Mixing
Evaluating $ int sqrt{x^2-1} dx$
$begingroup$
I've seen the $sec(theta)$ substitution, but can't I factor out an $i$ so that it becomes $isqrt{1-x^2}$? Then if I use the $sin(theta)=x$ substitution, it is comprised of only complex parts. If anyone could sort out the mistake, that'd be awesome!
calculus
edited Jan 14 at 13:36
amWhy
1
asked Jan 14 at 13:02
KoushalKoushal
1
$endgroup$
add a comment |
$begingroup$
I've seen the $sec(theta)$ substitution, but can't I factor out an $i$ so that it becomes $isqrt{1-x^2}$? Then if I use the $sin(theta)=x$ substitution, it is comprised of only complex parts. If anyone could sort out the mistake, that'd be awesome!
calculus
edited Jan 14 at 13:36
amWhy
1
asked Jan 14 at 13:02
KoushalKoushal
1
$endgroup$
$begingroup$
@OlivierOloa the expressions in the OP's statement are equivalent: $isqrt{1-x^2}=sqrt{x^2-1}$
$endgroup$
– Paras Khosla
Jan 14 at 13:10
$begingroup$
Why is it a problem that you obtain an imaginary result? Evaluate the integrand on [-1, 1].
$endgroup$
– Mariuslp
Jan 14 at 13:46
add a comment |
$begingroup$
I've seen the $sec(theta)$ substitution, but can't I factor out an $i$ so that it becomes $isqrt{1-x^2}$? Then if I use the $sin(theta)=x$ substitution, it is comprised of only complex parts. If anyone could sort out the mistake, that'd be awesome!
calculus
edited Jan 14 at 13:36
amWhy
1
asked Jan 14 at 13:02
KoushalKoushal
1
$endgroup$
I've seen the $sec(theta)$ substitution, but can't I factor out an $i$ so that it becomes $isqrt{1-x^2}$? Then if I use the $sin(theta)=x$ substitution, it is comprised of only complex parts. If anyone could sort out the mistake, that'd be awesome!
calculus
calculus
edited Jan 14 at 13:36
amWhy
1
asked Jan 14 at 13:02
KoushalKoushal
1
edited Jan 14 at 13:36
amWhy
1
asked Jan 14 at 13:02
KoushalKoushal
1
edited Jan 14 at 13:36
amWhy
1
edited Jan 14 at 13:36
amWhy
1
edited Jan 14 at 13:36
amWhy
1
1
asked Jan 14 at 13:02
KoushalKoushal
1
asked Jan 14 at 13:02
KoushalKoushal
1
asked Jan 14 at 13:02
KoushalKoushal
1
1
$begingroup$
@OlivierOloa the expressions in the OP's statement are equivalent: $isqrt{1-x^2}=sqrt{x^2-1}$
$endgroup$
– Paras Khosla
Jan 14 at 13:10
$begingroup$
Why is it a problem that you obtain an imaginary result? Evaluate the integrand on [-1, 1].
$endgroup$
– Mariuslp
Jan 14 at 13:46
add a comment |
$begingroup$
@OlivierOloa the expressions in the OP's statement are equivalent: $isqrt{1-x^2}=sqrt{x^2-1}$
$endgroup$
– Paras Khosla
Jan 14 at 13:10
$begingroup$
Why is it a problem that you obtain an imaginary result? Evaluate the integrand on [-1, 1].
$endgroup$
– Mariuslp
Jan 14 at 13:46
$begingroup$
@OlivierOloa the expressions in the OP's statement are equivalent: $isqrt{1-x^2}=sqrt{x^2-1}$
$endgroup$
– Paras Khosla
Jan 14 at 13:10
$begingroup$
@OlivierOloa the expressions in the OP's statement are equivalent: $isqrt{1-x^2}=sqrt{x^2-1}$
$endgroup$
– Paras Khosla
Jan 14 at 13:10
$begingroup$
Why is it a problem that you obtain an imaginary result? Evaluate the integrand on [-1, 1].
$endgroup$
– Mariuslp
Jan 14 at 13:46
$begingroup$
Why is it a problem that you obtain an imaginary result? Evaluate the integrand on [-1, 1].
$endgroup$
– Mariuslp
Jan 14 at 13:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can handle this integral with the hyperbolic cosine rather than a trigonometric function.
