Mixing
Compute $intfracdxsqrtx^2+16$ [duplicate]
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
This question already has an answer here:
Finding $int frac dxsqrt x^2 + 16$
3 answers
Here is what I have done so far:
$$I=int fracdxsqrtx^2+16 =frac14intsqrtfrac1left(fracx4 right)^2+1,dx$$
$fracx4=tan(u), dx=4sec^2(u)$
$$therefore I=intsqrtfrac1tan^2(u)+1sec^2(u),du=intsec^3(u),du$$
now by integration by parts:
$$I=tan(u)sec(u)-inttan^2(u)sec(u),du$$
I am sure $intsec^3(u)du$ is a standard integral but I am now sure if IBP is the best way to go. Is there another obvious way of doing it?
calculus integration
edited Jul 17 at 18:27
Michael Hardy
204k23186462
asked Jul 17 at 17:54
Henry Lee
49210
marked as duplicate by Jyrki Lahtonen, Parcly Taxel, Nosrati, Lord Shark the Unknown, Math Lover Jul 17 at 18:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
1
down vote
favorite
This question already has an answer here:
Finding $int frac dxsqrt x^2 + 16$
3 answers
Here is what I have done so far:
$$I=int fracdxsqrtx^2+16 =frac14intsqrtfrac1left(fracx4 right)^2+1,dx$$
$fracx4=tan(u), dx=4sec^2(u)$
$$therefore I=intsqrtfrac1tan^2(u)+1sec^2(u),du=intsec^3(u),du$$
now by integration by parts:
$$I=tan(u)sec(u)-inttan^2(u)sec(u),du$$
I am sure $intsec^3(u)du$ is a standard integral but I am now sure if IBP is the best way to go. Is there another obvious way of doing it?
calculus integration
edited Jul 17 at 18:27
Michael Hardy
204k23186462
asked Jul 17 at 17:54
Henry Lee
49210
marked as duplicate by Jyrki Lahtonen, Parcly Taxel, Nosrati, Lord Shark the Unknown, Math Lover Jul 17 at 18:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
With Approach0 we can find many, many integrals of the form $intdfracdxsqrtx^2+a^2$. At least two with $a=4$ explicitly, $a=3$ and $a=2$ also appear there...
– Jyrki Lahtonen
Jul 17 at 18:00
You've miscounted powers; you want to integrate the secant, not the cubed secant.
– J.G.
Jul 17 at 18:26
1 and 2...
– Jyrki Lahtonen
Jul 17 at 18:27
en.wikipedia.org/wiki/Integral_of_secant_cubed $qquad$
– Michael Hardy
Jul 17 at 18:28
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
Finding $int frac dxsqrt x^2 + 16$
3 answers
Here is what I have done so far:
$$I=int fracdxsqrtx^2+16 =frac14intsqrtfrac1left(fracx4 right)^2+1,dx$$
$fracx4=tan(u), dx=4sec^2(u)$
$$therefore I=intsqrtfrac1tan^2(u)+1sec^2(u),du=intsec^3(u),du$$
now by integration by parts:
$$I=tan(u)sec(u)-inttan^2(u)sec(u),du$$
I am sure $intsec^3(u)du$ is a standard integral but I am now sure if IBP is the best way to go. Is there another obvious way of doing it?
calculus integration
edited Jul 17 at 18:27
Michael Hardy
204k23186462
asked Jul 17 at 17:54
Henry Lee
49210
This question already has an answer here:
Finding $int frac dxsqrt x^2 + 16$
3 answers
Here is what I have done so far:
$$I=int fracdxsqrtx^2+16 =frac14intsqrtfrac1left(fracx4 right)^2+1,dx$$
$fracx4=tan(u), dx=4sec^2(u)$
$$therefore I=intsqrtfrac1tan^2(u)+1sec^2(u),du=intsec^3(u),du$$
now by integration by parts:
$$I=tan(u)sec(u)-inttan^2(u)sec(u),du$$
I am sure $intsec^3(u)du$ is a standard integral but I am now sure if IBP is the best way to go. Is there another obvious way of doing it?
