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Finding an indefinite integral

  1. Mar 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Calculate the indefinite integral of the function ## \int\frac{3x^3}{\sqrt{1-x^2}}##
    my book gives the answer ##-(2+x^2)\sqrt{1-x^2}+C##
    2. Relevant equations


    3. The attempt at a solution
    So I started trying to calculate this indefinite integral by using a substitution ##x=\sin(\theta)##

    so ##dx=\cos{\theta}d\theta##

    then I substituted this term into my integral and got

    ##3\int\frac{\sin^3\theta*\cos\theta*d\theta}{\cos\theta}##

    Then I canceled out the ## \cos\theta## and wrote ##\sin^2\theta=1-\cos^2\theta##

    ##3\int\sin\theta*(1-\cos^2\theta*)d\theta##

    here I substituted another new variable ##\cos\theta=t\Rightarrow dt=-\sin\theta d\theta## and I substituted that into my integral and got

    ##3\int(1-t^2)dt##

    which is equal to ##-3t+t^3+C##
    and when I inserted all my renamed variables back into the equation I got
    ## \int\frac{3x^3}{\sqrt{1-x^2}}=-3\cos{(\sin^{-1}x)}+(\cos({\sin^{-1}x}))^3##
    Which is nowhere near the answer that is given in my textbook so I don't even know where I made my mistake
    Thanks
     
  2. jcsd
  3. Mar 24, 2017 #2

    fresh_42

    Staff: Mentor

    What do you get, if you differentiate both solutions?
    How do you know, that they aren't the same?
     
    Last edited: Mar 24, 2017
  4. Mar 24, 2017 #3
    Try putting x=sinθ again and simplifying. Note that you seem to have made a sign error in your last integration.
     
  5. Mar 24, 2017 #4

    fresh_42

    Staff: Mentor

    No sign error in the integration result, only in the line before. But that's not the point. You simply have two expressions of possibly the same result.
     
  6. Mar 24, 2017 #5
    Thank you. I completely forgot that I could check my answer by just deriving what I got and checking if it is the same as the equation I started with
     
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