By-parts integration also works:
$$I:=intsqrt{x^2-1}dx=xsqrt{x^2-1}-intfrac{x^2}{sqrt{x^2-1}}dx.$$
Then
$$intfrac{x^2}{sqrt{x^2-1}}dx=intfrac{x^2-1+1}{sqrt{x^2-1}}dx=intsqrt{x^2-1},dx+intfrac{dx}{sqrt{x^2-1}}=I+text{arcosh}(x),$$
and you can draw $I$.
answered Jan 14 at 14:05
Yves DaoustYves Daoust
128k674226
$endgroup$
add a comment |
$begingroup$
The integral makes sense for $|x|ge1$, so generally your $theta$ will be complex and nonreal. So there's no real problem in getting a real function as the antiderivative, because the imaginary part will cancel out.
On the other hand, you will have problems in determining $costheta$ given $sintheta$, because you have to choose the correct branch.
If you go on, you will find the usual integral, multiplied by $i$,
$$
frac{i}{2}(theta-sinthetacostheta)=
frac{1}{2}(iarcsin x-ixsqrt{1-x^2})=
frac{1}{2}(iarcsin x+xsqrt{x^2-1})
$$
Note that, for $|x|>1$, $arcsin x$ is not real. You might enjoy determining it.
answered Jan 14 at 13:47
egregegreg
182k1485203
$endgroup$
1
$begingroup$
Who said that $|x|<1$ ?
$endgroup$
– Yves Daoust
Jan 14 at 14:08
$begingroup$
@YvesDaoust Surely I didn't.
$endgroup$
– egreg
Jan 14 at 15:05
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can handle this integral with the hyperbolic cosine rather than a trigonometric function.
By-parts integration also works:
$$I:=intsqrt{x^2-1}dx=xsqrt{x^2-1}-intfrac{x^2}{sqrt{x^2-1}}dx.$$
Then
$$intfrac{x^2}{sqrt{x^2-1}}dx=intfrac{x^2-1+1}{sqrt{x^2-1}}dx=intsqrt{x^2-1},dx+intfrac{dx}{sqrt{x^2-1}}=I+text{arcosh}(x),$$
and you can draw $I$.
answered Jan 14 at 14:05
Yves DaoustYves Daoust
128k674226
$endgroup$
add a comment |
$begingroup$
You can handle this integral with the hyperbolic cosine rather than a trigonometric function.
By-parts integration also works:
$$I:=intsqrt{x^2-1}dx=xsqrt{x^2-1}-intfrac{x^2}{sqrt{x^2-1}}dx.$$
Then
$$intfrac{x^2}{sqrt{x^2-1}}dx=intfrac{x^2-1+1}{sqrt{x^2-1}}dx=intsqrt{x^2-1},dx+intfrac{dx}{sqrt{x^2-1}}=I+text{arcosh}(x),$$
and you can draw $I$.
answered Jan 14 at 14:05
Yves DaoustYves Daoust
128k674226
$endgroup$
add a comment |
$begingroup$
You can handle this integral with the hyperbolic cosine rather than a trigonometric function.
By-parts integration also works:
$$I:=intsqrt{x^2-1}dx=xsqrt{x^2-1}-intfrac{x^2}{sqrt{x^2-1}}dx.$$
Then
$$intfrac{x^2}{sqrt{x^2-1}}dx=intfrac{x^2-1+1}{sqrt{x^2-1}}dx=intsqrt{x^2-1},dx+intfrac{dx}{sqrt{x^2-1}}=I+text{arcosh}(x),$$
and you can draw $I$.
answered Jan 14 at 14:05
Yves DaoustYves Daoust
128k674226
$endgroup$
You can handle this integral with the hyperbolic cosine rather than a trigonometric function.
By-parts integration also works:
$$I:=intsqrt{x^2-1}dx=xsqrt{x^2-1}-intfrac{x^2}{sqrt{x^2-1}}dx.$$
Then
$$intfrac{x^2}{sqrt{x^2-1}}dx=intfrac{x^2-1+1}{sqrt{x^2-1}}dx=intsqrt{x^2-1},dx+intfrac{dx}{sqrt{x^2-1}}=I+text{arcosh}(x),$$
and you can draw $I$.
answered Jan 14 at 14:05
Yves DaoustYves Daoust
128k674226
answered Jan 14 at 14:05
Yves DaoustYves Daoust
128k674226
answered Jan 14 at 14:05
Yves DaoustYves Daoust
128k674226
answered Jan 14 at 14:05
Yves DaoustYves Daoust
128k674226
128k674226
add a comment |
add a comment |
$begingroup$
The integral makes sense for $|x|ge1$, so generally your $theta$ will be complex and nonreal. So there's no real problem in getting a real function as the antiderivative, because the imaginary part will cancel out.
On the other hand, you will have problems in determining $costheta$ given $sintheta$, because you have to choose the correct branch.
