Help to solve: $int 1/(xsqrt{25-x^2}) dx$
up vote
3
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I'm a brand new student. Need some help to integrate this.
Perform this integration: $$int frac{1}{xsqrt{25-x^2}} dx$$
I'm able to obtain in theta terms like this:
$$frac 15 ln|cscθ-cotθ|+C$$
But I have problems to convert in terms of "$x$"
Thanks a lot.
This is what is suggested by this image:
Using the substitution $x=5sin{theta}$ and $dx=5cos{theta}$, we obtain:
$$int frac{1}{5sin{theta}(5cos{theta})}cdot 5cos{theta}~dtheta$$
Simplifying, we obtain:
$$frac{1}{5}int frac{dtheta}{sin{theta}}$$
Using trigonometric identities:
$$frac{1}{5}int csc{theta}~dtheta$$
Integrating:
$$frac{1}{5}ln|csc{theta}-cot{theta}|+C$$
integration
edited Mar 17 '17 at 16:36
projectilemotion
11.4k62041
asked Mar 17 '17 at 15:21
El_Master
162
|
show 3 more comments
up vote
3
down vote
favorite
I'm a brand new student. Need some help to integrate this.
Perform this integration: $$int frac{1}{xsqrt{25-x^2}} dx$$
I'm able to obtain in theta terms like this:
$$frac 15 ln|cscθ-cotθ|+C$$
But I have problems to convert in terms of "$x$"
Thanks a lot.
This is what is suggested by this image:
Using the substitution $x=5sin{theta}$ and $dx=5cos{theta}$, we obtain:
$$int frac{1}{5sin{theta}(5cos{theta})}cdot 5cos{theta}~dtheta$$
Simplifying, we obtain:
$$frac{1}{5}int frac{dtheta}{sin{theta}}$$
Using trigonometric identities:
$$frac{1}{5}int csc{theta}~dtheta$$
Integrating:
$$frac{1}{5}ln|csc{theta}-cot{theta}|+C$$
integration
edited Mar 17 '17 at 16:36
projectilemotion
11.4k62041
asked Mar 17 '17 at 15:21
El_Master
162
Please use: math.meta.stackexchange.com/questions/5020/…
– Arnaldo
Mar 17 '17 at 15:25
It helps if you explain the steps you took to get at the solution in $theta$, e.g. what was the substitution you used?
– StackTD
Mar 17 '17 at 15:27
Yes Exactly. See the problem image..I believe I have the half of the solution....thanks..
– El_Master
Mar 17 '17 at 15:27
@El_Master Did you use the substitution $x=5sin(theta)$ and $dx=5cos(theta)~dtheta$ ?
– projectilemotion
Mar 17 '17 at 15:33
you can use relationships between trigonometric functions to change it back, so if $x = sen theta$ then $cos theta = sqrt{1 - sen^2 theta} = sqrt{1 - x^2}$ and other identities that you can use
– Cato
Mar 17 '17 at 15:33
|
show 3 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm a brand new student. Need some help to integrate this.
Perform this integration: $$int frac{1}{xsqrt{25-x^2}} dx$$
I'm able to obtain in theta terms like this:
$$frac 15 ln|cscθ-cotθ|+C$$
But I have problems to convert in terms of "$x$"
Thanks a lot.
This is what is suggested by this image:
Using the substitution $x=5sin{theta}$ and $dx=5cos{theta}$, we obtain:
$$int frac{1}{5sin{theta}(5cos{theta})}cdot 5cos{theta}~dtheta$$
Simplifying, we obtain:
$$frac{1}{5}int frac{dtheta}{sin{theta}}$$
Using trigonometric identities:
$$frac{1}{5}int csc{theta}~dtheta$$
Integrating:
$$frac{1}{5}ln|csc{theta}-cot{theta}|+C$$
integration
edited Mar 17 '17 at 16:36
projectilemotion
11.4k62041
asked Mar 17 '17 at 15:21
El_Master
162
I'm a brand new student. Need some help to integrate this.
Perform this integration: $$int frac{1}{xsqrt{25-x^2}} dx$$
I'm able to obtain in theta terms like this:
$$frac 15 ln|cscθ-cotθ|+C$$
But I have problems to convert in terms of "$x$"
Thanks a lot.
