Mixing
Integration by substitition: replacing function of x by function of u
$begingroup$
I'm having trouble understanding a specific step used to solve the integral below, where instead of replacing $g(x)$ by $u$, it is replaced by a function of $u$: $tan(u)$
$$int sin(x)sqrt{1+cos^2(x)}dx$$
The following substitution is used:
$$cos(x) = tan(u)$$
$$mathbf{-sin(x)dx = sec^2(u) du}$$
Resulting in the standard integral which can be solved :
$$int sec^2(U)sqrt{1+tan^2(u)}du = int sec^2(u) sqrt{sec^2(u)} = int sec^3(u) du $$
The substition in boldface is the one I am having trouble understanding. The left part is the same as usual, i.e. $frac{du}{dx}dx = du$, yet it is set equal to the derivative of $tan(u)$. I understand this is due to replacing $cos(x)$ by $tan(u)$ instead of $u$, but I am having trouble understanding the exact steps. An explanation or name of the method would be greatly appreciated.
calculus integration substitution
asked Feb 1 at 19:56
RooseRoose
102
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding a specific step used to solve the integral below, where instead of replacing $g(x)$ by $u$, it is replaced by a function of $u$: $tan(u)$
$$int sin(x)sqrt{1+cos^2(x)}dx$$
The following substitution is used:
$$cos(x) = tan(u)$$
$$mathbf{-sin(x)dx = sec^2(u) du}$$
Resulting in the standard integral which can be solved :
$$int sec^2(U)sqrt{1+tan^2(u)}du = int sec^2(u) sqrt{sec^2(u)} = int sec^3(u) du $$
The substition in boldface is the one I am having trouble understanding. The left part is the same as usual, i.e. $frac{du}{dx}dx = du$, yet it is set equal to the derivative of $tan(u)$. I understand this is due to replacing $cos(x)$ by $tan(u)$ instead of $u$, but I am having trouble understanding the exact steps. An explanation or name of the method would be greatly appreciated.
calculus integration substitution
asked Feb 1 at 19:56
RooseRoose
102
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding a specific step used to solve the integral below, where instead of replacing $g(x)$ by $u$, it is replaced by a function of $u$: $tan(u)$
$$int sin(x)sqrt{1+cos^2(x)}dx$$
The following substitution is used:
$$cos(x) = tan(u)$$
$$mathbf{-sin(x)dx = sec^2(u) du}$$
Resulting in the standard integral which can be solved :
$$int sec^2(U)sqrt{1+tan^2(u)}du = int sec^2(u) sqrt{sec^2(u)} = int sec^3(u) du $$
The substition in boldface is the one I am having trouble understanding. The left part is the same as usual, i.e. $frac{du}{dx}dx = du$, yet it is set equal to the derivative of $tan(u)$. I understand this is due to replacing $cos(x)$ by $tan(u)$ instead of $u$, but I am having trouble understanding the exact steps. An explanation or name of the method would be greatly appreciated.
calculus integration substitution
asked Feb 1 at 19:56
RooseRoose
102
$endgroup$
I'm having trouble understanding a specific step used to solve the integral below, where instead of replacing $g(x)$ by $u$, it is replaced by a function of $u$: $tan(u)$
$$int sin(x)sqrt{1+cos^2(x)}dx$$
The following substitution is used:
$$cos(x) = tan(u)$$
$$mathbf{-sin(x)dx = sec^2(u) du}$$
Resulting in the standard integral which can be solved :
$$int sec^2(U)sqrt{1+tan^2(u)}du = int sec^2(u) sqrt{sec^2(u)} = int sec^3(u) du $$
The substition in boldface is the one I am having trouble understanding. The left part is the same as usual, i.e. $frac{du}{dx}dx = du$, yet it is set equal to the derivative of $tan(u)$. I understand this is due to replacing $cos(x)$ by $tan(u)$ instead of $u$, but I am having trouble understanding the exact steps. An explanation or name of the method would be greatly appreciated.
