Results 1 to 5 of 5

Math Help - Integral 16

  1. #1
    Super Member
    Joined
    Jun 2008
    Posts
    829

    Integral 16

    \int \frac{x^2dx}{\sqrt{(4-x^2)^5}}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    I learnt this from Krizalid aka the Inte-Killer. Try the substitution u = 1/x.

    I wouldn't be surprised if Kriz has another clever substitution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Trigonometric substitution, x=2 \sin t is a way.

    Another way is below.

    I assume that x>0,

    \frac{x^2}{(4-x^2)^{\frac{5}{2}}} = \frac{x^2}{x^5\left ( \frac{4}{x^2}-1 \right )^{\frac{5}{2}}} = \frac{1}{x^3 \left ( \frac{4}{x^2}-1 \right )^{\frac{5}{2}}}

    \int \frac{dx}{x^3 \left ( \frac{4}{x^2}-1 \right )^{\frac{5}{2}}}

    Let u = \sqrt{\frac{4}{x^2}-1}, then u^2 = \frac{4}{x^2}-1 and 2 u~du = -\frac{8}{x^3}~dx

    Applying the substitution gives us -\frac{1}{4}\int u^{-4}~du= \frac{1}{12u^3}+C = \frac{1}{12 \left ( \frac{4}{x^2}-1 \right )^{\frac{3}{2}}}+C

    For x < 0, the integral is -\frac{1}{12 \left ( \frac{4}{x^2}-1 \right )^{\frac{3}{2}}}+C.

    So we can say that, for any x, the integral is \frac{\sqrt{x^2}}{12x \left ( \frac{4}{x^2}-1 \right )^{\frac{3}{2}}}+C
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2008
    Posts
    792
    I thought I'd elaborate more on my post.

    \int \frac{x^2}{\sqrt{(4-x^2)^5}}dx

    Let  u = \frac{1}{x}

    du = -\frac{1}{x^2} dx

    - x^2 du = dx

    Replace:
    \int \frac{1}{u^4\sqrt{(4-\frac{1}{u^2})^5}}du

    \int \frac{1}{u^4\sqrt{(\frac{4u^2-1}{u^2})^5}}du

    \int \frac{u^5}{u^4\sqrt{(4u^2-1)^5}}du

    \int \frac{u}{\sqrt{(4u^2-1)^5}}du

    Now it became a matter of simple substition ( z = 4u^2 - 1).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,904
    Thanks
    765
    Hello, Apprentice123!

    Another approach . . .


    \int \frac{x^2\,dx}{\sqrt{(4-x^2)^5}}\;=\;\int\frac{x^2\,dx}{(4-x)^{\frac{5}{2}}}

    Let: x = 2\sin\theta \quad\Rightarrow\quad dx = 2\cos\theta\,d\theta

    Substitute: . \int\frac{(2\sin\theta)^2(2\cos\theta\,d\theta)}{(  4\cos^2\!\theta)^{\frac{5}{2}}} \;=\;\int\frac{4\sin^2\!\theta\cdot2\cos\theta\,d\  theta}{32\cos^5\!\theta} \;= \;\frac{1}{4}\int\frac{\sin^2\!\theta}{\cos^4\!\th  eta}\,d\theta

    . . = \;\frac{1}{4}\int \frac{\sin^2\!\theta}{\cos^2\!\theta}\cdot\frac{1}  {\cos^2\!\theta}\,d\theta \;=\;\frac{1}{4}\int\tan^2\!\theta(\sec^2\!\theta\  ,d\theta)



    Let u = \tan\theta \quad\Rightarrow\quad du = \sec^2\!\theta\,d\theta

    Substitute: . \frac{1}{4}\int u^2\,du \;=\;\frac{1}{12}u^3 + C


    Back-substitute: . \frac{1}{12}\tan^3\!\theta + C


    We had: . \sin\theta = \frac{x}{2}\quad\Rightarrow\quad \tan\theta = \frac{x}{\sqrt{4-x^2}}


    Answer: . \frac{1}{12}\left(\frac{x}{\sqrt{4-x^2}}\right)^3 + C \;=\;\frac{x^3}{12(4-x^2)^{\frac{3}{2}}} + C

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 08:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 03:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 02:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 08:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 07:37 AM

Search Tags


/mathhelpforum @mathhelpforum