Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$?
up vote
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I thought that $theta$ must be chosen such that $cosh(theta)$ has a range that is equal to the domain of $frac{1}{(x^2-1)^{3/2}}$. But this can't be done since the domain includes negative numbers, and $cosh(theta)$ is always positive. If this substitution is made, the answer seems to be valid for all values of x, why does it work?
calculus real-analysis integration indefinite-integrals
asked Nov 20 at 13:59
DinW
182
add a comment |
up vote
3
down vote
favorite
I thought that $theta$ must be chosen such that $cosh(theta)$ has a range that is equal to the domain of $frac{1}{(x^2-1)^{3/2}}$. But this can't be done since the domain includes negative numbers, and $cosh(theta)$ is always positive. If this substitution is made, the answer seems to be valid for all values of x, why does it work?
calculus real-analysis integration indefinite-integrals
asked Nov 20 at 13:59
DinW
182
It's better to use $x=sec t,sqrt{x^2-1}=|tan t|$
– lab bhattacharjee
Nov 20 at 14:15
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I thought that $theta$ must be chosen such that $cosh(theta)$ has a range that is equal to the domain of $frac{1}{(x^2-1)^{3/2}}$. But this can't be done since the domain includes negative numbers, and $cosh(theta)$ is always positive. If this substitution is made, the answer seems to be valid for all values of x, why does it work?
calculus real-analysis integration indefinite-integrals
asked Nov 20 at 13:59
DinW
182
I thought that $theta$ must be chosen such that $cosh(theta)$ has a range that is equal to the domain of $frac{1}{(x^2-1)^{3/2}}$. But this can't be done since the domain includes negative numbers, and $cosh(theta)$ is always positive. If this substitution is made, the answer seems to be valid for all values of x, why does it work?
calculus real-analysis integration indefinite-integrals
calculus real-analysis integration indefinite-integrals
asked Nov 20 at 13:59
DinW
182
asked Nov 20 at 13:59
DinW
182
asked Nov 20 at 13:59
DinW
182
asked Nov 20 at 13:59
DinW
182
asked Nov 20 at 13:59
DinW
182
182
It's better to use $x=sec t,sqrt{x^2-1}=|tan t|$
– lab bhattacharjee
Nov 20 at 14:15
add a comment |
It's better to use $x=sec t,sqrt{x^2-1}=|tan t|$
– lab bhattacharjee
Nov 20 at 14:15
It's better to use $x=sec t,sqrt{x^2-1}=|tan t|$
– lab bhattacharjee
Nov 20 at 14:15
It's better to use $x=sec t,sqrt{x^2-1}=|tan t|$
– lab bhattacharjee
Nov 20 at 14:15
add a comment |
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
So there are singularities at $pm 1$, you have to avoid those. So let's consider finding an antiderivative valid on $(-infty,-1)$.
In fact $x=cosh(theta)$ is not valid there. The substitution really being used there is $x=-cosh(theta)$, so $dx=-sinh(theta) d theta$, so the integral becomes
$$int frac{1}{(cosh^2(theta)-1)^{3/2}} (-sinh(theta)) d theta.$$
The catch comes when you rewrite $(cosh^2(theta)-1)^{3/2}$ as just $sinh^3(theta)$. This is not strictly correct, in fact it is $|sinh^3(theta)|$ in general. On $(-infty,-1)$, this absolute value reduces to $-sinh^3(theta)$. The formula collapses back to the formula you would get for an antiderivative on $(1,infty)$ because this minus sign cancels out with the minus sign that came with $dx$.
This is relatively general, but I'm not sure whether pinning down exactly what happens in general will be helpful to you at this point.
answered Nov 20 at 14:09
Ian
67.3k25285
Thank you. This response had a lot of detail and answered my question.