calculus integration substitution
calculus integration substitution
asked Feb 1 at 19:56
RooseRoose
102
asked Feb 1 at 19:56
RooseRoose
102
asked Feb 1 at 19:56
RooseRoose
102
asked Feb 1 at 19:56
RooseRoose
102
asked Feb 1 at 19:56
RooseRoose
102
102
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Implicitly;
$$frac{d}{du}(cos x) = -sin xfrac{dx}{du}$$
$$frac{d}{du}(tan u) = sec^2 u$$
They equate so:
$$-sin x frac{dx}{du} = sec^2 u$$
Then multiply by $du$ which is allowed.
$$-sin x dx = sec^2 u du$$
answered Feb 1 at 20:05
Rhys HughesRhys Hughes
7,0501630
$endgroup$
1
$begingroup$
Thank you for your answer. I'm having trouble understanding the first line. You're taking the derivative of $cos x$ with respect to u (assuming this is a typo)? Which results in: $frac{d}{dx}(cos x) frac{dx}{du} = -sin x frac{dx}{du}$? How does the $frac{dx}{du}$ come in?
$endgroup$
– Roose
Feb 1 at 20:11
$begingroup$
It's no typo, this is how you differentiate with respect to another variable, and it is because of the chain rule: $$frac{dy}{du}=frac{dy}{dx}cdotfrac{dx}{du}$$. In this case in my first line we take $y=cos(x)$ which leads to $frac{dy}{dx}=-sin x$, then we must multiply by $frac{dx}{du}$ to make this valid.
$endgroup$
– Rhys Hughes
Feb 1 at 20:16
1
$begingroup$
Ah the chain rule, my apologies. If I'm understanding this correctly, this is the reasoning: $$cos(x) = tan(u)$$ then $$frac{d}{du}(cos(x)) = frac{d}{du}(tan(u))$$ After which your next steps apply. Is this correct?
$endgroup$
– Roose
Feb 1 at 20:20
$begingroup$
Precisely that.
$endgroup$
– Rhys Hughes
Feb 1 at 20:33
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%2f3096657%2fintegration-by-substitition-replacing-function-of-x-by-function-of-u%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Implicitly;
$$frac{d}{du}(cos x) = -sin xfrac{dx}{du}$$
$$frac{d}{du}(tan u) = sec^2 u$$
They equate so:
$$-sin x frac{dx}{du} = sec^2 u$$
Then multiply by $du$ which is allowed.
$$-sin x dx = sec^2 u du$$
answered Feb 1 at 20:05
Rhys HughesRhys Hughes
7,0501630
$endgroup$
1
$begingroup$
Thank you for your answer. I'm having trouble understanding the first line. You're taking the derivative of $cos x$ with respect to u (assuming this is a typo)? Which results in: $frac{d}{dx}(cos x) frac{dx}{du} = -sin x frac{dx}{du}$? How does the $frac{dx}{du}$ come in?
$endgroup$
– Roose
Feb 1 at 20:11
$begingroup$
It's no typo, this is how you differentiate with respect to another variable, and it is because of the chain rule: $$frac{dy}{du}=frac{dy}{dx}cdotfrac{dx}{du}$$. In this case in my first line we take $y=cos(x)$ which leads to $frac{dy}{dx}=-sin x$, then we must multiply by $frac{dx}{du}$ to make this valid.
$endgroup$
– Rhys Hughes
Feb 1 at 20:16
1
$begingroup$
Ah the chain rule, my apologies. If I'm understanding this correctly, this is the reasoning: $$cos(x) = tan(u)$$ then $$frac{d}{du}(cos(x)) = frac{d}{du}(tan(u))$$ After which your next steps apply. Is this correct?
$endgroup$
– Roose
Feb 1 at 20:20
$begingroup$
Precisely that.
