Trig Subsitution When There's No Square Root
$begingroup$
I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$Ar int_a^infty frac{dx}{(r^2+x^2)^{(3/2)}}$
Anyway, so far, I have that:
$x = rtan theta$
$dx = rsec^2 theta$
$sqrt {(r^2+x^2)} = rsectheta$
Please click here to see the triangle I based the above values on.
.
Given that $(r^2+x^2)^{(3/2)}$ can be rewritten as $ (sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $lim limits_{b to infty}$ in front of every line please.
= $Ar int_a^b frac{rsec^2theta}{(rsectheta)^3}dtheta$
= $Ar int_a^b frac{rsec^2theta}{r^3sec^6theta}dtheta$
= $frac{A}{r} int_a^b frac{1}{sec^4theta}dtheta$
= $frac{A}{r} int_a^b cos^4theta dtheta$
= $frac{A}{r} int_a^b (cos^2theta)^2 dtheta$
= $frac{A}{r} int_a^b [ frac{1}{2}1+cos(2theta)) ]^2dtheta$
= $frac{A}{4r} int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$
= $frac{A}{4r} int_a^b 1 + 2cos(2theta) dtheta quad+quad frac{A}{4r} int_a^b cos^2(2theta) dtheta$
And from there it gets really messed up and I end up with a weird semi-final answer of $frac{A}{4r}[2theta+sin(2theta)] + frac{A}{32r} [4theta+sin(4theta)]$ which is wrong after I make substitutions.
.
I've done this a couple different ways (I keep messing up), but this is the most correct that I've been able to come up with and it's still wrong.
It would be soooo awesome if someone could tell me if I'm even on the right track. Like I said, if it was a simple square root instead of $quad ^{3/2}quad$, life would be awesome.
I already know that the final answer is $frac{A}{r}(1-frac{a}{sqrt{r^2+a^2}})$, but I really want to understand this. An explanation would be most welcome!
Thank you in advance for all your help!
calculus integration improper-integrals trigonometric-integrals
$endgroup$
add a comment |
$begingroup$
I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$Ar int_a^infty frac{dx}{(r^2+x^2)^{(3/2)}}$
Anyway, so far, I have that:
$x = rtan theta$
$dx = rsec^2 theta$
$sqrt {(r^2+x^2)} = rsectheta$
Please click here to see the triangle I based the above values on.
.
Given that $(r^2+x^2)^{(3/2)}$ can be rewritten as $ (sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $lim limits_{b to infty}$ in front of every line please.
= $Ar int_a^b frac{rsec^2theta}{(rsectheta)^3}dtheta$
= $Ar int_a^b frac{rsec^2theta}{r^3sec^6theta}dtheta$
= $frac{A}{r} int_a^b frac{1}{sec^4theta}dtheta$
= $frac{A}{r} int_a^b cos^4theta dtheta$
= $frac{A}{r} int_a^b (cos^2theta)^2 dtheta$
= $frac{A}{r} int_a^b [ frac{1}{2}1+cos(2theta)) ]^2dtheta$
= $frac{A}{4r} int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$
= $frac{A}{4r} int_a^b 1 + 2cos(2theta) dtheta quad+quad frac{A}{4r} int_a^b cos^2(2theta) dtheta$
And from there it gets really messed up and I end up with a weird semi-final answer of $frac{A}{4r}[2theta+sin(2theta)] + frac{A}{32r} [4theta+sin(4theta)]$ which is wrong after I make substitutions.
.
I've done this a couple different ways (I keep messing up), but this is the most correct that I've been able to come up with and it's still wrong.
It would be soooo awesome if someone could tell me if I'm even on the right track. Like I said, if it was a simple square root instead of $quad ^{3/2}quad$, life would be awesome.
I already know that the final answer is $frac{A}{r}(1-frac{a}{sqrt{r^2+a^2}})$, but I really want to understand this. An explanation would be most welcome!
Thank you in advance for all your help!
calculus integration improper-integrals trigonometric-integrals
$endgroup$
2
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
4 hours ago
add a comment |
$begingroup$
I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$Ar int_a^infty frac{dx}{(r^2+x^2)^{(3/2)}}$
Anyway, so far, I have that:
$x = rtan theta$
$dx = rsec^2 theta$
$sqrt {(r^2+x^2)} = rsectheta$
Please click here to see the triangle I based the above values on.