– DinW
Nov 20 at 21:31
add a comment |
up vote
3
down vote
If you look for a primitive in $(1,+infty)$,
put $$x=cosh(theta)$$
with $$dx=sinh(theta) dtheta$$
but if you want a primitive in $(-infty,-1)$, you should put
$$x=-cosh(theta)$$
with $$dx=-sinh(theta)dtheta$$
answered Nov 20 at 14:06
hamam_Abdallah
37.7k21634
add a comment |
up vote
0
down vote
Part of it's because we're presumably only integrating over values of $x$ for which the integrand is real, i.e. $|x|ge 1$. Any work with complex numbers that lets you pose the question otherwise in the first place also lets you get a complex $operatorname{arcosh}x$ with $|x|<1$.
answered Nov 20 at 14:17
J.G.
21.3k21933
add a comment |
up vote
0
down vote
$$I=intleft(x^2-1right)^{-3/2}dx$$
we want this term $x^2-1$ to be simplified to a single term, so think of trig identities.:
$$cos^2x+sin^2xequiv1,,cosh^2x-sinh^2xequiv1$$
are the ones you need to know. since it is of the form $x^2-b^2$ rather than $b^2-x^2$ or $x^2+b^2$ we know to use hyperbolic functions. Manipulating the right rule, we obtain:
$$sinh^2x=cosh^2x-1$$
now we can use this. let:
$x=cosh(t)$ and we get $dx=sinh(t)$ and rewrite the integral as:
$$I=intleft(cosh^2(t)-1right)^{-3/2}.sinh(t)dt=int(sinh(t))^{-3}.sinh(t)dt=inttext{csch}^2(t)dt=-coth(t)+C$$
answered Nov 20 at 23:51
Henry Lee
1,707218
Hi, thanks for the response. I asking more about the choice for the substitution than how to solve the integral itself. Since, the range of cosh(x) doesn't equal the domain of the function that is being integrated.
– DinW
Nov 21 at 1:22
It is a known option in examples like these where you have a fractional power of a second order polynomial
– Henry Lee
Nov 21 at 1:23
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
So there are singularities at $pm 1$, you have to avoid those. So let's consider finding an antiderivative valid on $(-infty,-1)$.
In fact $x=cosh(theta)$ is not valid there. The substitution really being used there is $x=-cosh(theta)$, so $dx=-sinh(theta) d theta$, so the integral becomes
$$int frac{1}{(cosh^2(theta)-1)^{3/2}} (-sinh(theta)) d theta.$$
The catch comes when you rewrite $(cosh^2(theta)-1)^{3/2}$ as just $sinh^3(theta)$. This is not strictly correct, in fact it is $|sinh^3(theta)|$ in general. On $(-infty,-1)$, this absolute value reduces to $-sinh^3(theta)$. The formula collapses back to the formula you would get for an antiderivative on $(1,infty)$ because this minus sign cancels out with the minus sign that came with $dx$.
This is relatively general, but I'm not sure whether pinning down exactly what happens in general will be helpful to you at this point.
answered Nov 20 at 14:09
Ian
67.3k25285
Thank you. This response had a lot of detail and answered my question.
– DinW
Nov 20 at 21:31
add a comment |
up vote
2
down vote
accepted
So there are singularities at $pm 1$, you have to avoid those. So let's consider finding an antiderivative valid on $(-infty,-1)$.
In fact $x=cosh(theta)$ is not valid there. The substitution really being used there is $x=-cosh(theta)$, so $dx=-sinh(theta) d theta$, so the integral becomes
$$int frac{1}{(cosh^2(theta)-1)^{3/2}} (-sinh(theta)) d theta.$$
The catch comes when you rewrite $(cosh^2(theta)-1)^{3/2}$ as just $sinh^3(theta)$. This is not strictly correct, in fact it is $|sinh^3(theta)|$ in general. On $(-infty,-1)$, this absolute value reduces to $-sinh^3(theta)$. The formula collapses back to the formula you would get for an antiderivative on $(1,infty)$ because this minus sign cancels out with the minus sign that came with $dx$.
This is relatively general, but I'm not sure whether pinning down exactly what happens in general will be helpful to you at this point.
answered Nov 20 at 14:09
Ian
67.3k25285
Thank you. This response had a lot of detail and answered my question.
– DinW
Nov 20 at 21:31
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
So there are singularities at $pm 1$, you have to avoid those. So let's consider finding an antiderivative valid on $(-infty,-1)$.
