Help to solve: $int 1/(xsqrt{25-x^2}) dx$
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
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
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
active
oldest
votes
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
223k15156274
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 |
$$x=5sintheta$$
$$frac x5=sintheta$$
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 '18 at 6:57
clathratus
3,243331
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2191043%2fhelp-to-solve-int-1-x-sqrt25-x2-dx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
223k15156274
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 |
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
223k15156274
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 |
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
223k15156274
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
223k15156274
answered Mar 17 '17 at 15:52
lab bhattacharjee
223k15156274
answered Mar 17 '17 at 15:52
lab bhattacharjee
223k15156274
answered Mar 17 '17 at 15:52
lab bhattacharjee
223k15156274
223k15156274
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 |
$$x=5sintheta$$
$$frac x5=sintheta$$
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 '18 at 6:57
clathratus
3,243331
add a comment |
$$x=5sintheta$$
$$frac x5=sintheta$$
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 '18 at 6:57
clathratus
3,243331
add a comment |
$$x=5sintheta$$
$$frac x5=sintheta$$
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 '18 at 6:57
clathratus
3,243331
$$x=5sintheta$$
$$frac x5=sintheta$$
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 '18 at 6:57
clathratus
3,243331
answered Nov 26 '18 at 6:57
clathratus
3,243331
answered Nov 26 '18 at 6:57
clathratus
3,243331
answered Nov 26 '18 at 6:57
clathratus
3,243331
3,243331
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2191043%2fhelp-to-solve-int-1-x-sqrt25-x2-dx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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