This is what is suggested by this image:
Using the substitution $x=5sin{theta}$ and $dx=5cos{theta}$, we obtain:
$$int frac{1}{5sin{theta}(5cos{theta})}cdot 5cos{theta}~dtheta$$
Simplifying, we obtain:
$$frac{1}{5}int frac{dtheta}{sin{theta}}$$
Using trigonometric identities:
$$frac{1}{5}int csc{theta}~dtheta$$
Integrating:
$$frac{1}{5}ln|csc{theta}-cot{theta}|+C$$
integration
integration
edited Mar 17 '17 at 16:36
projectilemotion
11.4k62041
asked Mar 17 '17 at 15:21
El_Master
162
edited Mar 17 '17 at 16:36
projectilemotion
11.4k62041
asked Mar 17 '17 at 15:21
El_Master
162
edited Mar 17 '17 at 16:36
projectilemotion
11.4k62041
edited Mar 17 '17 at 16:36
projectilemotion
11.4k62041
edited Mar 17 '17 at 16:36
projectilemotion
11.4k62041
11.4k62041
asked Mar 17 '17 at 15:21
El_Master
162
asked Mar 17 '17 at 15:21
El_Master
162
asked Mar 17 '17 at 15:21
El_Master
162
162
Please use: math.meta.stackexchange.com/questions/5020/…
– Arnaldo
Mar 17 '17 at 15:25
It helps if you explain the steps you took to get at the solution in $theta$, e.g. what was the substitution you used?
– StackTD
Mar 17 '17 at 15:27
Yes Exactly. See the problem image..I believe I have the half of the solution....thanks..
– El_Master
Mar 17 '17 at 15:27
@El_Master Did you use the substitution $x=5sin(theta)$ and $dx=5cos(theta)~dtheta$ ?
– projectilemotion
Mar 17 '17 at 15:33
you can use relationships between trigonometric functions to change it back, so if $x = sen theta$ then $cos theta = sqrt{1 - sen^2 theta} = sqrt{1 - x^2}$ and other identities that you can use
– Cato
Mar 17 '17 at 15:33
|
show 3 more comments
Please use: math.meta.stackexchange.com/questions/5020/…
– Arnaldo
Mar 17 '17 at 15:25
It helps if you explain the steps you took to get at the solution in $theta$, e.g. what was the substitution you used?
– StackTD
Mar 17 '17 at 15:27
Yes Exactly. See the problem image..I believe I have the half of the solution....thanks..
– El_Master
Mar 17 '17 at 15:27
@El_Master Did you use the substitution $x=5sin(theta)$ and $dx=5cos(theta)~dtheta$ ?
– projectilemotion
Mar 17 '17 at 15:33
you can use relationships between trigonometric functions to change it back, so if $x = sen theta$ then $cos theta = sqrt{1 - sen^2 theta} = sqrt{1 - x^2}$ and other identities that you can use
– Cato
Mar 17 '17 at 15:33
Please use: math.meta.stackexchange.com/questions/5020/…
– Arnaldo
Mar 17 '17 at 15:25
Please use: math.meta.stackexchange.com/questions/5020/…
– Arnaldo
Mar 17 '17 at 15:25
It helps if you explain the steps you took to get at the solution in $theta$, e.g. what was the substitution you used?
– StackTD
Mar 17 '17 at 15:27
It helps if you explain the steps you took to get at the solution in $theta$, e.g. what was the substitution you used?
– StackTD
Mar 17 '17 at 15:27
Yes Exactly. See the problem image..I believe I have the half of the solution....thanks..
– El_Master
Mar 17 '17 at 15:27
Yes Exactly. See the problem image..I believe I have the half of the solution....thanks..
– El_Master
Mar 17 '17 at 15:27
@El_Master Did you use the substitution $x=5sin(theta)$ and $dx=5cos(theta)~dtheta$ ?
– projectilemotion
Mar 17 '17 at 15:33
@El_Master Did you use the substitution $x=5sin(theta)$ and $dx=5cos(theta)~dtheta$ ?
– projectilemotion
Mar 17 '17 at 15:33
you can use relationships between trigonometric functions to change it back, so if $x = sen theta$ then $cos theta = sqrt{1 - sen^2 theta} = sqrt{1 - x^2}$ and other identities that you can use
– Cato
Mar 17 '17 at 15:33
you can use relationships between trigonometric functions to change it back, so if $x = sen theta$ then $cos theta = sqrt{1 - sen^2 theta} = sqrt{1 - x^2}$ and other identities that you can use
– Cato
Mar 17 '17 at 15:33
|
show 3 more comments
2 Answers
2
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up vote
0
down vote
HINT:
If trigonometric substitution is not mandatory, set $sqrt{a^2-x^2}=y$
$$intdfrac{2x dx}{2x^2sqrt{a^2-x^2}}=intdfrac{dy}{y^2-a^2}=?$$
Else setting $x=5sin y$ where $-dfracpi2le yledfracpi2impliescos yge0$
$cos y=sqrt{1-left(dfrac x5right)^2}=dfrac{sqrt{25-x^2}}5$
$$impliesintdfrac{dx}{xsqrt{25-x^2}}=intdfrac{5cos y dy}{5sin y(5cos y)}=dfrac{ln|csc y-cot y|}5+K$$
answered Mar 17 '17 at 15:52
lab bhattacharjee
222k15155273
I appreciete the help... I understand is not mandatory... but in this case I need it (rules are not mine).