If you go on, you will find the usual integral, multiplied by $i$,
$$
frac{i}{2}(theta-sinthetacostheta)=
frac{1}{2}(iarcsin x-ixsqrt{1-x^2})=
frac{1}{2}(iarcsin x+xsqrt{x^2-1})
$$
Note that, for $|x|>1$, $arcsin x$ is not real. You might enjoy determining it.
answered Jan 14 at 13:47
egregegreg
182k1485203
$endgroup$
1
$begingroup$
Who said that $|x|<1$ ?
$endgroup$
– Yves Daoust
Jan 14 at 14:08
$begingroup$
@YvesDaoust Surely I didn't.
$endgroup$
– egreg
Jan 14 at 15:05
add a comment |
$begingroup$
The integral makes sense for $|x|ge1$, so generally your $theta$ will be complex and nonreal. So there's no real problem in getting a real function as the antiderivative, because the imaginary part will cancel out.
On the other hand, you will have problems in determining $costheta$ given $sintheta$, because you have to choose the correct branch.
If you go on, you will find the usual integral, multiplied by $i$,
$$
frac{i}{2}(theta-sinthetacostheta)=
frac{1}{2}(iarcsin x-ixsqrt{1-x^2})=
frac{1}{2}(iarcsin x+xsqrt{x^2-1})
$$
Note that, for $|x|>1$, $arcsin x$ is not real. You might enjoy determining it.
answered Jan 14 at 13:47
egregegreg
182k1485203
$endgroup$
1
$begingroup$
Who said that $|x|<1$ ?
$endgroup$
– Yves Daoust
Jan 14 at 14:08
$begingroup$
@YvesDaoust Surely I didn't.
$endgroup$
– egreg
Jan 14 at 15:05
add a comment |
$begingroup$
The integral makes sense for $|x|ge1$, so generally your $theta$ will be complex and nonreal. So there's no real problem in getting a real function as the antiderivative, because the imaginary part will cancel out.
On the other hand, you will have problems in determining $costheta$ given $sintheta$, because you have to choose the correct branch.
If you go on, you will find the usual integral, multiplied by $i$,
$$
frac{i}{2}(theta-sinthetacostheta)=
frac{1}{2}(iarcsin x-ixsqrt{1-x^2})=
frac{1}{2}(iarcsin x+xsqrt{x^2-1})
$$
Note that, for $|x|>1$, $arcsin x$ is not real. You might enjoy determining it.
answered Jan 14 at 13:47
egregegreg
182k1485203
$endgroup$
The integral makes sense for $|x|ge1$, so generally your $theta$ will be complex and nonreal. So there's no real problem in getting a real function as the antiderivative, because the imaginary part will cancel out.
On the other hand, you will have problems in determining $costheta$ given $sintheta$, because you have to choose the correct branch.
If you go on, you will find the usual integral, multiplied by $i$,
$$
frac{i}{2}(theta-sinthetacostheta)=
frac{1}{2}(iarcsin x-ixsqrt{1-x^2})=
frac{1}{2}(iarcsin x+xsqrt{x^2-1})
$$
Note that, for $|x|>1$, $arcsin x$ is not real. You might enjoy determining it.
answered Jan 14 at 13:47
egregegreg
182k1485203
edited Jan 14 at 15:20
answered Jan 14 at 13:47
egregegreg
182k1485203
answered Jan 14 at 13:47
egregegreg
182k1485203
answered Jan 14 at 13:47
egregegreg
182k1485203
182k1485203
1
$begingroup$
Who said that $|x|<1$ ?
$endgroup$
– Yves Daoust
Jan 14 at 14:08
$begingroup$
@YvesDaoust Surely I didn't.
$endgroup$
– egreg
Jan 14 at 15:05
add a comment |
1
$begingroup$
Who said that $|x|<1$ ?
$endgroup$
– Yves Daoust
Jan 14 at 14:08
$begingroup$
@YvesDaoust Surely I didn't.
$endgroup$
– egreg
Jan 14 at 15:05
1
1
$begingroup$
Who said that $|x|<1$ ?
$endgroup$
– Yves Daoust
Jan 14 at 14:08
$begingroup$
Who said that $|x|<1$ ?
$endgroup$
– Yves Daoust
Jan 14 at 14:08
$begingroup$
@YvesDaoust Surely I didn't.
$endgroup$
– egreg
Jan 14 at 15:05
$begingroup$
@YvesDaoust Surely I didn't.
$endgroup$
– egreg
Jan 14 at 15:05
add a comment |
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$begingroup$
@OlivierOloa the expressions in the OP's statement are equivalent: $isqrt{1-x^2}=sqrt{x^2-1}$
$endgroup$
– Paras Khosla
Jan 14 at 13:10
$begingroup$
Why is it a problem that you obtain an imaginary result? Evaluate the integrand on [-1, 1].
$endgroup$
– Mariuslp
Jan 14 at 13:46