.
Given that $(r^2+x^2)^{(3/2)}$ can be rewritten as $ (sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $lim limits_{b to infty}$ in front of every line please.
= $Ar int_a^b frac{rsec^2theta}{(rsectheta)^3}dtheta$
= $Ar int_a^b frac{rsec^2theta}{r^3sec^6theta}dtheta$
= $frac{A}{r} int_a^b frac{1}{sec^4theta}dtheta$
= $frac{A}{r} int_a^b cos^4theta dtheta$
= $frac{A}{r} int_a^b (cos^2theta)^2 dtheta$
= $frac{A}{r} int_a^b [ frac{1}{2}1+cos(2theta)) ]^2dtheta$
= $frac{A}{4r} int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$
= $frac{A}{4r} int_a^b 1 + 2cos(2theta) dtheta quad+quad frac{A}{4r} int_a^b cos^2(2theta) dtheta$
And from there it gets really messed up and I end up with a weird semi-final answer of $frac{A}{4r}[2theta+sin(2theta)] + frac{A}{32r} [4theta+sin(4theta)]$ which is wrong after I make substitutions.
.
I've done this a couple different ways (I keep messing up), but this is the most correct that I've been able to come up with and it's still wrong.
It would be soooo awesome if someone could tell me if I'm even on the right track. Like I said, if it was a simple square root instead of $quad ^{3/2}quad$, life would be awesome.
I already know that the final answer is $frac{A}{r}(1-frac{a}{sqrt{r^2+a^2}})$, but I really want to understand this. An explanation would be most welcome!
Thank you in advance for all your help!
calculus integration improper-integrals trigonometric-integrals
$endgroup$
I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$Ar int_a^infty frac{dx}{(r^2+x^2)^{(3/2)}}$
Anyway, so far, I have that:
$x = rtan theta$
$dx = rsec^2 theta$
$sqrt {(r^2+x^2)} = rsectheta$
Please click here to see the triangle I based the above values on.
.
Given that $(r^2+x^2)^{(3/2)}$ can be rewritten as $ (sqrt{r^2+x^2})^3$, I begin to solve.
Please pretend I have $lim limits_{b to infty}$ in front of every line please.
= $Ar int_a^b frac{rsec^2theta}{(rsectheta)^3}dtheta$
= $Ar int_a^b frac{rsec^2theta}{r^3sec^6theta}dtheta$
= $frac{A}{r} int_a^b frac{1}{sec^4theta}dtheta$
= $frac{A}{r} int_a^b cos^4theta dtheta$
= $frac{A}{r} int_a^b (cos^2theta)^2 dtheta$
= $frac{A}{r} int_a^b [ frac{1}{2}1+cos(2theta)) ]^2dtheta$
= $frac{A}{4r} int_a^b 1 + 2cos(2theta) + cos^2(2theta) dtheta$
= $frac{A}{4r} int_a^b 1 + 2cos(2theta) dtheta quad+quad frac{A}{4r} int_a^b cos^2(2theta) dtheta$
And from there it gets really messed up and I end up with a weird semi-final answer of $frac{A}{4r}[2theta+sin(2theta)] + frac{A}{32r} [4theta+sin(4theta)]$ which is wrong after I make substitutions.
.
I've done this a couple different ways (I keep messing up), but this is the most correct that I've been able to come up with and it's still wrong.
It would be soooo awesome if someone could tell me if I'm even on the right track. Like I said, if it was a simple square root instead of $quad ^{3/2}quad$, life would be awesome.
I already know that the final answer is $frac{A}{r}(1-frac{a}{sqrt{r^2+a^2}})$, but I really want to understand this. An explanation would be most welcome!
