Where did I go wrong in my working? - integration via substitution $int_0^frac12ln3frac1e^x+e^-xdx$.

Clash Royale CLAN TAG#URR8PPP

up vote
2
down vote

favorite

1

Here is the question:

Using the substitution $u=e^x$, find the exact value of
$int_0^frac12ln3dfrac1e^x+e^-xdx$

The actual to this answer is $dfracpi12$, and in the official working out it seems they took the arctan of two numbers:
(official working)

This surprised me because I've never seen arctan used in these questions before so I didn't even think about doing that: could anyone also explain when to take the arctan in these sort of questions please?

My working:

$u=e^x, x=ln(u)$

$int_0^frac12ln3dfrac1e^x+e^-xdx = int_0^frac12ln3dfrac1u+u^-1dx $

$fracdudx = e^x therefore fracdue^x=dx$

$int_1^sqrt3frac1u^2+1 du = int_1^sqrt3(u^2+1)^-1 du$

$int_1^sqrt3(u^-2+1) du = [-u^-1+u]_1^sqrt3$

$(-sqrt3^-1+sqrt3) - (-1 + 1) = frac2sqrt3$

calculus integration derivatives proof-verification substitution

share|cite|improve this question

edited Jul 27 at 2:56

Nosrati

19.2k41544

asked Jul 27 at 2:21

Cheks Nweze

627

add a comment | 

up vote
2
down vote

favorite

1

Here is the question:

Using the substitution $u=e^x$, find the exact value of
$int_0^frac12ln3dfrac1e^x+e^-xdx$

The actual to this answer is $dfracpi12$, and in the official working out it seems they took the arctan of two numbers:
(official working)

This surprised me because I've never seen arctan used in these questions before so I didn't even think about doing that: could anyone also explain when to take the arctan in these sort of questions please?

My working:

$u=e^x, x=ln(u)$

$int_0^frac12ln3dfrac1e^x+e^-xdx = int_0^frac12ln3dfrac1u+u^-1dx $

$fracdudx = e^x therefore fracdue^x=dx$

$int_1^sqrt3frac1u^2+1 du = int_1^sqrt3(u^2+1)^-1 du$

$int_1^sqrt3(u^-2+1) du = [-u^-1+u]_1^sqrt3$

$(-sqrt3^-1+sqrt3) - (-1 + 1) = frac2sqrt3$

calculus integration derivatives proof-verification substitution

share|cite|improve this question

edited Jul 27 at 2:56

Nosrati

19.2k41544

asked Jul 27 at 2:21

Cheks Nweze

627

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Here is the question:

Using the substitution $u=e^x$, find the exact value of
$int_0^frac12ln3dfrac1e^x+e^-xdx$

The actual to this answer is $dfracpi12$, and in the official working out it seems they took the arctan of two numbers:
(official working)

This surprised me because I've never seen arctan used in these questions before so I didn't even think about doing that: could anyone also explain when to take the arctan in these sort of questions please?

My working:

$u=e^x, x=ln(u)$

$int_0^frac12ln3dfrac1e^x+e^-xdx = int_0^frac12ln3dfrac1u+u^-1dx $

$fracdudx = e^x therefore fracdue^x=dx$

$int_1^sqrt3frac1u^2+1 du = int_1^sqrt3(u^2+1)^-1 du$

$int_1^sqrt3(u^-2+1) du = [-u^-1+u]_1^sqrt3$

$(-sqrt3^-1+sqrt3) - (-1 + 1) = frac2sqrt3$

calculus integration derivatives proof-verification substitution

share|cite|improve this question

edited Jul 27 at 2:56

Nosrati

19.2k41544

asked Jul 27 at 2:21

Cheks Nweze

627

Here is the question:

Using the substitution $u=e^x$, find the exact value of
$int_0^frac12ln3dfrac1e^x+e^-xdx$

The actual to this answer is $dfracpi12$, and in the official working out it seems they took the arctan of two numbers:
(official working)

This surprised me because I've never seen arctan used in these questions before so I didn't even think about doing that: could anyone also explain when to take the arctan in these sort of questions please?

My working:

$u=e^x, x=ln(u)$

$int_0^frac12ln3dfrac1e^x+e^-xdx = int_0^frac12ln3dfrac1u+u^-1dx $

$fracdudx = e^x therefore fracdue^x=dx$

$int_1^sqrt3frac1u^2+1 du = int_1^sqrt3(u^2+1)^-1 du$

$int_1^sqrt3(u^-2+1) du = [-u^-1+u]_1^sqrt3$

$(-sqrt3^-1+sqrt3) - (-1 + 1) = frac2sqrt3$

calculus integration derivatives proof-verification substitution

share|cite|improve this question

edited Jul 27 at 2:56

Nosrati

19.2k41544

asked Jul 27 at 2:21

Cheks Nweze

627

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 2:56

Nosrati

19.2k41544

edited Jul 27 at 2:56

Nosrati

19.2k41544

edited Jul 27 at 2:56

Nosrati

19.2k41544

19.2k41544

asked Jul 27 at 2:21

Cheks Nweze

627

asked Jul 27 at 2:21

Cheks Nweze

627

asked Jul 27 at 2:21

Cheks Nweze

627

627

add a comment | 

add a comment | 

4 Answers
4

active

oldest

votes

up vote
5
down vote



accepted

Hint: See that $(u^2+1)^-1neq u^-2+1$, also
$$intdfrac1u^2+1du=arctan u+C$$

share|cite|improve this answer

answered Jul 27 at 2:26

Nosrati

19.2k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh right, okay - so does this mean that any expression of the form $int_^ frac1u du$ can be rewritten as $arctan(u)$?
  – Cheks Nweze
  Jul 27 at 13:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, why do we use $arctan$ and not $arccos$ or $arcsin$?
  – Cheks Nweze
  Jul 27 at 14:23
  

  
  
  
  

add a comment | 

up vote
3
down vote

HINT:

$$intfrac1u^2+1,du=arctan(u)+Cne -frac1u + u +C$$

share|cite|improve this answer

answered Jul 27 at 2:26

Mark Viola

126k1172167

add a comment | 

up vote
1
down vote

$$I=int_0^frac12ln(3)frac1e^x+e^-xdx=int_0^frac12ln(3)frac1e^x+frac1e^xdx=int_0^frac12ln(3)frace^x(e^x)^2+1dx$$
$u=e^x,,fracdudx=e^x$
$$therefore I=int_1^sqrt3frac1u^2+1du=arctan(sqrt3)-arctan(1)=fracpi3-fracpi4=fracpi12$$

share|cite|improve this answer

answered Jul 27 at 11:15

Henry Lee

49210

add a comment | 

up vote
0
down vote

Hint: Try multiplying the numerator and denominator by $e^x$ and see what happens. Then make the appropiate substitution after doing that.

Hope this really helps! :)

share|cite|improve this answer

answered Jul 27 at 2:26

user573497

2009

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863986%2fwhere-did-i-go-wrong-in-my-working-integration-via-substitution-int-0-f%23new-answer', 'question_page');

);

Post as a guest

Name Email

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

Hint: See that $(u^2+1)^-1neq u^-2+1$, also
$$intdfrac1u^2+1du=arctan u+C$$

share|cite|improve this answer

answered Jul 27 at 2:26

Nosrati

19.2k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh right, okay - so does this mean that any expression of the form $int_^ frac1u du$ can be rewritten as $arctan(u)$?
  – Cheks Nweze
  Jul 27 at 13:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, why do we use $arctan$ and not $arccos$ or $arcsin$?
  – Cheks Nweze
  Jul 27 at 14:23
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

Hint: See that $(u^2+1)^-1neq u^-2+1$, also
$$intdfrac1u^2+1du=arctan u+C$$

share|cite|improve this answer

answered Jul 27 at 2:26

Nosrati

19.2k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh right, okay - so does this mean that any expression of the form $int_^ frac1u du$ can be rewritten as $arctan(u)$?
  – Cheks Nweze
  Jul 27 at 13:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, why do we use $arctan$ and not $arccos$ or $arcsin$?
  – Cheks Nweze
  Jul 27 at 14:23
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

Hint: See that $(u^2+1)^-1neq u^-2+1$, also
$$intdfrac1u^2+1du=arctan u+C$$

share|cite|improve this answer

answered Jul 27 at 2:26

Nosrati

19.2k41544

Hint: See that $(u^2+1)^-1neq u^-2+1$, also
$$intdfrac1u^2+1du=arctan u+C$$

share|cite|improve this answer

answered Jul 27 at 2:26

Nosrati

19.2k41544

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 2:26

Nosrati

19.2k41544

answered Jul 27 at 2:26

Nosrati

19.2k41544

answered Jul 27 at 2:26

Nosrati

19.2k41544

19.2k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh right, okay - so does this mean that any expression of the form $int_^ frac1u du$ can be rewritten as $arctan(u)$?
  – Cheks Nweze
  Jul 27 at 13:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, why do we use $arctan$ and not $arccos$ or $arcsin$?
  – Cheks Nweze
  Jul 27 at 14:23
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh right, okay - so does this mean that any expression of the form $int_^ frac1u du$ can be rewritten as $arctan(u)$?
  – Cheks Nweze
  Jul 27 at 13:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, why do we use $arctan$ and not $arccos$ or $arcsin$?
  – Cheks Nweze
  Jul 27 at 14:23
  

  
  
  
  

Oh right, okay - so does this mean that any expression of the form $int_^ frac1u du$ can be rewritten as $arctan(u)$?
– Cheks Nweze
Jul 27 at 13:52

Oh right, okay - so does this mean that any expression of the form $int_^ frac1u du$ can be rewritten as $arctan(u)$?
– Cheks Nweze
Jul 27 at 13:52

Also, why do we use $arctan$ and not $arccos$ or $arcsin$?
– Cheks Nweze
Jul 27 at 14:23

