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How to solve $int fracdxsqrt9x^2+18x+2$?
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up vote
0
down vote
favorite
How would I solve
$$int fracdxsqrt9x^2+18x+2 ?$$
I feel like I'm missing a basic step? Something to do with substitution?
calculus integration
edited Jul 24 at 13:45
Ethan Bolker
35.7k54199
asked Jul 24 at 13:24
K.M.
17912
 |Â
show 3 more comments
up vote
0
down vote
favorite
How would I solve
$$int fracdxsqrt9x^2+18x+2 ?$$
I feel like I'm missing a basic step? Something to do with substitution?
calculus integration
edited Jul 24 at 13:45
Ethan Bolker
35.7k54199
asked Jul 24 at 13:24
K.M.
17912
2
Do completing the square then use a $tan(u)$ substitution
– Henry Lee
Jul 24 at 13:27
If you need more just ask
– Henry Lee
Jul 24 at 13:27
@HenryLee what is the $tan(u)$ substitution?
– K.M.
Jul 24 at 13:34
$x=tan(u)$, its a trig sub
– Sorfosh
Jul 24 at 13:38
Let $x=tan(u)$ with some rearrangement go make sure that you get $tan^2(x)+1$
– Henry Lee
Jul 24 at 13:39
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How would I solve
$$int fracdxsqrt9x^2+18x+2 ?$$
I feel like I'm missing a basic step? Something to do with substitution?
calculus integration
edited Jul 24 at 13:45
Ethan Bolker
35.7k54199
asked Jul 24 at 13:24
K.M.
17912
How would I solve
$$int fracdxsqrt9x^2+18x+2 ?$$
I feel like I'm missing a basic step? Something to do with substitution?
calculus integration
edited Jul 24 at 13:45
Ethan Bolker
35.7k54199
asked Jul 24 at 13:24
K.M.
17912
edited Jul 24 at 13:45
Ethan Bolker
35.7k54199
edited Jul 24 at 13:45
Ethan Bolker
35.7k54199
edited Jul 24 at 13:45
Ethan Bolker
35.7k54199
35.7k54199
asked Jul 24 at 13:24
K.M.
17912
asked Jul 24 at 13:24
K.M.
17912
asked Jul 24 at 13:24
K.M.
17912
17912
2
Do completing the square then use a $tan(u)$ substitution
– Henry Lee
Jul 24 at 13:27
If you need more just ask
– Henry Lee
Jul 24 at 13:27
@HenryLee what is the $tan(u)$ substitution?
– K.M.
Jul 24 at 13:34
$x=tan(u)$, its a trig sub
– Sorfosh
Jul 24 at 13:38
Let $x=tan(u)$ with some rearrangement go make sure that you get $tan^2(x)+1$
– Henry Lee
Jul 24 at 13:39
 |Â
show 3 more comments
2
Do completing the square then use a $tan(u)$ substitution
– Henry Lee
Jul 24 at 13:27
If you need more just ask
– Henry Lee
Jul 24 at 13:27
@HenryLee what is the $tan(u)$ substitution?
– K.M.
Jul 24 at 13:34
$x=tan(u)$, its a trig sub
– Sorfosh
Jul 24 at 13:38
Let $x=tan(u)$ with some rearrangement go make sure that you get $tan^2(x)+1$
– Henry Lee
Jul 24 at 13:39
2
2
Do completing the square then use a $tan(u)$ substitution
– Henry Lee
Jul 24 at 13:27
Do completing the square then use a $tan(u)$ substitution
– Henry Lee
Jul 24 at 13:27
If you need more just ask
– Henry Lee
Jul 24 at 13:27
If you need more just ask
– Henry Lee
Jul 24 at 13:27
@HenryLee what is the $tan(u)$ substitution?
– K.M.
Jul 24 at 13:34
@HenryLee what is the $tan(u)$ substitution?
– K.M.