$endgroup$
– Rhys Hughes
Feb 1 at 20:33
add a comment |
$begingroup$
Implicitly;
$$frac{d}{du}(cos x) = -sin xfrac{dx}{du}$$
$$frac{d}{du}(tan u) = sec^2 u$$
They equate so:
$$-sin x frac{dx}{du} = sec^2 u$$
Then multiply by $du$ which is allowed.
$$-sin x dx = sec^2 u du$$
answered Feb 1 at 20:05
Rhys HughesRhys Hughes
7,0501630
$endgroup$
1
$begingroup$
Thank you for your answer. I'm having trouble understanding the first line. You're taking the derivative of $cos x$ with respect to u (assuming this is a typo)? Which results in: $frac{d}{dx}(cos x) frac{dx}{du} = -sin x frac{dx}{du}$? How does the $frac{dx}{du}$ come in?
$endgroup$
– Roose
Feb 1 at 20:11
$begingroup$
It's no typo, this is how you differentiate with respect to another variable, and it is because of the chain rule: $$frac{dy}{du}=frac{dy}{dx}cdotfrac{dx}{du}$$. In this case in my first line we take $y=cos(x)$ which leads to $frac{dy}{dx}=-sin x$, then we must multiply by $frac{dx}{du}$ to make this valid.
$endgroup$
– Rhys Hughes
Feb 1 at 20:16
1
$begingroup$
Ah the chain rule, my apologies. If I'm understanding this correctly, this is the reasoning: $$cos(x) = tan(u)$$ then $$frac{d}{du}(cos(x)) = frac{d}{du}(tan(u))$$ After which your next steps apply. Is this correct?
$endgroup$
– Roose
Feb 1 at 20:20
$begingroup$
Precisely that.
$endgroup$
– Rhys Hughes
Feb 1 at 20:33
add a comment |
$begingroup$
Implicitly;
$$frac{d}{du}(cos x) = -sin xfrac{dx}{du}$$
$$frac{d}{du}(tan u) = sec^2 u$$
They equate so:
$$-sin x frac{dx}{du} = sec^2 u$$
Then multiply by $du$ which is allowed.
$$-sin x dx = sec^2 u du$$
answered Feb 1 at 20:05
Rhys HughesRhys Hughes
7,0501630
$endgroup$
Implicitly;
$$frac{d}{du}(cos x) = -sin xfrac{dx}{du}$$
$$frac{d}{du}(tan u) = sec^2 u$$
They equate so:
$$-sin x frac{dx}{du} = sec^2 u$$
Then multiply by $du$ which is allowed.
$$-sin x dx = sec^2 u du$$
answered Feb 1 at 20:05
Rhys HughesRhys Hughes
7,0501630
answered Feb 1 at 20:05
Rhys HughesRhys Hughes
7,0501630
answered Feb 1 at 20:05
Rhys HughesRhys Hughes
7,0501630
answered Feb 1 at 20:05
Rhys HughesRhys Hughes
7,0501630
7,0501630
1
$begingroup$
Thank you for your answer. I'm having trouble understanding the first line. You're taking the derivative of $cos x$ with respect to u (assuming this is a typo)? Which results in: $frac{d}{dx}(cos x) frac{dx}{du} = -sin x frac{dx}{du}$? How does the $frac{dx}{du}$ come in?
$endgroup$
– Roose
Feb 1 at 20:11
$begingroup$
It's no typo, this is how you differentiate with respect to another variable, and it is because of the chain rule: $$frac{dy}{du}=frac{dy}{dx}cdotfrac{dx}{du}$$. In this case in my first line we take $y=cos(x)$ which leads to $frac{dy}{dx}=-sin x$, then we must multiply by $frac{dx}{du}$ to make this valid.
$endgroup$
– Rhys Hughes
Feb 1 at 20:16
1
$begingroup$
Ah the chain rule, my apologies. If I'm understanding this correctly, this is the reasoning: $$cos(x) = tan(u)$$ then $$frac{d}{du}(cos(x)) = frac{d}{du}(tan(u))$$ After which your next steps apply. Is this correct?
$endgroup$
– Roose
Feb 1 at 20:20
$begingroup$
Precisely that.