Thank you in advance for all your help!
calculus integration improper-integrals trigonometric-integrals
calculus integration improper-integrals trigonometric-integrals
asked 4 hours ago
CodingMeeCodingMee
204
204
2
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
4 hours ago
add a comment |
2
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
4 hours ago
2
2
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
4 hours ago
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
frac{A}{r}int_{a/r}^{infty}frac{1}{(1+u^2)^{3/2}},du
$$
Now let's concentrate on the antiderivative
$$
intfrac{1}{(1+u^2)^{3/2}},du=
intfrac{1+u^2-u^2}{(1+u^2)^{3/2}},du=
intfrac{1}{(1+u^2)^{1/2}},du-intfrac{u^2}{(1+u^2)^{3/2}},du
$$
Do the second term by parts
$$
int ufrac{u}{(1+u^2)^{3/2}},du=
-frac{u}{(1+u^2)^{1/2}}+intfrac{1}{(1+u^2)^{1/2}},du
$$
See what happens?
$$
intfrac{1}{(1+u^2)^{3/2}},du=frac{u}{(1+u^2)^{1/2}}+c
$$
which we can verify by direct differentiation.
Now
$$
left[frac{u}{(1+u^2)^{1/2}}right]_{a/r}^{infty}=1-frac{a/r}{(1+(a/r)^2)^{1/2}}
=1-frac{a}{(r^2+a^2)^{1/2}}
$$
and your integral is indeed
$$
frac{A}{r}left(1-frac{a}{sqrt{r^2+a^2}}right)
$$
$endgroup$
add a comment |
$begingroup$
Firstly you made an error in the first line of working
$$(rsec{(theta)})^3=r^3sec^3{(theta)}$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan{(frac{x}{r})}$ then the limits should change as $x=a implies theta=arctan{(frac{a}{r})}$ also $x=infty implies theta=frac{pi}2$.
$endgroup$
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 are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
frac{A}{r}int_{a/r}^{infty}frac{1}{(1+u^2)^{3/2}},du
$$
Now let's concentrate on the antiderivative
$$
intfrac{1}{(1+u^2)^{3/2}},du=
intfrac{1+u^2-u^2}{(1+u^2)^{3/2}},du=
intfrac{1}{(1+u^2)^{1/2}},du-intfrac{u^2}{(1+u^2)^{3/2}},du
$$
Do the second term by parts
$$
int ufrac{u}{(1+u^2)^{3/2}},du=
-frac{u}{(1+u^2)^{1/2}}+intfrac{1}{(1+u^2)^{1/2}},du
$$
See what happens?
$$
intfrac{1}{(1+u^2)^{3/2}},du=frac{u}{(1+u^2)^{1/2}}+c
$$
which we can verify by direct differentiation.
Now
$$
left[frac{u}{(1+u^2)^{1/2}}right]_{a/r}^{infty}=1-frac{a/r}{(1+(a/r)^2)^{1/2}}
=1-frac{a}{(r^2+a^2)^{1/2}}
$$
and your integral is indeed
$$
frac{A}{r}left(1-frac{a}{sqrt{r^2+a^2}}right)
$$
$endgroup$
add a comment |
$begingroup$
You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
frac{A}{r}int_{a/r}^{infty}frac{1}{(1+u^2)^{3/2}},du
$$
Now let's concentrate on the antiderivative
$$
intfrac{1}{(1+u^2)^{3/2}},du=
intfrac{1+u^2-u^2}{(1+u^2)^{3/2}},du=
intfrac{1}{(1+u^2)^{1/2}},du-intfrac{u^2}{(1+u^2)^{3/2}},du
$$
Do the second term by parts
$$
int ufrac{u}{(1+u^2)^{3/2}},du=
-frac{u}{(1+u^2)^{1/2}}+intfrac{1}{(1+u^2)^{1/2}},du
$$
See what happens?
$$
intfrac{1}{(1+u^2)^{3/2}},du=frac{u}{(1+u^2)^{1/2}}+c
$$
which we can verify by direct differentiation.