– El_Master
Mar 17 '17 at 15:59
add a comment |
up vote
0
down vote
$$x=5sintheta$$
$$frac x5=sintheta$$
Therefore
$$frac5x=csctheta$$
And since $sintheta=frac x5$, we know that
$$costheta=sqrt{1-sin^2theta}=sqrt{1-frac{x^2}{25}}$$
And since $cottheta=frac{costheta}{sintheta}$,
$$cottheta=frac{5sqrt{1-frac{x^2}{25}}}{x}=frac{sqrt{25-x^2}}x$$
Which gives your result as
$$frac15lnbigg|frac{5-sqrt{25-x^2}}xbigg|+C$$
answered Nov 26 at 6:57
clathratus
2,434324
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
up vote
0
down vote
HINT:
If trigonometric substitution is not mandatory, set $sqrt{a^2-x^2}=y$
$$intdfrac{2x dx}{2x^2sqrt{a^2-x^2}}=intdfrac{dy}{y^2-a^2}=?$$
Else setting $x=5sin y$ where $-dfracpi2le yledfracpi2impliescos yge0$
$cos y=sqrt{1-left(dfrac x5right)^2}=dfrac{sqrt{25-x^2}}5$
$$impliesintdfrac{dx}{xsqrt{25-x^2}}=intdfrac{5cos y dy}{5sin y(5cos y)}=dfrac{ln|csc y-cot y|}5+K$$
answered Mar 17 '17 at 15:52
lab bhattacharjee
222k15155273
I appreciete the help... I understand is not mandatory... but in this case I need it (rules are not mine).
– El_Master
Mar 17 '17 at 15:59
add a comment |
up vote
0
down vote
HINT:
If trigonometric substitution is not mandatory, set $sqrt{a^2-x^2}=y$
$$intdfrac{2x dx}{2x^2sqrt{a^2-x^2}}=intdfrac{dy}{y^2-a^2}=?$$
Else setting $x=5sin y$ where $-dfracpi2le yledfracpi2impliescos yge0$
$cos y=sqrt{1-left(dfrac x5right)^2}=dfrac{sqrt{25-x^2}}5$
$$impliesintdfrac{dx}{xsqrt{25-x^2}}=intdfrac{5cos y dy}{5sin y(5cos y)}=dfrac{ln|csc y-cot y|}5+K$$
answered Mar 17 '17 at 15:52
lab bhattacharjee
222k15155273
I appreciete the help... I understand is not mandatory... but in this case I need it (rules are not mine).
– El_Master
Mar 17 '17 at 15:59
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT:
If trigonometric substitution is not mandatory, set $sqrt{a^2-x^2}=y$
$$intdfrac{2x dx}{2x^2sqrt{a^2-x^2}}=intdfrac{dy}{y^2-a^2}=?$$
Else setting $x=5sin y$ where $-dfracpi2le yledfracpi2impliescos yge0$
$cos y=sqrt{1-left(dfrac x5right)^2}=dfrac{sqrt{25-x^2}}5$
$$impliesintdfrac{dx}{xsqrt{25-x^2}}=intdfrac{5cos y dy}{5sin y(5cos y)}=dfrac{ln|csc y-cot y|}5+K$$
answered Mar 17 '17 at 15:52
lab bhattacharjee
222k15155273
HINT:
If trigonometric substitution is not mandatory, set $sqrt{a^2-x^2}=y$
$$intdfrac{2x dx}{2x^2sqrt{a^2-x^2}}=intdfrac{dy}{y^2-a^2}=?$$
Else setting $x=5sin y$ where $-dfracpi2le yledfracpi2impliescos yge0$
$cos y=sqrt{1-left(dfrac x5right)^2}=dfrac{sqrt{25-x^2}}5$
$$impliesintdfrac{dx}{xsqrt{25-x^2}}=intdfrac{5cos y dy}{5sin y(5cos y)}=dfrac{ln|csc y-cot y|}5+K$$
answered Mar 17 '17 at 15:52
lab bhattacharjee
222k15155273
answered Mar 17 '17 at 15:52
lab bhattacharjee
222k15155273
answered Mar 17 '17 at 15:52
lab bhattacharjee
222k15155273
answered Mar 17 '17 at 15:52
lab bhattacharjee
222k15155273
222k15155273
I appreciete the help... I understand is not mandatory... but in this case I need it (rules are not mine).
