∫ x² / √(1 - 4x²) dx : What am I doing wrong?

Question

∫ x² / √(1 - 4x²) dx = {x√(1-4x²) + √(1-4x²)³ / 6x} / (-4)

According to GraphCalc it's wrong, and my graphing calculator can't give me the answer. What am I doing wrong?

∫ x² / √(1 - 4x²) dx

I want to integrate by parts

let u = x²   →   du = 2x dx

dv = (1 - 4x²)^(-1/2) dx   →   v = √(1-4x²) / -4x

∫ u dv = uv - ∫ v du

∫ x²/√(1 - 4x²) dx = (x²) * [√(1-4x²)]/(-4x) - ∫ [√(1-4x²)]/(-4x) 2x dx

∫ x²/√(1 - 4x²) dx = (x) * [√(1-4x²)]/(-4) + ½ ∫ √(1-4x²) dx

∫ x²/√(1 - 4x²) dx = (x) * [√(1-4x²)]/(-4) + ½ [(1-4x²)^(3/2)]/(-12x)

∫ x²/√(1 - 4x²) dx = (x) * [√(1-4x²)]/(-4) - [(1-4x²)^(3/2)]/(24x)

∫ x²/√(1 - 4x²) dx = {x√(1-4x²) + √(1-4x²)³ / 6x} / (-4)

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It becomes worse if

u = 1/√(1 - 4x²) , and

dv = x² dx

why have I made a mistake? I try to follow what I am taught, yet I still err

Answer 1

Ferroci is right: you found out v uncorrectly (finding v is nothing but a mere integration); in fact

v = ∫ 1/√(1 - 4x²) dx = ∫ 1/√[1 - (2x)²] dx = (1/2) ∫ 2dx /√[1 - (2x)²] =

(1/2) ∫ d(2x) /√[1 - (2x)²] = (1/2) arcsin(2x)

nevertheless, if by parts method is choosen, without resorting to trig substitution,

I would rather set u and dv as follows:

∫ x² / √(1 - 4x²) dx =

rewrite it as:

∫ x [x/ √(1 - 4x²)] dx =

divide and multiply by - 4 and rearrange the integrand as:

(-1/4) ∫ x [(- 4x) / √(1 - 4x²)] dx =

let:

x = u → dx = du

[(- 4x) / √(1 - 4x²)] dx = dv → (integrating) →

(1/2)(2)[(- 4x) / √(1 - 4x²)] dx = dv →

(1/2)[(- 8x) / √(1 - 4x²)] dx = dv →

(1/2)[d(1 - 4x²)] / √(1 - 4x²) = dv →

(1/2)[(1 - 4x²)^(-1/2)] d(1 - 4x²) = dv →

(1/2){(1 - 4x²)^[(-1/2) +1]} /[(-1/2) +1] = v →

(1/2)[(1 - 4x²)^(1/2)] /(1/2) = v →

(1/2)(2) √(1 - 4x²) = v →

√(1 - 4x²) = v

then let's integrate by parts:

(-1/4) ∫ x [(- 4x) / √(1 - 4x²)] dx = (-1/4) [x √(1 - 4x²) - ∫ √(1 - 4x²) dx] →

(-1/4) ∫ x [(- 4x) / √(1 - 4x²)] dx = (-1/4)x √(1 - 4x²) + (1/4) ∫ √(1 - 4x²) dx →

multiply and divide the remaining integral by √(1 - 4x²):

(-1/4) ∫ x [(- 4x) / √(1 - 4x²)] dx = (-1/4)x √(1 - 4x²) +

(1/4) ∫ {[√(1 - 4x²)√(1 - 4x²)]/√(1 - 4x²)} dx →

(-1/4) ∫ x [(- 4x) / √(1 - 4x²)] dx = (-1/4)x √(1 - 4x²) +

(1/4) ∫ [(1 - 4x²) /√(1 - 4x²) dx →

split it into:

(-1/4) ∫ x [(- 4x) / √(1 - 4x²)] dx = (-1/4)x √(1 - 4x²) +

(1/4) ∫ {[1/√(1 - 4x²)] - [4x² /√(1 - 4x²)]} dx →

(-1/4) ∫ x [(- 4x) / √(1 - 4x²)] dx = (-1/4)x √(1 - 4x²) +

(1/4) ∫ [1/√(1 - 4x²)] dx - (1/4) ∫ [4x² /√(1 - 4x²)] dx →

(-1/4) ∫ x [(- 4x) / √(1 - 4x²)] dx = (-1/4)x √(1 - 4x²) +

(1/4) ∫ [1/√(1 - 4x²)] dx - ∫ [x² /√(1 - 4x²)] dx →

now rewrite the integral at left side into its initial form:

∫ [x² / √(1 - 4x²)] dx = (-1/4)x √(1 - 4x²) + (1/4) ∫ [1/√(1 - 4x²)] dx -

∫ [x² /√(1 - 4x²)] dx →

now note that the unknown integral appears at both sides with opposite signs;

thus, let's shift it to left side, yielding:

∫ [x² / √(1 - 4x²)] dx + ∫ [x² /√(1 - 4x²)] dx = (-1/4)x √(1 - 4x²) +

(1/4) ∫ [1/√(1 - 4x²)] dx  Ã¢Â†Â’

2 ∫ [x² /√(1 - 4x²)] dx = (-1/4)x √(1 - 4x²) + (1/4) ∫ [1/√(1 - 4x²)] dx  Ã¢Â†Â’

∫ [x² /√(1 - 4x²)] dx = (1/2){(-1/4)x √(1 - 4x²) + (1/4) ∫ [1/√(1 - 4x²)] dx}  Ã¢Â†Â’

∫ [x² /√(1 - 4x²)] dx = (-1/8)x √(1 - 4x²) + (1/8) ∫ [1/√(1 - 4x²)] dx  Ã¢Â†Â’

as for the remaining integral, rewrite the square as:

∫ [x² /√(1 - 4x²)] dx = (-1/8)x √(1 - 4x²) + (1/8) ∫ {1/√[1 - (2x)²]} dx  Ã¢Â†Â’

divide and multiply by 2, in order to turn the numerator into the derivative

of (2x) (i.e. 2):

∫ [x² /√(1 - 4x²)] dx = (-1/8)x √(1 - 4x²) + (1/8)(1/2) ∫ 2 dx /√[1 - (2x)²]   →

∫ [x² /√(1 - 4x²)] dx = (-1/8)x √(1 - 4x²) + (1/16) ∫ d(2x) /√[1 - (2x)²] →

finally:

∫ [x² /√(1 - 4x²)] dx = (-1/8)x √(1 - 4x²) + (1/16) arcsin(2x) + C

I hope it helps...

Bye!!

Answer 2

Have you learned about using trig substitutions yet? Whenever you have an integral with something that looks like √(a-bx^2) in the denominator, you should try a substitution of x = √(a/b) cos(u). Then it just becomes a matter of simplifying things using trig identities.

So for your integral, try the substition x = 1/2 sin(u). Then you have dx = 1/2 cos(u). Plugging all this into your integral gives you the new integral

∫ (1/4 sin^2 (u))(1/2 cos(u)) /  ÃƒÂ¢Ã‚ˆÂš(1 - 4(1/4 sin^2 (u))) du

1/8 ∫  sin^2(u)cos(u) / √(1 - sin^2 (u)) du

Now remember that sin^2 u + cos^2 u = 1, so

√(1 - sin^2 (u)) = √(cos^2 (u)) = cos(u)

So your integral becomes

1/8 ∫ sin^2(u) du

Assuming you know what the integral of sin^2 (u) du is, you get

1/16 ( u - sin(u)cos(u) ) + C

You're not quite done though, since your original question was in terms of x, and this answer is in terms of u. We need a way to go back to u.

Remember our substitution, x = 1/2 sin(u). Solving for u gives

u = arcsin(2x)

Also, we can solve that equation for sin(u) to find that sin(u) = 2x.

Now we just need to know what cos(u) is. But that's easy since sin^2(u) + cos^2(u) = 1. Solve this equation to find that

cos(u) = √(1 - sin^2(u)) = √(1 - 4x^2)

Now we can plug these three in for u, sin(u), and cos(u) into what we got for our integral to get

1/16 ( arcsin(2x) - 2x√(1 - 4x^2) ) + C

Hope this all makes sense.

Answer 3

your mistake is when you find v.

v should be (1/2)arcsin(2x).

I'd prefer to use trig substitution instead of integration by parts because you'll use, i believe, 2 more times of integration by parts again, which is gonna make it long and messy.

∫x² dx / √(1-4x²)

http://i34.tinypic.com/ngrmet.jpg

from the diagram, we have:

sinu = 2x/1

sinu = 2x

x = (1/2)sinu

dx = (1/2)cosu du

cosu = √(1-4x²)/1

√(1-4x²) = cosu

now substitute:

∫[(1/2)sinu]² * (1/2)cosu du / cosu

∫(1/8) sin²u du

trig identity:

sin²u = 1/2 - (1/2)cos(2u)

subsitute:

∫(1/8) [1/2 - (1/2)cos(2u)] du

∫1/16 - (1/16) cos(2u) du

(1/16)u - (1/16) (1/2) sin(2u) + C

trig identity:

sin(2u) = 2cosu sinu

(1/16)u - (1/16)(1/2) (2sinu cosu) + C

(1/16)u - (1/16) sinu cosu + C

re-substitute:

(1/16)arcsin(2x) - (1/16) * 2x * √(1-4x²) + C

(1/16)arcsin(2x) - (1/8)x √(1-4x²) + C

hope it helps!

Answer 4

I hate math, so my answer will prove to be pointless, but I just wanted to say: Cool that you could do all the symbols for this equation online!

Answer 5

I would suggest you to start with partial fractions ... by parts would not be a good idea.

Do long division first and then partial fractions...

hope this helps ::)

* Editted*

Oops! sorry..I didnt see the root sign :)