Now
$$
left[frac{u}{(1+u^2)^{1/2}}right]_{a/r}^{infty}=1-frac{a/r}{(1+(a/r)^2)^{1/2}}
=1-frac{a}{(r^2+a^2)^{1/2}}
$$
and your integral is indeed
$$
frac{A}{r}left(1-frac{a}{sqrt{r^2+a^2}}right)
$$
$endgroup$
add a comment |
$begingroup$
You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
frac{A}{r}int_{a/r}^{infty}frac{1}{(1+u^2)^{3/2}},du
$$
Now let's concentrate on the antiderivative
$$
intfrac{1}{(1+u^2)^{3/2}},du=
intfrac{1+u^2-u^2}{(1+u^2)^{3/2}},du=
intfrac{1}{(1+u^2)^{1/2}},du-intfrac{u^2}{(1+u^2)^{3/2}},du
$$
Do the second term by parts
$$
int ufrac{u}{(1+u^2)^{3/2}},du=
-frac{u}{(1+u^2)^{1/2}}+intfrac{1}{(1+u^2)^{1/2}},du
$$
See what happens?
$$
intfrac{1}{(1+u^2)^{3/2}},du=frac{u}{(1+u^2)^{1/2}}+c
$$
which we can verify by direct differentiation.
Now
$$
left[frac{u}{(1+u^2)^{1/2}}right]_{a/r}^{infty}=1-frac{a/r}{(1+(a/r)^2)^{1/2}}
=1-frac{a}{(r^2+a^2)^{1/2}}
$$
and your integral is indeed
$$
frac{A}{r}left(1-frac{a}{sqrt{r^2+a^2}}right)
$$
$endgroup$
You are doing $(rsectheta)^3=r^6sec^6theta$. Oops! ;-)
There's a slicker way to do it.
Get rid of the $r$ with $x=ru$ to begin with, so your integral becomes
$$
frac{A}{r}int_{a/r}^{infty}frac{1}{(1+u^2)^{3/2}},du
$$
Now let's concentrate on the antiderivative
$$
intfrac{1}{(1+u^2)^{3/2}},du=
intfrac{1+u^2-u^2}{(1+u^2)^{3/2}},du=
intfrac{1}{(1+u^2)^{1/2}},du-intfrac{u^2}{(1+u^2)^{3/2}},du
$$
Do the second term by parts
$$
int ufrac{u}{(1+u^2)^{3/2}},du=
-frac{u}{(1+u^2)^{1/2}}+intfrac{1}{(1+u^2)^{1/2}},du
$$
See what happens?
$$
intfrac{1}{(1+u^2)^{3/2}},du=frac{u}{(1+u^2)^{1/2}}+c
$$
which we can verify by direct differentiation.
Now
$$
left[frac{u}{(1+u^2)^{1/2}}right]_{a/r}^{infty}=1-frac{a/r}{(1+(a/r)^2)^{1/2}}
=1-frac{a}{(r^2+a^2)^{1/2}}
$$
and your integral is indeed
$$
frac{A}{r}left(1-frac{a}{sqrt{r^2+a^2}}right)
$$
answered 3 hours ago
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
$begingroup$
Firstly you made an error in the first line of working
$$(rsec{(theta)})^3=r^3sec^3{(theta)}$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan{(frac{x}{r})}$ then the limits should change as $x=a implies theta=arctan{(frac{a}{r})}$ also $x=infty implies theta=frac{pi}2$.
$endgroup$
add a comment |
$begingroup$
Firstly you made an error in the first line of working
$$(rsec{(theta)})^3=r^3sec^3{(theta)}$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan{(frac{x}{r})}$ then the limits should change as $x=a implies theta=arctan{(frac{a}{r})}$ also $x=infty implies theta=frac{pi}2$.
$endgroup$
add a comment |
$begingroup$
Firstly you made an error in the first line of working
$$(rsec{(theta)})^3=r^3sec^3{(theta)}$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan{(frac{x}{r})}$ then the limits should change as $x=a implies theta=arctan{(frac{a}{r})}$ also $x=infty implies theta=frac{pi}2$.
$endgroup$
Firstly you made an error in the first line of working
$$(rsec{(theta)})^3=r^3sec^3{(theta)}$$
Secondly, you need to change the range of integration after performing a substitution. If $theta=arctan{(frac{x}{r})}$ then the limits should change as $x=a implies theta=arctan{(frac{a}{r})}$ also $x=infty implies theta=frac{pi}2$.
answered 4 hours ago
Peter ForemanPeter Foreman
3,4521216
3,4521216
add a comment |
add a comment |
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2
$begingroup$
The denominator in the 2nd line is $r^3sec^3theta$ instead of $r^3sec^6theta$.
$endgroup$
– Kay K.
4 hours ago