Trying to $int_0^pi/2cos(2x)ln^2(tan(x/2)) dx$

Clash Royale CLAN TAG#URR8PPP

up vote
3
down vote

favorite

1

I would like to evaluate this integral

$$large int_0^pi/2cos(2x)ln^2(tan(x/2))mathrm dx$$

Choosing

$u=fracx2$

$mathrm dx=2mathrm du$

$$2int_0^pi/4cos(4u)ln^2(tan u)mathrm du$$

Choosing $cos(4u)=sin^4(u)-6cos^2(u)sin^2(u)+cos^4(4)$

transforming into

$$2int_0^pi/4sec^2(u)cdot frac[tan^2(u)-2tan(u)-1][tan^2(u)+2tan(u)-1]ln^2(tan(u))(1+tan^2(u))^3du$$

Choosing

$v=tan(u)$

$mathrm du=frac1sec^2(u)mathrm dv$

$$2int_0^inftyfrac(v^2-2v-1)(v^2+2v-1)ln^2(v)(1+v^2)^3mathrm dv$$

$$2int_0^inftyfracln^2(v)1+v^2dv-4int_0^inftyfracln^2(v)(1+v^2)^2dv+4int_0^inftyfracln^2(v)(1+v^2)^3dv$$

I was thinking of applying binomial series, but it can't be don't

how would I continue this?

calculus integration definite-integrals trigonometric-integrals

share|cite|improve this question

edited Jul 24 at 13:04

Richard

1829

asked Jul 24 at 8:43

user565198

add a comment | 

up vote
3
down vote

favorite

1

I would like to evaluate this integral

$$large int_0^pi/2cos(2x)ln^2(tan(x/2))mathrm dx$$

Choosing

$u=fracx2$

$mathrm dx=2mathrm du$

$$2int_0^pi/4cos(4u)ln^2(tan u)mathrm du$$

Choosing $cos(4u)=sin^4(u)-6cos^2(u)sin^2(u)+cos^4(4)$

transforming into

$$2int_0^pi/4sec^2(u)cdot frac[tan^2(u)-2tan(u)-1][tan^2(u)+2tan(u)-1]ln^2(tan(u))(1+tan^2(u))^3du$$

Choosing

$v=tan(u)$

$mathrm du=frac1sec^2(u)mathrm dv$

$$2int_0^inftyfrac(v^2-2v-1)(v^2+2v-1)ln^2(v)(1+v^2)^3mathrm dv$$

$$2int_0^inftyfracln^2(v)1+v^2dv-4int_0^inftyfracln^2(v)(1+v^2)^2dv+4int_0^inftyfracln^2(v)(1+v^2)^3dv$$

I was thinking of applying binomial series, but it can't be don't

how would I continue this?

calculus integration definite-integrals trigonometric-integrals

share|cite|improve this question

edited Jul 24 at 13:04

Richard

1829

asked Jul 24 at 8:43

user565198

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

I would like to evaluate this integral

$$large int_0^pi/2cos(2x)ln^2(tan(x/2))mathrm dx$$

Choosing

$u=fracx2$

$mathrm dx=2mathrm du$

$$2int_0^pi/4cos(4u)ln^2(tan u)mathrm du$$

Choosing $cos(4u)=sin^4(u)-6cos^2(u)sin^2(u)+cos^4(4)$

transforming into

$$2int_0^pi/4sec^2(u)cdot frac[tan^2(u)-2tan(u)-1][tan^2(u)+2tan(u)-1]ln^2(tan(u))(1+tan^2(u))^3du$$

Choosing

$v=tan(u)$

$mathrm du=frac1sec^2(u)mathrm dv$

$$2int_0^inftyfrac(v^2-2v-1)(v^2+2v-1)ln^2(v)(1+v^2)^3mathrm dv$$

$$2int_0^inftyfracln^2(v)1+v^2dv-4int_0^inftyfracln^2(v)(1+v^2)^2dv+4int_0^inftyfracln^2(v)(1+v^2)^3dv$$

I was thinking of applying binomial series, but it can't be don't

how would I continue this?

calculus integration definite-integrals trigonometric-integrals

share|cite|improve this question

edited Jul 24 at 13:04

Richard

1829

asked Jul 24 at 8:43

user565198

I would like to evaluate this integral

$$large int_0^pi/2cos(2x)ln^2(tan(x/2))mathrm dx$$

Choosing

$u=fracx2$

$mathrm dx=2mathrm du$

$$2int_0^pi/4cos(4u)ln^2(tan u)mathrm du$$

Choosing $cos(4u)=sin^4(u)-6cos^2(u)sin^2(u)+cos^4(4)$

transforming into

$$2int_0^pi/4sec^2(u)cdot frac[tan^2(u)-2tan(u)-1][tan^2(u)+2tan(u)-1]ln^2(tan(u))(1+tan^2(u))^3du$$