Jul 24 at 13:34
$x=tan(u)$, its a trig sub
– Sorfosh
Jul 24 at 13:38
$x=tan(u)$, its a trig sub
– Sorfosh
Jul 24 at 13:38
Let $x=tan(u)$ with some rearrangement go make sure that you get $tan^2(x)+1$
– Henry Lee
Jul 24 at 13:39
Let $x=tan(u)$ with some rearrangement go make sure that you get $tan^2(x)+1$
– Henry Lee
Jul 24 at 13:39
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
$$I=intfrac1sqrt9x^2+18x+2,dx$$
$$I=intfrac1sqrt(3x+3)^2-7,dx$$
$$=frac1sqrt7intfrac1sqrtfrac(3x+3)^27-1,dx$$
But in this case since it is a minus sign I think you would use a hyperbolic substitution
Sorry for the photos I got bored of typing out mathjax on a phone
answered Jul 24 at 13:56
Henry Lee
49210
add a comment |Â
up vote
1
down vote
You can use the fact that$$9x^2+18x+2=(3x+3)^2-7.$$In other words, do the substitution $3x+3=y$ and $3,mathrm dx=mathrm dy$. Can you take it from here?
answered Jul 24 at 13:27
José Carlos Santos
113k1697176
This should be a comment, or a small hint.
– Lolita
Jul 24 at 19:09
add a comment |Â
up vote
1
down vote
$$int fracdxsqrt9x^2+18x+2=int fracdxsqrt9(x+1)^2-7$$
then let $3(x+1)=sqrt7cosh u$. Therfore
$$int dfrac13sqrt7fracsqrt7sinh usinh udu=dfrac13u+C=colorbluedfrac13operatornamearccoshdfrac3x+3sqrt7+C$$
Another substitution is $3(x+1)=sqrt7sec u$. Then
beginalign
int dfrac13sqrt7fracsqrt7tan usec utan udu
&= dfrac13int sec u du+C \
&= dfrac13ln|sec u+tan u|+C \
&= colorblue+C
endalign
answered Jul 24 at 13:31
Nosrati
19.3k41544
Is there any difference one would do a hyperbolic substitution rather than a normal trig sub? Extra points for obscurity?
– Sorfosh
Jul 24 at 13:41
It doesn't make difference when you use hyperbolic function or trigs.
– Nosrati
Jul 24 at 13:47
@Sorfosh IMO, hyperbolic functions are entire functions and $sinh$ is bijective, so using hyperbolic functions has a lower chance of leading to errors due to the problem of injectivity of substitution or singularities.
– Szeto
Jul 24 at 14:06
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$I=intfrac1sqrt9x^2+18x+2,dx$$
$$I=intfrac1sqrt(3x+3)^2-7,dx$$
$$=frac1sqrt7intfrac1sqrtfrac(3x+3)^27-1,dx$$
But in this case since it is a minus sign I think you would use a hyperbolic substitution
Sorry for the photos I got bored of typing out mathjax on a phone
answered Jul 24 at 13:56
Henry Lee
49210
add a comment |Â
up vote
1
down vote
accepted
$$I=intfrac1sqrt9x^2+18x+2,dx$$
$$I=intfrac1sqrt(3x+3)^2-7,dx$$
$$=frac1sqrt7intfrac1sqrtfrac(3x+3)^27-1,dx$$
But in this case since it is a minus sign I think you would use a hyperbolic substitution
Sorry for the photos I got bored of typing out mathjax on a phone
answered Jul 24 at 13:56
Henry Lee
49210
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$I=intfrac1sqrt9x^2+18x+2,dx$$
$$I=intfrac1sqrt(3x+3)^2-7,dx$$
$$=frac1sqrt7intfrac1sqrtfrac(3x+3)^27-1,dx$$
But in this case since it is a minus sign I think you would use a hyperbolic substitution
Sorry for the photos I got bored of typing out mathjax on a phone
answered Jul 24 at 13:56
Henry Lee
49210
$$I=intfrac1sqrt9x^2+18x+2,dx$$
$$I=intfrac1sqrt(3x+3)^2-7,dx$$
$$=frac1sqrt7intfrac1sqrtfrac(3x+3)^27-1,dx$$
But in this case since it is a minus sign I think you would use a hyperbolic substitution
Sorry for the photos I got bored of typing out mathjax on a phone
answered Jul 24 at 13:56
Henry Lee
49210
edited Jul 24 at 14:12
answered Jul 24 at 13:56
Henry Lee
49210
answered Jul 24 at 13:56
Henry Lee
49210
answered Jul 24 at 13:56
Henry Lee
49210
49210
add a comment |Â
add a comment |Â
up vote
1
down vote
You can use the fact that$$9x^2+18x+2=(3x+3)^2-7.$$In other words, do the substitution $3x+3=y$ and $3,mathrm dx=mathrm dy$. Can you take it from here?
answered Jul 24 at 13:27
José Carlos Santos
113k1697176
This should be a comment, or a small hint.
