comment |Â
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
edited Jul 30 at 8:24
user 108128
19k41544
asked Jul 30 at 7:25
Piano Land
35812
$$int frac1x^2-a^2dx$$
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
edited Jul 30 at 8:24
user 108128
19k41544
asked Jul 30 at 7:25
Piano Land
35812
edited Jul 30 at 8:24
user 108128
19k41544
edited Jul 30 at 8:24
user 108128
19k41544
edited Jul 30 at 8:24
user 108128
19k41544
19k41544
asked Jul 30 at 7:25
Piano Land
35812
asked Jul 30 at 7:25
Piano Land
35812
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
add a comment |Â
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
add a comment |Â
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
answered Jul 30 at 7:47
Karn Watcharasupat
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
add a comment |Â
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.
answered Jul 30 at 7:47
denklo
913
(+1) It leads to the shortest solution.
– Bernard
Jul 30 at 8:52
add a comment |Â
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
$$
frac1x^2-a^2=frac12aleft(frac1x-a-frac1x+aright)
$$
answered Jul 30 at 7:56
egreg
164k1180187
add a comment |Â
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
answered Jul 30 at 7:47
Karn Watcharasupat
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
add a comment |Â
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
answered Jul 30 at 7:47
Karn Watcharasupat
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
add a comment |Â
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
answered Jul 30 at 7:47
Karn Watcharasupat
3,7992426
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
answered Jul 30 at 7:47
Karn Watcharasupat
3,7992426
answered Jul 30 at 7:47
Karn Watcharasupat
3,7992426
answered Jul 30 at 7:47
Karn Watcharasupat
3,7992426
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
add a comment |Â
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
add a comment |Â
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.
answered Jul 30 at 7:47
denklo
913
(+1) It leads to the shortest solution.
– Bernard
Jul 30 at 8:52
add a comment |Â
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.
answered Jul 30 at 7:47
denklo
913
(+1) It leads to the shortest solution.
– Bernard
Jul 30 at 8:52
add a comment |Â
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.
answered Jul 30 at 7:47
denklo
913
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.
answered Jul 30 at 7:47
denklo
913
answered Jul 30 at 7:47
denklo
913
answered Jul 30 at 7:47
denklo
913
answered Jul 30 at 7:47
denklo
913
913
(+1) It leads to the shortest solution.
– Bernard
Jul 30 at 8:52
add a comment |Â
(+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
add a comment |Â
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
$$
frac1x^2-a^2=frac12aleft(frac1x-a-frac1x+aright)
$$
answered Jul 30 at 7:56
egreg
164k1180187
add a comment |Â
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
$$
frac1x^2-a^2=frac12aleft(frac1x-a-frac1x+aright)
$$
answered Jul 30 at 7:56
egreg
164k1180187
add a comment |Â
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
$$
frac1x^2-a^2=frac12aleft(frac1x-a-frac1x+aright)
$$
answered Jul 30 at 7:56
egreg
164k1180187
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
$$
frac1x^2-a^2=frac12aleft(frac1x-a-frac1x+aright)
$$
answered Jul 30 at 7:56
egreg
164k1180187
answered Jul 30 at 7:56
egreg
164k1180187
answered Jul 30 at 7:56
egreg
164k1180187
answered Jul 30 at 7:56
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866731%2fintegration-of-frac1x2-a2-by-trigonometric-substitution%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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