Choosing

$v=tan(u)$

$mathrm du=frac1sec^2(u)mathrm dv$

$$2int_0^inftyfrac(v^2-2v-1)(v^2+2v-1)ln^2(v)(1+v^2)^3mathrm dv$$

$$2int_0^inftyfracln^2(v)1+v^2dv-4int_0^inftyfracln^2(v)(1+v^2)^2dv+4int_0^inftyfracln^2(v)(1+v^2)^3dv$$

I was thinking of applying binomial series, but it can't be don't

how would I continue this?

calculus integration definite-integrals trigonometric-integrals

share|cite|improve this question

edited Jul 24 at 13:04

Richard

1829

asked Jul 24 at 8:43

user565198

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 13:04

Richard

1829

edited Jul 24 at 13:04

Richard

1829

edited Jul 24 at 13:04

Richard

1829

1829

asked Jul 24 at 8:43

user565198

asked Jul 24 at 8:43

user565198

asked Jul 24 at 8:43

user565198

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote

Considering $$I=intcos (2 x) log ^2left(tan left(fracx2right)right),dx$$ use integration by parts
$$u=log ^2left(tan left(fracx2right)right)implies du=2 csc (x) log left(tan left(fracx2right)right),dx$$
$$dv=cos(2x),dx implies v=frac12 sin (2 x)$$
$$I=frac12 sin (2 x) log ^2left(tan left(fracx2right)right)-2int cos (x) log left(tan left(fracx2right)right),dx$$ One more integration by parts to get the antiderivative and a beautiful result for the integral.

share|cite|improve this answer

answered Jul 24 at 8:57

Claude Leibovici

111k1055126

add a comment | 

up vote
3
down vote

Although an antiderivative of the integrand is easy to find viz. Claude Leibovici's method, the limits given present a challenge. The integral is $$[frac12sin 2xln^2tanfracx2-2sin xlntanfracx2+2x]^pi/2_0.$$For $n>0$, $lim_xtoinftyx^n e^-x=0$ so $lim_yto 0^+yln^n y=0$. Trigonometric small-angle approximations then reduce the integral to $pi+lim_xto 0(-xln^2fracx2+2xlnfracx2)=pi$.

share|cite|improve this answer

answered Jul 24 at 12:53

J.G.

13.2k11424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree with you. The series expansion built at $x=0$ is "quite" nice. By the way, $+1$. Cheers.
  – Claude Leibovici
  Jul 24 at 13:15
  

  
  
  
  

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%2f2861122%2ftrying-to-int-0-pi-2-cos2x-ln2-tanx-2-dx%23new-answer', 'question_page');

);

Post as a guest

Name Email

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Considering $$I=intcos (2 x) log ^2left(tan left(fracx2right)right),dx$$ use integration by parts
$$u=log ^2left(tan left(fracx2right)right)implies du=2 csc (x) log left(tan left(fracx2right)right),dx$$
$$dv=cos(2x),dx implies v=frac12 sin (2 x)$$
$$I=frac12 sin (2 x) log ^2left(tan left(fracx2right)right)-2int cos (x) log left(tan left(fracx2right)right),dx$$ One more integration by parts to get the antiderivative and a beautiful result for the integral.

share|cite|improve this answer

answered Jul 24 at 8:57

Claude Leibovici

111k1055126

add a comment | 

up vote
3
down vote

Considering $$I=intcos (2 x) log ^2left(tan left(fracx2right)right),dx$$ use integration by parts
$$u=log ^2left(tan left(fracx2right)right)implies du=2 csc (x) log left(tan left(fracx2right)right),dx$$
$$dv=cos(2x),dx implies v=frac12 sin (2 x)$$
$$I=frac12 sin (2 x) log ^2left(tan left(fracx2right)right)-2int cos (x) log left(tan left(fracx2right)right),dx$$ One more integration by parts to get the antiderivative and a beautiful result for the integral.

share|cite|improve this answer

answered Jul 24 at 8:57

Claude Leibovici

111k1055126

add a comment | 

up vote
3
down vote

up vote
3
down vote

Considering $$I=intcos (2 x) log ^2left(tan left(fracx2right)right),dx$$ use integration by parts
$$u=log ^2left(tan left(fracx2right)right)implies du=2 csc (x) log left(tan left(fracx2right)right),dx$$
$$dv=cos(2x),dx implies v=frac12 sin (2 x)$$
$$I=frac12 sin (2 x) log ^2left(tan left(fracx2right)right)-2int cos (x) log left(tan left(fracx2right)right),dx$$ One more integration by parts to get the antiderivative and a beautiful result for the integral.