In fact $x=cosh(theta)$ is not valid there. The substitution really being used there is $x=-cosh(theta)$, so $dx=-sinh(theta) d theta$, so the integral becomes
$$int frac{1}{(cosh^2(theta)-1)^{3/2}} (-sinh(theta)) d theta.$$
The catch comes when you rewrite $(cosh^2(theta)-1)^{3/2}$ as just $sinh^3(theta)$. This is not strictly correct, in fact it is $|sinh^3(theta)|$ in general. On $(-infty,-1)$, this absolute value reduces to $-sinh^3(theta)$. The formula collapses back to the formula you would get for an antiderivative on $(1,infty)$ because this minus sign cancels out with the minus sign that came with $dx$.
This is relatively general, but I'm not sure whether pinning down exactly what happens in general will be helpful to you at this point.
answered Nov 20 at 14:09
Ian
67.3k25285
So there are singularities at $pm 1$, you have to avoid those. So let's consider finding an antiderivative valid on $(-infty,-1)$.
In fact $x=cosh(theta)$ is not valid there. The substitution really being used there is $x=-cosh(theta)$, so $dx=-sinh(theta) d theta$, so the integral becomes
$$int frac{1}{(cosh^2(theta)-1)^{3/2}} (-sinh(theta)) d theta.$$
The catch comes when you rewrite $(cosh^2(theta)-1)^{3/2}$ as just $sinh^3(theta)$. This is not strictly correct, in fact it is $|sinh^3(theta)|$ in general. On $(-infty,-1)$, this absolute value reduces to $-sinh^3(theta)$. The formula collapses back to the formula you would get for an antiderivative on $(1,infty)$ because this minus sign cancels out with the minus sign that came with $dx$.
This is relatively general, but I'm not sure whether pinning down exactly what happens in general will be helpful to you at this point.
answered Nov 20 at 14:09
Ian
67.3k25285
edited Nov 20 at 14:15
answered Nov 20 at 14:09
Ian
67.3k25285
answered Nov 20 at 14:09
Ian
67.3k25285
answered Nov 20 at 14:09
Ian
67.3k25285
67.3k25285
Thank you. This response had a lot of detail and answered my question.
– DinW
Nov 20 at 21:31
add a comment |
Thank you. This response had a lot of detail and answered my question.
– DinW
Nov 20 at 21:31
Thank you. This response had a lot of detail and answered my question.
– DinW
Nov 20 at 21:31
Thank you. This response had a lot of detail and answered my question.
– DinW
Nov 20 at 21:31
add a comment |
up vote
3
down vote
If you look for a primitive in $(1,+infty)$,
put $$x=cosh(theta)$$
with $$dx=sinh(theta) dtheta$$
but if you want a primitive in $(-infty,-1)$, you should put
$$x=-cosh(theta)$$
with $$dx=-sinh(theta)dtheta$$
answered Nov 20 at 14:06
hamam_Abdallah
37.7k21634
add a comment |
up vote
3
down vote
If you look for a primitive in $(1,+infty)$,
put $$x=cosh(theta)$$
with $$dx=sinh(theta) dtheta$$
but if you want a primitive in $(-infty,-1)$, you should put
$$x=-cosh(theta)$$
with $$dx=-sinh(theta)dtheta$$
answered Nov 20 at 14:06
hamam_Abdallah
37.7k21634
add a comment |
up vote
3
down vote
up vote
3
down vote
If you look for a primitive in $(1,+infty)$,
put $$x=cosh(theta)$$
with $$dx=sinh(theta) dtheta$$
but if you want a primitive in $(-infty,-1)$, you should put
$$x=-cosh(theta)$$
with $$dx=-sinh(theta)dtheta$$
answered Nov 20 at 14:06
hamam_Abdallah
37.7k21634
If you look for a primitive in $(1,+infty)$,
put $$x=cosh(theta)$$
with $$dx=sinh(theta) dtheta$$
but if you want a primitive in $(-infty,-1)$, you should put
$$x=-cosh(theta)$$
with $$dx=-sinh(theta)dtheta$$
answered Nov 20 at 14:06
hamam_Abdallah
37.7k21634
answered Nov 20 at 14:06
hamam_Abdallah
37.7k21634
answered Nov 20 at 14:06
hamam_Abdallah
37.7k21634
answered Nov 20 at 14:06
hamam_Abdallah
37.7k21634
37.7k21634
add a comment |
add a comment |
up vote
0
down vote
Part of it's because we're presumably only integrating over values of $x$ for which the integrand is real, i.e. $|x|ge 1$. Any work with complex numbers that lets you pose the question otherwise in the first place also lets you get a complex $operatorname{arcosh}x$ with $|x|<1$.
