Dgdrxrt

Keep going mode for require-package

Who can trigger ship-wide alerts in Star Trek?

What to do with post with dry rot?

Losing the Initialization Vector in Cipher Block Chaining

Mortgage adviser recommends a longer term than necessary combined with overpayments

Single author papers against my advisor's will?

How does modal jazz use chord progressions?

What is the largest species of polychaete?

Limit for e and 1/e

Can a monk deflect thrown melee weapons?

Stars Make Stars

If I can make up priors, why can't I make up posteriors?

How do you clear the ApexPages.getMessages() collection in a test?

When is phishing education going too far?

Cold is to Refrigerator as warm is to?

When communicating altitude with a '9' in it, should it be pronounced "nine hundred" or "niner hundred"?

What LEGO pieces have "real-world" functionality?

Array/tabular for long multiplication

Statistical model of ligand substitution

Estimated State payment too big --> money back; + 2018 Tax Reform

I'm having difficulty getting my players to do stuff in a sandbox campaign

Can the prologue be the backstory of your main character?

Fishing simulator

Complexity of many constant time steps with occasional logarithmic steps

How to evaluate $int_0^frac1sqrt2fracdx(1+x^2)sqrt1-x^2$

5

$begingroup$

$int_0^frac1sqrt2fracdx(1+x^2)sqrt1-x^2$

I tried using substitute, $x=sintheta$

But I ended up with $int_0^fracpi4fracdtheta1+sin^2theta$

Is my substitution correct? Please give me a hint to work this out! Thank you very much.

calculus integration

share|cite|improve this question

edited Apr 1 at 2:39

clathratus

5,1241439

asked Mar 31 at 18:34

emilemil

465410

$endgroup$

add a comment | 

5

$begingroup$

$int_0^frac1sqrt2fracdx(1+x^2)sqrt1-x^2$

I tried using substitute, $x=sintheta$

But I ended up with $int_0^fracpi4fracdtheta1+sin^2theta$

Is my substitution correct? Please give me a hint to work this out! Thank you very much.

calculus integration

share|cite|improve this question

edited Apr 1 at 2:39

clathratus

5,1241439

asked Mar 31 at 18:34

emilemil

465410

$endgroup$

add a comment | 

$begingroup$

$int_0^frac1sqrt2fracdx(1+x^2)sqrt1-x^2$

I tried using substitute, $x=sintheta$

But I ended up with $int_0^fracpi4fracdtheta1+sin^2theta$

Is my substitution correct? Please give me a hint to work this out! Thank you very much.

calculus integration

share|cite|improve this question

edited Apr 1 at 2:39

clathratus

5,1241439

asked Mar 31 at 18:34

emilemil

465410

$endgroup$

share|cite|improve this question

edited Apr 1 at 2:39

clathratus

5,1241439

asked Mar 31 at 18:34

emilemil

465410

share|cite|improve this question

edited Apr 1 at 2:39

clathratus

5,1241439

asked Mar 31 at 18:34

emilemil

465410

edited Apr 1 at 2:39

clathratus

5,1241439

edited Apr 1 at 2:39

clathratus

5,1241439

asked Mar 31 at 18:34

emilemil

465410

asked Mar 31 at 18:34

emilemil

465410

3 Answers
3

active

oldest

votes

6

$begingroup$

hint:
$$frac11+sin^2 x=fracsec^2 x1+2tan^2 x$$

share|cite|improve this answer

answered Mar 31 at 18:41

E.H.EE.H.E

16.4k11969

$endgroup$

add a comment | 

3

$begingroup$

We see that
$$frac11+sin(x)^2=fracsec(x)^21+2tan(x)^2$$
as was hinted by @E.H.E. We then preform the sub $u=tan(x)$ to get
$$int_0^1 fracdu1+2u^2$$
Then we may in fact compute the integral
$$I(x;a,b,c)=intfrac dxax^2+bx+c=intfracdxa(x+fracb2a)^2+g$$
Here $g=c-fracb^24a$. If we assume that $4ac>b^2$, then we may make the substitution $x+fracb2a=sqrtfracgatan u$ which gives
$$I(x;a,b,c)=sqrtfracgaintfracsec^2u, dugtan^2u+g$$
$$I(x;a,b,c)=fracusqrtag$$
$$I(x;a,b,c)=frac2sqrt4ac-b^2arctanfrac2ax+bsqrt4ac-b^2+C$$
And by noting that your integral is given by $I(1;2,0,1)-I(0;2,0,1)$ we have your integral at the value
$$fracarctansqrt2sqrt2approx 0.675510858856$$