share|cite|improve this answer

answered Jul 24 at 8:57

Claude Leibovici

111k1055126

Considering $$I=intcos (2 x) log ^2left(tan left(fracx2right)right),dx$$ use integration by parts
$$u=log ^2left(tan left(fracx2right)right)implies du=2 csc (x) log left(tan left(fracx2right)right),dx$$
$$dv=cos(2x),dx implies v=frac12 sin (2 x)$$
$$I=frac12 sin (2 x) log ^2left(tan left(fracx2right)right)-2int cos (x) log left(tan left(fracx2right)right),dx$$ One more integration by parts to get the antiderivative and a beautiful result for the integral.

share|cite|improve this answer

answered Jul 24 at 8:57

Claude Leibovici

111k1055126

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 8:57

Claude Leibovici

111k1055126

answered Jul 24 at 8:57

Claude Leibovici

111k1055126

answered Jul 24 at 8:57

Claude Leibovici

111k1055126

111k1055126

add a comment | 

add a comment | 

up vote
3
down vote

Although an antiderivative of the integrand is easy to find viz. Claude Leibovici's method, the limits given present a challenge. The integral is $$[frac12sin 2xln^2tanfracx2-2sin xlntanfracx2+2x]^pi/2_0.$$For $n>0$, $lim_xtoinftyx^n e^-x=0$ so $lim_yto 0^+yln^n y=0$. Trigonometric small-angle approximations then reduce the integral to $pi+lim_xto 0(-xln^2fracx2+2xlnfracx2)=pi$.

share|cite|improve this answer

answered Jul 24 at 12:53

J.G.

13.2k11424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree with you. The series expansion built at $x=0$ is "quite" nice. By the way, $+1$. Cheers.
  – Claude Leibovici
  Jul 24 at 13:15
  

  
  
  
  

add a comment | 

up vote
3
down vote

Although an antiderivative of the integrand is easy to find viz. Claude Leibovici's method, the limits given present a challenge. The integral is $$[frac12sin 2xln^2tanfracx2-2sin xlntanfracx2+2x]^pi/2_0.$$For $n>0$, $lim_xtoinftyx^n e^-x=0$ so $lim_yto 0^+yln^n y=0$. Trigonometric small-angle approximations then reduce the integral to $pi+lim_xto 0(-xln^2fracx2+2xlnfracx2)=pi$.

share|cite|improve this answer

answered Jul 24 at 12:53

J.G.

13.2k11424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree with you. The series expansion built at $x=0$ is "quite" nice. By the way, $+1$. Cheers.
  – Claude Leibovici
  Jul 24 at 13:15
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Although an antiderivative of the integrand is easy to find viz. Claude Leibovici's method, the limits given present a challenge. The integral is $$[frac12sin 2xln^2tanfracx2-2sin xlntanfracx2+2x]^pi/2_0.$$For $n>0$, $lim_xtoinftyx^n e^-x=0$ so $lim_yto 0^+yln^n y=0$. Trigonometric small-angle approximations then reduce the integral to $pi+lim_xto 0(-xln^2fracx2+2xlnfracx2)=pi$.

share|cite|improve this answer

answered Jul 24 at 12:53

J.G.

13.2k11424

Although an antiderivative of the integrand is easy to find viz. Claude Leibovici's method, the limits given present a challenge. The integral is $$[frac12sin 2xln^2tanfracx2-2sin xlntanfracx2+2x]^pi/2_0.$$For $n>0$, $lim_xtoinftyx^n e^-x=0$ so $lim_yto 0^+yln^n y=0$. Trigonometric small-angle approximations then reduce the integral to $pi+lim_xto 0(-xln^2fracx2+2xlnfracx2)=pi$.

share|cite|improve this answer

answered Jul 24 at 12:53

J.G.

13.2k11424

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 12:53

J.G.

13.2k11424

answered Jul 24 at 12:53

J.G.

13.2k11424

answered Jul 24 at 12:53

J.G.

13.2k11424

13.2k11424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree with you. The series expansion built at $x=0$ is "quite" nice. By the way, $+1$. Cheers.
  – Claude Leibovici
  Jul 24 at 13:15
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I agree with you. The series expansion built at $x=0$ is "quite" nice. By the way, $+1$. Cheers.
  – Claude Leibovici
  Jul 24 at 13:15
  

  
  
  
  

I agree with you. The series expansion built at $x=0$ is "quite" nice. By the way, $+1$. Cheers.
– Claude Leibovici
Jul 24 at 13:15

I agree with you. The series expansion built at $x=0$ is "quite" nice. By the way, $+1$. Cheers.
– Claude Leibovici
Jul 24 at 13:15

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%2f2861122%2ftrying-to-int-0-pi-2-cos2x-ln2-tanx-2-dx%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