$endgroup$
– Rhys Hughes
Feb 1 at 20:33
add a comment |
1
$begingroup$
Thank you for your answer. I'm having trouble understanding the first line. You're taking the derivative of $cos x$ with respect to u (assuming this is a typo)? Which results in: $frac{d}{dx}(cos x) frac{dx}{du} = -sin x frac{dx}{du}$? How does the $frac{dx}{du}$ come in?
$endgroup$
– Roose
Feb 1 at 20:11
$begingroup$
It's no typo, this is how you differentiate with respect to another variable, and it is because of the chain rule: $$frac{dy}{du}=frac{dy}{dx}cdotfrac{dx}{du}$$. In this case in my first line we take $y=cos(x)$ which leads to $frac{dy}{dx}=-sin x$, then we must multiply by $frac{dx}{du}$ to make this valid.
$endgroup$
– Rhys Hughes
Feb 1 at 20:16
1
$begingroup$
Ah the chain rule, my apologies. If I'm understanding this correctly, this is the reasoning: $$cos(x) = tan(u)$$ then $$frac{d}{du}(cos(x)) = frac{d}{du}(tan(u))$$ After which your next steps apply. Is this correct?
$endgroup$
– Roose
Feb 1 at 20:20
$begingroup$
Precisely that.
$endgroup$
– Rhys Hughes
Feb 1 at 20:33
1
1
$begingroup$
Thank you for your answer. I'm having trouble understanding the first line. You're taking the derivative of $cos x$ with respect to u (assuming this is a typo)? Which results in: $frac{d}{dx}(cos x) frac{dx}{du} = -sin x frac{dx}{du}$? How does the $frac{dx}{du}$ come in?
$endgroup$
– Roose
Feb 1 at 20:11
$begingroup$
Thank you for your answer. I'm having trouble understanding the first line. You're taking the derivative of $cos x$ with respect to u (assuming this is a typo)? Which results in: $frac{d}{dx}(cos x) frac{dx}{du} = -sin x frac{dx}{du}$? How does the $frac{dx}{du}$ come in?
$endgroup$
– Roose
Feb 1 at 20:11
$begingroup$
It's no typo, this is how you differentiate with respect to another variable, and it is because of the chain rule: $$frac{dy}{du}=frac{dy}{dx}cdotfrac{dx}{du}$$. In this case in my first line we take $y=cos(x)$ which leads to $frac{dy}{dx}=-sin x$, then we must multiply by $frac{dx}{du}$ to make this valid.
$endgroup$
– Rhys Hughes
Feb 1 at 20:16
$begingroup$
It's no typo, this is how you differentiate with respect to another variable, and it is because of the chain rule: $$frac{dy}{du}=frac{dy}{dx}cdotfrac{dx}{du}$$. In this case in my first line we take $y=cos(x)$ which leads to $frac{dy}{dx}=-sin x$, then we must multiply by $frac{dx}{du}$ to make this valid.
$endgroup$
– Rhys Hughes
Feb 1 at 20:16
1
1
$begingroup$
Ah the chain rule, my apologies. If I'm understanding this correctly, this is the reasoning: $$cos(x) = tan(u)$$ then $$frac{d}{du}(cos(x)) = frac{d}{du}(tan(u))$$ After which your next steps apply. Is this correct?
$endgroup$
– Roose
Feb 1 at 20:20
$begingroup$
Ah the chain rule, my apologies. If I'm understanding this correctly, this is the reasoning: $$cos(x) = tan(u)$$ then $$frac{d}{du}(cos(x)) = frac{d}{du}(tan(u))$$ After which your next steps apply. Is this correct?
$endgroup$
– Roose
Feb 1 at 20:20
$begingroup$
Precisely that.
$endgroup$
– Rhys Hughes
Feb 1 at 20:33
$begingroup$
Precisely that.
$endgroup$
– Rhys Hughes
Feb 1 at 20:33
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.
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%2f3096657%2fintegration-by-substitition-replacing-function-of-x-by-function-of-u%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