share|cite|improve this answer

answered Apr 1 at 2:01

clathratusclathratus

5,1241439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why was the hint @EHE gave not enough? Why do the calculation for the OP?
  $endgroup$
  – JavaMan
  Apr 1 at 2:43
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @JavaMan Because it was an opportunity to demonstrate something more than the OP asked, namely $I(x;a,b,c)$. If I were a beginning calc. student I would find it very helpful, even interesting, to see such an answer. IMO, a good answer helps you in more ways than $1$, and helps you connect concepts/ideas (and in this case integrals).
  $endgroup$
  – clathratus
  Apr 1 at 3:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course I found it very useful. Using a second substitution is somewhat challenging task. Thank you for pointing it out!
  $endgroup$
  – emil
  Apr 1 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @emil You are very welcome :)
  $endgroup$
  – clathratus
  Apr 1 at 15:45
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint

Divide numerator & denominator by $sin^2theta$

and set $cottheta =u$

Or divide numerator & denominator by $cos^2theta$

and set $tantheta=v$

share|cite|improve this answer

answered Mar 31 at 18:39

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

add a comment | 

Your Answer

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
);



);

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

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3169743%2fhow-to-evaluate-int-0-frac1-sqrt2-fracdx1x2-sqrt1-x2%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

hint:
$$frac11+sin^2 x=fracsec^2 x1+2tan^2 x$$

share|cite|improve this answer

answered Mar 31 at 18:41

E.H.EE.H.E

16.4k11969

$endgroup$

add a comment | 

6

$begingroup$

hint:
$$frac11+sin^2 x=fracsec^2 x1+2tan^2 x$$

share|cite|improve this answer

answered Mar 31 at 18:41

E.H.EE.H.E

16.4k11969

$endgroup$

add a comment | 

$begingroup$

hint:
$$frac11+sin^2 x=fracsec^2 x1+2tan^2 x$$

share|cite|improve this answer

answered Mar 31 at 18:41

E.H.EE.H.E

16.4k11969

$endgroup$

share|cite|improve this answer

answered Mar 31 at 18:41

E.H.EE.H.E

16.4k11969

answered Mar 31 at 18:41

E.H.EE.H.E

16.4k11969

answered Mar 31 at 18:41

E.H.EE.H.E

16.4k11969

3

$begingroup$

We see that
$$frac11+sin(x)^2=fracsec(x)^21+2tan(x)^2$$
as was hinted by @E.H.E. We then preform the sub $u=tan(x)$ to get
$$int_0^1 fracdu1+2u^2$$
Then we may in fact compute the integral
$$I(x;a,b,c)=intfrac dxax^2+bx+c=intfracdxa(x+fracb2a)^2+g$$
Here $g=c-fracb^24a$. If we assume that $4ac>b^2$, then we may make the substitution $x+fracb2a=sqrtfracgatan u$ which gives
$$I(x;a,b,c)=sqrtfracgaintfracsec^2u, dugtan^2u+g$$
$$I(x;a,b,c)=fracusqrtag$$
$$I(x;a,b,c)=frac2sqrt4ac-b^2arctanfrac2ax+bsqrt4ac-b^2+C$$
And by noting that your integral is given by $I(1;2,0,1)-I(0;2,0,1)$ we have your integral at the value
$$fracarctansqrt2sqrt2approx 0.675510858856$$

share|cite|improve this answer

answered Apr 1 at 2:01

clathratusclathratus

5,1241439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why was the hint @EHE gave not enough? Why do the calculation for the OP?
  $endgroup$
  – JavaMan
  Apr 1 at 2:43
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @JavaMan Because it was an opportunity to demonstrate something more than the OP asked, namely $I(x;a,b,c)$. If I were a beginning calc. student I would find it very helpful, even interesting, to see such an answer. IMO, a good answer helps you in more ways than $1$, and helps you connect concepts/ideas (and in this case integrals).
  $endgroup$
  – clathratus
  Apr 1 at 3:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course I found it very useful. Using a second substitution is somewhat challenging task. Thank you for pointing it out!
  $endgroup$
  – emil
  Apr 1 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @emil You are very welcome :)
  $endgroup$
  – clathratus
  Apr 1 at 15:45
  

  
  