This question already has an answer here:
Finding $int frac dxsqrt x^2 + 16$
3 answers
calculus integration
edited Jul 17 at 18:27
Michael Hardy
204k23186462
asked Jul 17 at 17:54
Henry Lee
49210
edited Jul 17 at 18:27
Michael Hardy
204k23186462
edited Jul 17 at 18:27
Michael Hardy
204k23186462
edited Jul 17 at 18:27
Michael Hardy
204k23186462
204k23186462
asked Jul 17 at 17:54
Henry Lee
49210
asked Jul 17 at 17:54
Henry Lee
49210
asked Jul 17 at 17:54
Henry Lee
49210
49210
marked as duplicate by Jyrki Lahtonen, Parcly Taxel, Nosrati, Lord Shark the Unknown, Math Lover Jul 17 at 18:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jyrki Lahtonen, Parcly Taxel, Nosrati, Lord Shark the Unknown, Math Lover Jul 17 at 18:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
With Approach0 we can find many, many integrals of the form $intdfracdxsqrtx^2+a^2$. At least two with $a=4$ explicitly, $a=3$ and $a=2$ also appear there...
– Jyrki Lahtonen
Jul 17 at 18:00
You've miscounted powers; you want to integrate the secant, not the cubed secant.
– J.G.
Jul 17 at 18:26
1 and 2...
– Jyrki Lahtonen
Jul 17 at 18:27
en.wikipedia.org/wiki/Integral_of_secant_cubed $qquad$
– Michael Hardy
Jul 17 at 18:28
add a comment |Â
With Approach0 we can find many, many integrals of the form $intdfracdxsqrtx^2+a^2$. At least two with $a=4$ explicitly, $a=3$ and $a=2$ also appear there...
– Jyrki Lahtonen
Jul 17 at 18:00
You've miscounted powers; you want to integrate the secant, not the cubed secant.
– J.G.
Jul 17 at 18:26
1 and 2...
– Jyrki Lahtonen
Jul 17 at 18:27
en.wikipedia.org/wiki/Integral_of_secant_cubed $qquad$
– Michael Hardy
Jul 17 at 18:28
With Approach0 we can find many, many integrals of the form $intdfracdxsqrtx^2+a^2$. At least two with $a=4$ explicitly, $a=3$ and $a=2$ also appear there...
– Jyrki Lahtonen
Jul 17 at 18:00
With Approach0 we can find many, many integrals of the form $intdfracdxsqrtx^2+a^2$. At least two with $a=4$ explicitly, $a=3$ and $a=2$ also appear there...
– Jyrki Lahtonen
Jul 17 at 18:00
You've miscounted powers; you want to integrate the secant, not the cubed secant.
– J.G.
Jul 17 at 18:26
You've miscounted powers; you want to integrate the secant, not the cubed secant.
– J.G.
Jul 17 at 18:26
1 and 2...
– Jyrki Lahtonen
Jul 17 at 18:27
1 and 2...
– Jyrki Lahtonen
Jul 17 at 18:27
en.wikipedia.org/wiki/Integral_of_secant_cubed $qquad$
– Michael Hardy
Jul 17 at 18:28
en.wikipedia.org/wiki/Integral_of_secant_cubed $qquad$
– Michael Hardy
Jul 17 at 18:28
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
Hint: Substitute
$$sqrtx^2+16=xt+4$$ this is the so-called Euler substitution.
See here https://en.wikipedia.org/wiki/Euler_substitution
answered Jul 17 at 18:00
Dr. Sonnhard Graubner
66.8k32659
add a comment |Â
up vote
0
down vote
- Either you've learnt about the inverse hyperbolic functions, and you know that
$$intfracmathrm d xsqrtx^2+1=argsinh x=lnbigl(x+sqrtx^2+1bigr),$$
from which one deduces readily with the substitution $x=at$ ($a>0$) that
$$intfracmathrm d xsqrtx^2+a^2=argsinh frac xa=lnbigl(x+sqrtx^2+a^2bigr).$$ - Or you've only learnt about the hyperbolic functions. In this case, set $x=asinh t$, $mathrm d x=acosh t,mathrm d t$, so
$$intfracmathrm d xsqrtx^2+a^2=intfracacosh t,mathrm d tasqrtsinh^2 t+1=intfraccosh t,mathrm d tcosh t=t.$$
There remains to solve for $x$ the equation
$$t=sinh x=fracmathrm e^2x-12mathrm e^x.$$
answered Jul 17 at 18:22
Bernard
110k635103
add a comment |Â
up vote
0
down vote
Use the trigonometric substitution as follows:
Let $$x = 4times tan(theta)$$
By doing the necessary operations you will end up with $$int sec (theta), dtheta $$
Which i guess you can easily solve
edited Jul 17 at 18:35
Bernard
110k635103
answered Jul 17 at 18:25
JFC
103
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint: Substitute
$$sqrtx^2+16=xt+4$$ this is the so-called Euler substitution.
