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How to calculate $int sqrt1+frac4x^2(1-2x)^2dx = $?

0

$begingroup$

I want to find an arc length of y = $ln(1-x^2)$ on the $xin<0,frac12>$ interval?

$y' = frac-2x1-x^2$

So:

$$int sqrt1+frac4x^2(1-2x)^2dx = $$

$$= int sqrtfrac(1-2x)^2(1-2x)^2+frac4x^2(1-2x)^2dx = $$

$$= int sqrtfrac8x^2-4x+1(1-2x)^2dx = $$

$$= int fracsqrt8x^2-4x+11-2xdx = $$

$$= int frac8x^2-4x+1(1-2x)sqrt8x^2-4x+1dx $$

And now I have tried from one of Euler's substitution, which was total nightmare and algebraically impossible for me.

Is there any simpler way to calculate this integral?

calculus definite-integrals indefinite-integrals arc-length

share|cite|improve this question

asked Apr 8 at 20:39

wenoweno

44311

$endgroup$

add a comment | 

0

$begingroup$

I want to find an arc length of y = $ln(1-x^2)$ on the $xin<0,frac12>$ interval?

$y' = frac-2x1-x^2$

So:

$$int sqrt1+frac4x^2(1-2x)^2dx = $$

$$= int sqrtfrac(1-2x)^2(1-2x)^2+frac4x^2(1-2x)^2dx = $$

$$= int sqrtfrac8x^2-4x+1(1-2x)^2dx = $$

$$= int fracsqrt8x^2-4x+11-2xdx = $$

$$= int frac8x^2-4x+1(1-2x)sqrt8x^2-4x+1dx $$

And now I have tried from one of Euler's substitution, which was total nightmare and algebraically impossible for me.

Is there any simpler way to calculate this integral?

calculus definite-integrals indefinite-integrals arc-length

share|cite|improve this question

asked Apr 8 at 20:39

wenoweno

44311

$endgroup$

add a comment | 

$begingroup$

I want to find an arc length of y = $ln(1-x^2)$ on the $xin<0,frac12>$ interval?

$y' = frac-2x1-x^2$

So:

$$int sqrt1+frac4x^2(1-2x)^2dx = $$

$$= int sqrtfrac(1-2x)^2(1-2x)^2+frac4x^2(1-2x)^2dx = $$

$$= int sqrtfrac8x^2-4x+1(1-2x)^2dx = $$

$$= int fracsqrt8x^2-4x+11-2xdx = $$

$$= int frac8x^2-4x+1(1-2x)sqrt8x^2-4x+1dx $$

And now I have tried from one of Euler's substitution, which was total nightmare and algebraically impossible for me.

Is there any simpler way to calculate this integral?

calculus definite-integrals indefinite-integrals arc-length

share|cite|improve this question

asked Apr 8 at 20:39

wenoweno

44311

$endgroup$

share|cite|improve this question

asked Apr 8 at 20:39

wenoweno

44311

share|cite|improve this question

asked Apr 8 at 20:39

wenoweno

44311

asked Apr 8 at 20:39

wenoweno

44311

asked Apr 8 at 20:39

wenoweno

44311

2 Answers
2

active

oldest

votes

3

$begingroup$

In the first equation after "So:", in the numerator, you should have $(1-x^2)^2$ instead of $(1-2x)^2$. Then your function to integrate becomes $$sqrt1+frac4x^2(1-x^2)^2=sqrtfrac(1+x^2)^2(1-x^2)^2$$
You can now take the square root, since $0lt 1-x^2$ and $0lt 1+x^2$

share|cite|improve this answer

edited Apr 8 at 20:49

answered Apr 8 at 20:43

AndreiAndrei

13.8k21230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, that makes it trivial. I'll accept in 4 minutes.
  $endgroup$
  – weno
  Apr 8 at 20:49
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint:

Use the substitution
$$frac2x1-2x=sinh t,qquad frac2,mathrm d x(1-2x)^2==cosh t,mathrm dt.$$

share|cite|improve this answer

answered Apr 8 at 20:52

BernardBernard

124k741117

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add a comment | 

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3

$begingroup$

In the first equation after "So:", in the numerator, you should have $(1-x^2)^2$ instead of $(1-2x)^2$. Then your function to integrate becomes $$sqrt1+frac4x^2(1-x^2)^2=sqrtfrac(1+x^2)^2(1-x^2)^2$$
You can now take the square root, since $0lt 1-x^2$ and $0lt 1+x^2$

share|cite|improve this answer

edited Apr 8 at 20:49

answered Apr 8 at 20:43

AndreiAndrei

13.8k21230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, that makes it trivial. I'll accept in 4 minutes.
  $endgroup$
  – weno
  Apr 8 at 20:49
  

  
  
  
  

add a comment | 

3

$begingroup$

In the first equation after "So:", in the numerator, you should have $(1-x^2)^2$ instead of $(1-2x)^2$. Then your function to integrate becomes $$sqrt1+frac4x^2(1-x^2)^2=sqrtfrac(1+x^2)^2(1-x^2)^2$$
You can now take the square root, since $0lt 1-x^2$ and $0lt 1+x^2$

share|cite|improve this answer

edited Apr 8 at 20:49

answered Apr 8 at 20:43

AndreiAndrei

13.8k21230

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, that makes it trivial. I'll accept in 4 minutes.
  $endgroup$
  – weno
  Apr 8 at 20:49
  

  
  
  
  

add a comment | 

$begingroup$

In the first equation after "So:", in the numerator, you should have $(1-x^2)^2$ instead of $(1-2x)^2$. Then your function to integrate becomes $$sqrt1+frac4x^2(1-x^2)^2=sqrtfrac(1+x^2)^2(1-x^2)^2$$
You can now take the square root, since $0lt 1-x^2$ and $0lt 1+x^2$

share|cite|improve this answer

edited Apr 8 at 20:49

answered Apr 8 at 20:43

AndreiAndrei

13.8k21230

$endgroup$

share|cite|improve this answer

edited Apr 8 at 20:49

answered Apr 8 at 20:43

AndreiAndrei

13.8k21230

answered Apr 8 at 20:43

AndreiAndrei

13.8k21230

answered Apr 8 at 20:43

AndreiAndrei

13.8k21230

0

$begingroup$

Hint:

Use the substitution
$$frac2x1-2x=sinh t,qquad frac2,mathrm d x(1-2x)^2==cosh t,mathrm dt.$$

share|cite|improve this answer

answered Apr 8 at 20:52

BernardBernard

124k741117

$endgroup$

add a comment | 

0

$begingroup$

Hint:

Use the substitution
$$frac2x1-2x=sinh t,qquad frac2,mathrm d x(1-2x)^2==cosh t,mathrm dt.$$

share|cite|improve this answer

answered Apr 8 at 20:52

BernardBernard

124k741117

$endgroup$

add a comment | 

$begingroup$

Hint:

Use the substitution
$$frac2x1-2x=sinh t,qquad frac2,mathrm d x(1-2x)^2==cosh t,mathrm dt.$$

share|cite|improve this answer

answered Apr 8 at 20:52

BernardBernard

124k741117

$endgroup$

share|cite|improve this answer

answered Apr 8 at 20:52

BernardBernard

124k741117

answered Apr 8 at 20:52

BernardBernard

124k741117

answered Apr 8 at 20:52

BernardBernard

124k741117