– Lolita
Jul 24 at 19:09
add a comment |Â
up vote
1
down vote
You can use the fact that$$9x^2+18x+2=(3x+3)^2-7.$$In other words, do the substitution $3x+3=y$ and $3,mathrm dx=mathrm dy$. Can you take it from here?
answered Jul 24 at 13:27
José Carlos Santos
113k1697176
This should be a comment, or a small hint.
– Lolita
Jul 24 at 19:09
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can use the fact that$$9x^2+18x+2=(3x+3)^2-7.$$In other words, do the substitution $3x+3=y$ and $3,mathrm dx=mathrm dy$. Can you take it from here?
answered Jul 24 at 13:27
José Carlos Santos
113k1697176
You can use the fact that$$9x^2+18x+2=(3x+3)^2-7.$$In other words, do the substitution $3x+3=y$ and $3,mathrm dx=mathrm dy$. Can you take it from here?
answered Jul 24 at 13:27
José Carlos Santos
113k1697176
answered Jul 24 at 13:27
José Carlos Santos
113k1697176
answered Jul 24 at 13:27
José Carlos Santos
113k1697176
answered Jul 24 at 13:27
José Carlos Santos
113k1697176
113k1697176
This should be a comment, or a small hint.
– Lolita
Jul 24 at 19:09
add a comment |Â
This should be a comment, or a small hint.
– Lolita
Jul 24 at 19:09
This should be a comment, or a small hint.
– Lolita
Jul 24 at 19:09
This should be a comment, or a small hint.
– Lolita
Jul 24 at 19:09
add a comment |Â
up vote
1
down vote
$$int fracdxsqrt9x^2+18x+2=int fracdxsqrt9(x+1)^2-7$$
then let $3(x+1)=sqrt7cosh u$. Therfore
$$int dfrac13sqrt7fracsqrt7sinh usinh udu=dfrac13u+C=colorbluedfrac13operatornamearccoshdfrac3x+3sqrt7+C$$
Another substitution is $3(x+1)=sqrt7sec u$. Then
beginalign
int dfrac13sqrt7fracsqrt7tan usec utan udu
&= dfrac13int sec u du+C \
&= dfrac13ln|sec u+tan u|+C \
&= colorblue+C
endalign
answered Jul 24 at 13:31
Nosrati
19.3k41544
Is there any difference one would do a hyperbolic substitution rather than a normal trig sub? Extra points for obscurity?
– Sorfosh
Jul 24 at 13:41
It doesn't make difference when you use hyperbolic function or trigs.
– Nosrati
Jul 24 at 13:47
@Sorfosh IMO, hyperbolic functions are entire functions and $sinh$ is bijective, so using hyperbolic functions has a lower chance of leading to errors due to the problem of injectivity of substitution or singularities.
– Szeto
Jul 24 at 14:06
add a comment |Â
up vote
1
down vote
$$int fracdxsqrt9x^2+18x+2=int fracdxsqrt9(x+1)^2-7$$
then let $3(x+1)=sqrt7cosh u$. Therfore
$$int dfrac13sqrt7fracsqrt7sinh usinh udu=dfrac13u+C=colorbluedfrac13operatornamearccoshdfrac3x+3sqrt7+C$$
Another substitution is $3(x+1)=sqrt7sec u$. Then
beginalign
int dfrac13sqrt7fracsqrt7tan usec utan udu
&= dfrac13int sec u du+C \
&= dfrac13ln|sec u+tan u|+C \
&= colorblue+C
endalign
answered Jul 24 at 13:31
Nosrati
19.3k41544
Is there any difference one would do a hyperbolic substitution rather than a normal trig sub? Extra points for obscurity?
– Sorfosh
Jul 24 at 13:41
It doesn't make difference when you use hyperbolic function or trigs.
– Nosrati
Jul 24 at 13:47
@Sorfosh IMO, hyperbolic functions are entire functions and $sinh$ is bijective, so using hyperbolic functions has a lower chance of leading to errors due to the problem of injectivity of substitution or singularities.