See here https://en.wikipedia.org/wiki/Euler_substitution
answered Jul 17 at 18:00
Dr. Sonnhard Graubner
66.8k32659
add a comment |Â
up vote
0
down vote
Hint: Substitute
$$sqrtx^2+16=xt+4$$ this is the so-called Euler substitution.
See here https://en.wikipedia.org/wiki/Euler_substitution
answered Jul 17 at 18:00
Dr. Sonnhard Graubner
66.8k32659
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: Substitute
$$sqrtx^2+16=xt+4$$ this is the so-called Euler substitution.
See here https://en.wikipedia.org/wiki/Euler_substitution
answered Jul 17 at 18:00
Dr. Sonnhard Graubner
66.8k32659
Hint: Substitute
$$sqrtx^2+16=xt+4$$ this is the so-called Euler substitution.
See here https://en.wikipedia.org/wiki/Euler_substitution
answered Jul 17 at 18:00
Dr. Sonnhard Graubner
66.8k32659
answered Jul 17 at 18:00
Dr. Sonnhard Graubner
66.8k32659
answered Jul 17 at 18:00
Dr. Sonnhard Graubner
66.8k32659
answered Jul 17 at 18:00
Dr. Sonnhard Graubner
66.8k32659
66.8k32659
add a comment |Â
add a comment |Â
up vote
0
down vote
- Either you've learnt about the inverse hyperbolic functions, and you know that
$$intfracmathrm d xsqrtx^2+1=argsinh x=lnbigl(x+sqrtx^2+1bigr),$$
from which one deduces readily with the substitution $x=at$ ($a>0$) that
$$intfracmathrm d xsqrtx^2+a^2=argsinh frac xa=lnbigl(x+sqrtx^2+a^2bigr).$$ - Or you've only learnt about the hyperbolic functions. In this case, set $x=asinh t$, $mathrm d x=acosh t,mathrm d t$, so
$$intfracmathrm d xsqrtx^2+a^2=intfracacosh t,mathrm d tasqrtsinh^2 t+1=intfraccosh t,mathrm d tcosh t=t.$$
There remains to solve for $x$ the equation
$$t=sinh x=fracmathrm e^2x-12mathrm e^x.$$
answered Jul 17 at 18:22
Bernard
110k635103
add a comment |Â
up vote
0
down vote
- Either you've learnt about the inverse hyperbolic functions, and you know that
$$intfracmathrm d xsqrtx^2+1=argsinh x=lnbigl(x+sqrtx^2+1bigr),$$
from which one deduces readily with the substitution $x=at$ ($a>0$) that
$$intfracmathrm d xsqrtx^2+a^2=argsinh frac xa=lnbigl(x+sqrtx^2+a^2bigr).$$ - Or you've only learnt about the hyperbolic functions. In this case, set $x=asinh t$, $mathrm d x=acosh t,mathrm d t$, so
$$intfracmathrm d xsqrtx^2+a^2=intfracacosh t,mathrm d tasqrtsinh^2 t+1=intfraccosh t,mathrm d tcosh t=t.$$
There remains to solve for $x$ the equation
$$t=sinh x=fracmathrm e^2x-12mathrm e^x.$$
answered Jul 17 at 18:22
Bernard
110k635103
add a comment |Â
up vote
0
down vote
up vote
0
down vote
- Either you've learnt about the inverse hyperbolic functions, and you know that
$$intfracmathrm d xsqrtx^2+1=argsinh x=lnbigl(x+sqrtx^2+1bigr),$$
from which one deduces readily with the substitution $x=at$ ($a>0$) that
$$intfracmathrm d xsqrtx^2+a^2=argsinh frac xa=lnbigl(x+sqrtx^2+a^2bigr).$$ - Or you've only learnt about the hyperbolic functions. In this case, set $x=asinh t$, $mathrm d x=acosh t,mathrm d t$, so
$$intfracmathrm d xsqrtx^2+a^2=intfracacosh t,mathrm d tasqrtsinh^2 t+1=intfraccosh t,mathrm d tcosh t=t.$$