answered Nov 20 at 14:17
J.G.
21.3k21933
add a comment |
up vote
0
down vote
Part of it's because we're presumably only integrating over values of $x$ for which the integrand is real, i.e. $|x|ge 1$. Any work with complex numbers that lets you pose the question otherwise in the first place also lets you get a complex $operatorname{arcosh}x$ with $|x|<1$.
answered Nov 20 at 14:17
J.G.
21.3k21933
add a comment |
up vote
0
down vote
up vote
0
down vote
Part of it's because we're presumably only integrating over values of $x$ for which the integrand is real, i.e. $|x|ge 1$. Any work with complex numbers that lets you pose the question otherwise in the first place also lets you get a complex $operatorname{arcosh}x$ with $|x|<1$.
answered Nov 20 at 14:17
J.G.
21.3k21933
Part of it's because we're presumably only integrating over values of $x$ for which the integrand is real, i.e. $|x|ge 1$. Any work with complex numbers that lets you pose the question otherwise in the first place also lets you get a complex $operatorname{arcosh}x$ with $|x|<1$.
answered Nov 20 at 14:17
J.G.
21.3k21933
answered Nov 20 at 14:17
J.G.
21.3k21933
answered Nov 20 at 14:17
J.G.
21.3k21933
answered Nov 20 at 14:17
J.G.
21.3k21933
21.3k21933
add a comment |
add a comment |
up vote
0
down vote
$$I=intleft(x^2-1right)^{-3/2}dx$$
we want this term $x^2-1$ to be simplified to a single term, so think of trig identities.:
$$cos^2x+sin^2xequiv1,,cosh^2x-sinh^2xequiv1$$
are the ones you need to know. since it is of the form $x^2-b^2$ rather than $b^2-x^2$ or $x^2+b^2$ we know to use hyperbolic functions. Manipulating the right rule, we obtain:
$$sinh^2x=cosh^2x-1$$
now we can use this. let:
$x=cosh(t)$ and we get $dx=sinh(t)$ and rewrite the integral as:
$$I=intleft(cosh^2(t)-1right)^{-3/2}.sinh(t)dt=int(sinh(t))^{-3}.sinh(t)dt=inttext{csch}^2(t)dt=-coth(t)+C$$
answered Nov 20 at 23:51
Henry Lee
1,707218
Hi, thanks for the response. I asking more about the choice for the substitution than how to solve the integral itself. Since, the range of cosh(x) doesn't equal the domain of the function that is being integrated.
– DinW
Nov 21 at 1:22
It is a known option in examples like these where you have a fractional power of a second order polynomial
– Henry Lee
Nov 21 at 1:23
add a comment |
up vote
0
down vote
$$I=intleft(x^2-1right)^{-3/2}dx$$
we want this term $x^2-1$ to be simplified to a single term, so think of trig identities.:
$$cos^2x+sin^2xequiv1,,cosh^2x-sinh^2xequiv1$$
are the ones you need to know. since it is of the form $x^2-b^2$ rather than $b^2-x^2$ or $x^2+b^2$ we know to use hyperbolic functions. Manipulating the right rule, we obtain:
$$sinh^2x=cosh^2x-1$$
now we can use this. let:
$x=cosh(t)$ and we get $dx=sinh(t)$ and rewrite the integral as:
$$I=intleft(cosh^2(t)-1right)^{-3/2}.sinh(t)dt=int(sinh(t))^{-3}.sinh(t)dt=inttext{csch}^2(t)dt=-coth(t)+C$$
answered Nov 20 at 23:51
Henry Lee
1,707218
Hi, thanks for the response. I asking more about the choice for the substitution than how to solve the integral itself. Since, the range of cosh(x) doesn't equal the domain of the function that is being integrated.