– Szeto
Jul 24 at 14:06
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$int fracdxsqrt9x^2+18x+2=int fracdxsqrt9(x+1)^2-7$$
then let $3(x+1)=sqrt7cosh u$. Therfore
$$int dfrac13sqrt7fracsqrt7sinh usinh udu=dfrac13u+C=colorbluedfrac13operatornamearccoshdfrac3x+3sqrt7+C$$
Another substitution is $3(x+1)=sqrt7sec u$. Then
beginalign
int dfrac13sqrt7fracsqrt7tan usec utan udu
&= dfrac13int sec u du+C \
&= dfrac13ln|sec u+tan u|+C \
&= colorblue+C
endalign
answered Jul 24 at 13:31
Nosrati
19.3k41544
$$int fracdxsqrt9x^2+18x+2=int fracdxsqrt9(x+1)^2-7$$
then let $3(x+1)=sqrt7cosh u$. Therfore
$$int dfrac13sqrt7fracsqrt7sinh usinh udu=dfrac13u+C=colorbluedfrac13operatornamearccoshdfrac3x+3sqrt7+C$$
Another substitution is $3(x+1)=sqrt7sec u$. Then
beginalign
int dfrac13sqrt7fracsqrt7tan usec utan udu
&= dfrac13int sec u du+C \
&= dfrac13ln|sec u+tan u|+C \
&= colorblue+C
endalign
answered Jul 24 at 13:31
Nosrati
19.3k41544
edited Jul 24 at 14:41
answered Jul 24 at 13:31
Nosrati
19.3k41544
answered Jul 24 at 13:31
Nosrati
19.3k41544
answered Jul 24 at 13:31
Nosrati
19.3k41544
19.3k41544
Is there any difference one would do a hyperbolic substitution rather than a normal trig sub? Extra points for obscurity?
– Sorfosh
Jul 24 at 13:41
It doesn't make difference when you use hyperbolic function or trigs.
– Nosrati
Jul 24 at 13:47
@Sorfosh IMO, hyperbolic functions are entire functions and $sinh$ is bijective, so using hyperbolic functions has a lower chance of leading to errors due to the problem of injectivity of substitution or singularities.
– Szeto
Jul 24 at 14:06
add a comment |Â
Is there any difference one would do a hyperbolic substitution rather than a normal trig sub? Extra points for obscurity?
– Sorfosh
Jul 24 at 13:41
It doesn't make difference when you use hyperbolic function or trigs.
– Nosrati
Jul 24 at 13:47
@Sorfosh IMO, hyperbolic functions are entire functions and $sinh$ is bijective, so using hyperbolic functions has a lower chance of leading to errors due to the problem of injectivity of substitution or singularities.
– Szeto
Jul 24 at 14:06
Is there any difference one would do a hyperbolic substitution rather than a normal trig sub? Extra points for obscurity?
– Sorfosh
Jul 24 at 13:41
Is there any difference one would do a hyperbolic substitution rather than a normal trig sub? Extra points for obscurity?
– Sorfosh
Jul 24 at 13:41
It doesn't make difference when you use hyperbolic function or trigs.
– Nosrati
Jul 24 at 13:47
It doesn't make difference when you use hyperbolic function or trigs.
– Nosrati
Jul 24 at 13:47
@Sorfosh IMO, hyperbolic functions are entire functions and $sinh$ is bijective, so using hyperbolic functions has a lower chance of leading to errors due to the problem of injectivity of substitution or singularities.
– Szeto
Jul 24 at 14:06
@Sorfosh IMO, hyperbolic functions are entire functions and $sinh$ is bijective, so using hyperbolic functions has a lower chance of leading to errors due to the problem of injectivity of substitution or singularities.
– Szeto
Jul 24 at 14:06
add a comment |Â
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2
Do completing the square then use a $tan(u)$ substitution
– Henry Lee
Jul 24 at 13:27
If you need more just ask
– Henry Lee
Jul 24 at 13:27
@HenryLee what is the $tan(u)$ substitution?
– K.M.
Jul 24 at 13:34
$x=tan(u)$, its a trig sub
– Sorfosh
Jul 24 at 13:38
Let $x=tan(u)$ with some rearrangement go make sure that you get $tan^2(x)+1$
– Henry Lee
Jul 24 at 13:39