There remains to solve for $x$ the equation
$$t=sinh x=fracmathrm e^2x-12mathrm e^x.$$
answered Jul 17 at 18:22
Bernard
110k635103
- Either you've learnt about the inverse hyperbolic functions, and you know that
$$intfracmathrm d xsqrtx^2+1=argsinh x=lnbigl(x+sqrtx^2+1bigr),$$
from which one deduces readily with the substitution $x=at$ ($a>0$) that
$$intfracmathrm d xsqrtx^2+a^2=argsinh frac xa=lnbigl(x+sqrtx^2+a^2bigr).$$ - Or you've only learnt about the hyperbolic functions. In this case, set $x=asinh t$, $mathrm d x=acosh t,mathrm d t$, so
$$intfracmathrm d xsqrtx^2+a^2=intfracacosh t,mathrm d tasqrtsinh^2 t+1=intfraccosh t,mathrm d tcosh t=t.$$
There remains to solve for $x$ the equation
$$t=sinh x=fracmathrm e^2x-12mathrm e^x.$$
answered Jul 17 at 18:22
Bernard
110k635103
answered Jul 17 at 18:22
Bernard
110k635103
answered Jul 17 at 18:22
Bernard
110k635103
answered Jul 17 at 18:22
Bernard
110k635103
110k635103
add a comment |Â
add a comment |Â
up vote
0
down vote
Use the trigonometric substitution as follows:
Let $$x = 4times tan(theta)$$
By doing the necessary operations you will end up with $$int sec (theta), dtheta $$
Which i guess you can easily solve
edited Jul 17 at 18:35
Bernard
110k635103
answered Jul 17 at 18:25
JFC
103
add a comment |Â
up vote
0
down vote
Use the trigonometric substitution as follows:
Let $$x = 4times tan(theta)$$
By doing the necessary operations you will end up with $$int sec (theta), dtheta $$
Which i guess you can easily solve
edited Jul 17 at 18:35
Bernard
110k635103
answered Jul 17 at 18:25
JFC
103
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Use the trigonometric substitution as follows:
Let $$x = 4times tan(theta)$$
By doing the necessary operations you will end up with $$int sec (theta), dtheta $$
Which i guess you can easily solve
edited Jul 17 at 18:35
Bernard
110k635103
answered Jul 17 at 18:25
JFC
103
Use the trigonometric substitution as follows:
Let $$x = 4times tan(theta)$$
By doing the necessary operations you will end up with $$int sec (theta), dtheta $$
Which i guess you can easily solve
edited Jul 17 at 18:35
Bernard
110k635103
answered Jul 17 at 18:25
JFC
103
edited Jul 17 at 18:35
Bernard
110k635103
edited Jul 17 at 18:35
Bernard
110k635103
edited Jul 17 at 18:35
Bernard
110k635103
110k635103
answered Jul 17 at 18:25
JFC
103
answered Jul 17 at 18:25
JFC
103
answered Jul 17 at 18:25
JFC
103
103
add a comment |Â
add a comment |Â
With Approach0 we can find many, many integrals of the form $intdfracdxsqrtx^2+a^2$. At least two with $a=4$ explicitly, $a=3$ and $a=2$ also appear there...
– Jyrki Lahtonen
Jul 17 at 18:00
You've miscounted powers; you want to integrate the secant, not the cubed secant.
– J.G.
Jul 17 at 18:26
1 and 2...
– Jyrki Lahtonen
Jul 17 at 18:27
en.wikipedia.org/wiki/Integral_of_secant_cubed $qquad$
– Michael Hardy
Jul 17 at 18:28