– DinW
Nov 21 at 1:22
It is a known option in examples like these where you have a fractional power of a second order polynomial
– Henry Lee
Nov 21 at 1:23
add a comment |
up vote
0
down vote
up vote
0
down vote
$$I=intleft(x^2-1right)^{-3/2}dx$$
we want this term $x^2-1$ to be simplified to a single term, so think of trig identities.:
$$cos^2x+sin^2xequiv1,,cosh^2x-sinh^2xequiv1$$
are the ones you need to know. since it is of the form $x^2-b^2$ rather than $b^2-x^2$ or $x^2+b^2$ we know to use hyperbolic functions. Manipulating the right rule, we obtain:
$$sinh^2x=cosh^2x-1$$
now we can use this. let:
$x=cosh(t)$ and we get $dx=sinh(t)$ and rewrite the integral as:
$$I=intleft(cosh^2(t)-1right)^{-3/2}.sinh(t)dt=int(sinh(t))^{-3}.sinh(t)dt=inttext{csch}^2(t)dt=-coth(t)+C$$
answered Nov 20 at 23:51
Henry Lee
1,707218
$$I=intleft(x^2-1right)^{-3/2}dx$$
we want this term $x^2-1$ to be simplified to a single term, so think of trig identities.:
$$cos^2x+sin^2xequiv1,,cosh^2x-sinh^2xequiv1$$
are the ones you need to know. since it is of the form $x^2-b^2$ rather than $b^2-x^2$ or $x^2+b^2$ we know to use hyperbolic functions. Manipulating the right rule, we obtain:
$$sinh^2x=cosh^2x-1$$
now we can use this. let:
$x=cosh(t)$ and we get $dx=sinh(t)$ and rewrite the integral as:
$$I=intleft(cosh^2(t)-1right)^{-3/2}.sinh(t)dt=int(sinh(t))^{-3}.sinh(t)dt=inttext{csch}^2(t)dt=-coth(t)+C$$
answered Nov 20 at 23:51
Henry Lee
1,707218
answered Nov 20 at 23:51
Henry Lee
1,707218
answered Nov 20 at 23:51
Henry Lee
1,707218
answered Nov 20 at 23:51
Henry Lee
1,707218
1,707218
Hi, thanks for the response. I asking more about the choice for the substitution than how to solve the integral itself. Since, the range of cosh(x) doesn't equal the domain of the function that is being integrated.
– DinW
Nov 21 at 1:22
It is a known option in examples like these where you have a fractional power of a second order polynomial
– Henry Lee
Nov 21 at 1:23
add a comment |
Hi, thanks for the response. I asking more about the choice for the substitution than how to solve the integral itself. Since, the range of cosh(x) doesn't equal the domain of the function that is being integrated.
– DinW
Nov 21 at 1:22
It is a known option in examples like these where you have a fractional power of a second order polynomial
– Henry Lee
Nov 21 at 1:23
Hi, thanks for the response. I asking more about the choice for the substitution than how to solve the integral itself. Since, the range of cosh(x) doesn't equal the domain of the function that is being integrated.
– DinW
Nov 21 at 1:22
Hi, thanks for the response. I asking more about the choice for the substitution than how to solve the integral itself. Since, the range of cosh(x) doesn't equal the domain of the function that is being integrated.
– DinW
Nov 21 at 1:22
It is a known option in examples like these where you have a fractional power of a second order polynomial
– Henry Lee
Nov 21 at 1:23
It is a known option in examples like these where you have a fractional power of a second order polynomial
– Henry Lee
Nov 21 at 1:23
add a comment |
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It's better to use $x=sec t,sqrt{x^2-1}=|tan t|$
– lab bhattacharjee
Nov 20 at 14:15