– El_Master
Mar 17 '17 at 15:59
add a comment |
I appreciete the help... I understand is not mandatory... but in this case I need it (rules are not mine).
– El_Master
Mar 17 '17 at 15:59
I appreciete the help... I understand is not mandatory... but in this case I need it (rules are not mine).
– El_Master
Mar 17 '17 at 15:59
I appreciete the help... I understand is not mandatory... but in this case I need it (rules are not mine).
– El_Master
Mar 17 '17 at 15:59
add a comment |
up vote
0
down vote
$$x=5sintheta$$
$$frac x5=sintheta$$
Therefore
$$frac5x=csctheta$$
And since $sintheta=frac x5$, we know that
$$costheta=sqrt{1-sin^2theta}=sqrt{1-frac{x^2}{25}}$$
And since $cottheta=frac{costheta}{sintheta}$,
$$cottheta=frac{5sqrt{1-frac{x^2}{25}}}{x}=frac{sqrt{25-x^2}}x$$
Which gives your result as
$$frac15lnbigg|frac{5-sqrt{25-x^2}}xbigg|+C$$
answered Nov 26 at 6:57
clathratus
2,434324
add a comment |
up vote
0
down vote
$$x=5sintheta$$
$$frac x5=sintheta$$
Therefore
$$frac5x=csctheta$$
And since $sintheta=frac x5$, we know that
$$costheta=sqrt{1-sin^2theta}=sqrt{1-frac{x^2}{25}}$$
And since $cottheta=frac{costheta}{sintheta}$,
$$cottheta=frac{5sqrt{1-frac{x^2}{25}}}{x}=frac{sqrt{25-x^2}}x$$
Which gives your result as
$$frac15lnbigg|frac{5-sqrt{25-x^2}}xbigg|+C$$
answered Nov 26 at 6:57
clathratus
2,434324
add a comment |
up vote
0
down vote
up vote
0
down vote
$$x=5sintheta$$
$$frac x5=sintheta$$
Therefore
$$frac5x=csctheta$$
And since $sintheta=frac x5$, we know that
$$costheta=sqrt{1-sin^2theta}=sqrt{1-frac{x^2}{25}}$$
And since $cottheta=frac{costheta}{sintheta}$,
$$cottheta=frac{5sqrt{1-frac{x^2}{25}}}{x}=frac{sqrt{25-x^2}}x$$
Which gives your result as
$$frac15lnbigg|frac{5-sqrt{25-x^2}}xbigg|+C$$
answered Nov 26 at 6:57
clathratus
2,434324
$$x=5sintheta$$
$$frac x5=sintheta$$
Therefore
$$frac5x=csctheta$$
And since $sintheta=frac x5$, we know that
$$costheta=sqrt{1-sin^2theta}=sqrt{1-frac{x^2}{25}}$$
And since $cottheta=frac{costheta}{sintheta}$,
$$cottheta=frac{5sqrt{1-frac{x^2}{25}}}{x}=frac{sqrt{25-x^2}}x$$
Which gives your result as
$$frac15lnbigg|frac{5-sqrt{25-x^2}}xbigg|+C$$
answered Nov 26 at 6:57
clathratus
2,434324
answered Nov 26 at 6:57
clathratus
2,434324
answered Nov 26 at 6:57
clathratus
2,434324
answered Nov 26 at 6:57
clathratus
2,434324
2,434324
add a comment |
add a comment |
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Please use: math.meta.stackexchange.com/questions/5020/…
– Arnaldo
Mar 17 '17 at 15:25
It helps if you explain the steps you took to get at the solution in $theta$, e.g. what was the substitution you used?
– StackTD
Mar 17 '17 at 15:27
Yes Exactly. See the problem image..I believe I have the half of the solution....thanks..
– El_Master
Mar 17 '17 at 15:27
@El_Master Did you use the substitution $x=5sin(theta)$ and $dx=5cos(theta)~dtheta$ ?
– projectilemotion
Mar 17 '17 at 15:33
you can use relationships between trigonometric functions to change it back, so if $x = sen theta$ then $cos theta = sqrt{1 - sen^2 theta} = sqrt{1 - x^2}$ and other identities that you can use
– Cato
Mar 17 '17 at 15:33