  
  

add a comment | 

3

$begingroup$

We see that
$$frac11+sin(x)^2=fracsec(x)^21+2tan(x)^2$$
as was hinted by @E.H.E. We then preform the sub $u=tan(x)$ to get
$$int_0^1 fracdu1+2u^2$$
Then we may in fact compute the integral
$$I(x;a,b,c)=intfrac dxax^2+bx+c=intfracdxa(x+fracb2a)^2+g$$
Here $g=c-fracb^24a$. If we assume that $4ac>b^2$, then we may make the substitution $x+fracb2a=sqrtfracgatan u$ which gives
$$I(x;a,b,c)=sqrtfracgaintfracsec^2u, dugtan^2u+g$$
$$I(x;a,b,c)=fracusqrtag$$
$$I(x;a,b,c)=frac2sqrt4ac-b^2arctanfrac2ax+bsqrt4ac-b^2+C$$
And by noting that your integral is given by $I(1;2,0,1)-I(0;2,0,1)$ we have your integral at the value
$$fracarctansqrt2sqrt2approx 0.675510858856$$

share|cite|improve this answer

answered Apr 1 at 2:01

clathratusclathratus

5,1241439

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why was the hint @EHE gave not enough? Why do the calculation for the OP?
  $endgroup$
  – JavaMan
  Apr 1 at 2:43
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  @JavaMan Because it was an opportunity to demonstrate something more than the OP asked, namely $I(x;a,b,c)$. If I were a beginning calc. student I would find it very helpful, even interesting, to see such an answer. IMO, a good answer helps you in more ways than $1$, and helps you connect concepts/ideas (and in this case integrals).
  $endgroup$
  – clathratus
  Apr 1 at 3:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course I found it very useful. Using a second substitution is somewhat challenging task. Thank you for pointing it out!
  $endgroup$
  – emil
  Apr 1 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @emil You are very welcome :)
  $endgroup$
  – clathratus
  Apr 1 at 15:45
  

  
  
  
  

add a comment | 

$begingroup$

We see that
$$frac11+sin(x)^2=fracsec(x)^21+2tan(x)^2$$
as was hinted by @E.H.E. We then preform the sub $u=tan(x)$ to get
$$int_0^1 fracdu1+2u^2$$
Then we may in fact compute the integral
$$I(x;a,b,c)=intfrac dxax^2+bx+c=intfracdxa(x+fracb2a)^2+g$$
Here $g=c-fracb^24a$. If we assume that $4ac>b^2$, then we may make the substitution $x+fracb2a=sqrtfracgatan u$ which gives
$$I(x;a,b,c)=sqrtfracgaintfracsec^2u, dugtan^2u+g$$
$$I(x;a,b,c)=fracusqrtag$$
$$I(x;a,b,c)=frac2sqrt4ac-b^2arctanfrac2ax+bsqrt4ac-b^2+C$$
And by noting that your integral is given by $I(1;2,0,1)-I(0;2,0,1)$ we have your integral at the value
$$fracarctansqrt2sqrt2approx 0.675510858856$$

share|cite|improve this answer

answered Apr 1 at 2:01

clathratusclathratus

5,1241439

$endgroup$

share|cite|improve this answer

answered Apr 1 at 2:01

clathratusclathratus

5,1241439

answered Apr 1 at 2:01

clathratusclathratus

5,1241439

answered Apr 1 at 2:01

clathratusclathratus

5,1241439

2

$begingroup$

Hint

Divide numerator & denominator by $sin^2theta$

and set $cottheta =u$

Or divide numerator & denominator by $cos^2theta$

and set $tantheta=v$

share|cite|improve this answer

answered Mar 31 at 18:39

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

add a comment | 

2

$begingroup$

Hint

Divide numerator & denominator by $sin^2theta$

and set $cottheta =u$

Or divide numerator & denominator by $cos^2theta$

and set $tantheta=v$

share|cite|improve this answer

answered Mar 31 at 18:39

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

add a comment | 

$begingroup$

Hint

Divide numerator & denominator by $sin^2theta$

and set $cottheta =u$

Or divide numerator & denominator by $cos^2theta$

and set $tantheta=v$

share|cite|improve this answer

answered Mar 31 at 18:39

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

share|cite|improve this answer

answered Mar 31 at 18:39

lab bhattacharjeelab bhattacharjee

229k15159279

answered Mar 31 at 18:39

lab bhattacharjeelab bhattacharjee

229k15159279

answered Mar 31 at 18:39

lab bhattacharjeelab bhattacharjee

229k15159279