Also, why do we use $arctan$ and not $arccos$ or $arcsin$?
– Cheks Nweze
Jul 27 at 14:23

add a comment | 

up vote
3
down vote

HINT:

$$intfrac1u^2+1,du=arctan(u)+Cne -frac1u + u +C$$

share|cite|improve this answer

answered Jul 27 at 2:26

Mark Viola

126k1172167

add a comment | 

up vote
3
down vote

HINT:

$$intfrac1u^2+1,du=arctan(u)+Cne -frac1u + u +C$$

share|cite|improve this answer

answered Jul 27 at 2:26

Mark Viola

126k1172167

add a comment | 

up vote
3
down vote

up vote
3
down vote

HINT:

$$intfrac1u^2+1,du=arctan(u)+Cne -frac1u + u +C$$

share|cite|improve this answer

answered Jul 27 at 2:26

Mark Viola

126k1172167

HINT:

$$intfrac1u^2+1,du=arctan(u)+Cne -frac1u + u +C$$

share|cite|improve this answer

answered Jul 27 at 2:26

Mark Viola

126k1172167

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 2:26

Mark Viola

126k1172167

answered Jul 27 at 2:26

Mark Viola

126k1172167

answered Jul 27 at 2:26

Mark Viola

126k1172167

126k1172167

add a comment | 

add a comment | 

up vote
1
down vote

$$I=int_0^frac12ln(3)frac1e^x+e^-xdx=int_0^frac12ln(3)frac1e^x+frac1e^xdx=int_0^frac12ln(3)frace^x(e^x)^2+1dx$$
$u=e^x,,fracdudx=e^x$
$$therefore I=int_1^sqrt3frac1u^2+1du=arctan(sqrt3)-arctan(1)=fracpi3-fracpi4=fracpi12$$

share|cite|improve this answer

answered Jul 27 at 11:15

Henry Lee

49210

add a comment | 

up vote
1
down vote

$$I=int_0^frac12ln(3)frac1e^x+e^-xdx=int_0^frac12ln(3)frac1e^x+frac1e^xdx=int_0^frac12ln(3)frace^x(e^x)^2+1dx$$
$u=e^x,,fracdudx=e^x$
$$therefore I=int_1^sqrt3frac1u^2+1du=arctan(sqrt3)-arctan(1)=fracpi3-fracpi4=fracpi12$$

share|cite|improve this answer

answered Jul 27 at 11:15

Henry Lee

49210

add a comment | 

up vote
1
down vote

up vote
1
down vote

$$I=int_0^frac12ln(3)frac1e^x+e^-xdx=int_0^frac12ln(3)frac1e^x+frac1e^xdx=int_0^frac12ln(3)frace^x(e^x)^2+1dx$$
$u=e^x,,fracdudx=e^x$
$$therefore I=int_1^sqrt3frac1u^2+1du=arctan(sqrt3)-arctan(1)=fracpi3-fracpi4=fracpi12$$

share|cite|improve this answer

answered Jul 27 at 11:15

Henry Lee

49210

$$I=int_0^frac12ln(3)frac1e^x+e^-xdx=int_0^frac12ln(3)frac1e^x+frac1e^xdx=int_0^frac12ln(3)frace^x(e^x)^2+1dx$$
$u=e^x,,fracdudx=e^x$
$$therefore I=int_1^sqrt3frac1u^2+1du=arctan(sqrt3)-arctan(1)=fracpi3-fracpi4=fracpi12$$

share|cite|improve this answer

answered Jul 27 at 11:15

Henry Lee

49210

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 11:15

Henry Lee

49210

answered Jul 27 at 11:15

Henry Lee

49210

answered Jul 27 at 11:15

Henry Lee

49210

49210

add a comment | 

add a comment | 

up vote
0
down vote

Hint: Try multiplying the numerator and denominator by $e^x$ and see what happens. Then make the appropiate substitution after doing that.

Hope this really helps! :)

share|cite|improve this answer

answered Jul 27 at 2:26

user573497

2009

add a comment | 

up vote
0
down vote

Hint: Try multiplying the numerator and denominator by $e^x$ and see what happens. Then make the appropiate substitution after doing that.

Hope this really helps! :)

share|cite|improve this answer

answered Jul 27 at 2:26

user573497

2009

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint: Try multiplying the numerator and denominator by $e^x$ and see what happens. Then make the appropiate substitution after doing that.

Hope this really helps! :)

share|cite|improve this answer

answered Jul 27 at 2:26

user573497

2009

Hint: Try multiplying the numerator and denominator by $e^x$ and see what happens. Then make the appropiate substitution after doing that.

Hope this really helps! :)

share|cite|improve this answer

answered Jul 27 at 2:26

user573497

2009

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 2:26

user573497

2009

answered Jul 27 at 2:26

user573497

2009

answered Jul 27 at 2:26

user573497

2009

2009

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863986%2fwhere-did-i-go-wrong-in-my-working-integration-via-substitution-int-0-f%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Name Email

Name Email

Name

Email

Name Email

Name

Email