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Math Help - Trigonometric substitution, integral of 8dx/(4x^2+1)^2 basic error somewhere

  1. #1
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    Trigonometric substitution, integral of 8dx/(4x^2+1)^2 basic error somewhere

    Alright...been at this on and off all day!

    The problem is the integral of 8dx/(4x^2+1)^2.

    I let x=1/2*tanT

    dx=sec^2(T)/2

    I work this problem, going through the integral of 4sec^2T/sec^4T

    Push through to integral of 4*cos^2T, skipping forward I end up with 2t+sin(2t)

    I build my triangle. Side opposite is 2x. Side adjacent is 1. Hypotenuse is 4x^2+1.

    T=arctan(2x). Sin(2t) = side opposite (2x) over hypotenuse ( 4x^2+1 ).

    So my answer is 2arctan(2x)+2x/(4x^2+1)+C However I have a problem. In the second term there, that should be 4x.

    Any idea where I missed it?

    Edit:

    Alright, second problem. Integrating x^2/(1-x^2)^(5/2)dx

    Substitutions are x=sint
    dx=cost(t)

    I work it all the way down to the integration of tan^2tsec^2t dt=tan^3t/3.

    My triangle is side opposite=x, side adjacent= (1-x^2), hypotenuse 1. I substitute in and get (1/3)*(x/(1-x^2))^3 + C

    Once again my side adjacent is messed up. The side adjacent in my answer should be the square root of what I have.

    I think the problems are related.
    Last edited by Wolvenmoon; February 18th 2011 at 07:26 PM.
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  2. #2
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    OK, for starters, you're mixing up your x's and t's...

    \displaystyle \int{\frac{8\,dx}{(4x^2 + 1)^2}}

    Make the substitution \displaystyle x = \frac{1}{2}\tan{t} \implies dx = \frac{1}{2}\sec^2{t}\,dt and the integral becomes

    \displaystyle \int{\frac{8\cdot \frac{1}{2}\sec^2{t}\,dt}{\left[4\left(\frac{1}{2}\tan{t}\right)^2 + 1\right]^2}}

    \displaystyle = \int{\frac{4\sec^2{t}\,dt}{(\tan^2{t} + 1)^2}}

    \displaystyle = \int{\frac{4\sec^2{t}\,dt}{\sec^4{t}}}

    \displaystyle = \int{\frac{4\,dt}{\sec^2{t}}}

    \displaystyle = \int{4\cos^2{t}\,dt}

    \displaystyle = \int{2 + 2\cos{2t}\,dt}

    \displaystyle = 2t + \sin{2t} + C

    \displaystyle = 2t + \frac{2\tan{t}}{1 + \tan^2{t}} + C

    \displaystyle = 2\arctan{2x} + \frac{4x}{1 + (2x)^2} + C

    \displaystyle = 2\arctan{2x} + \frac{4x}{1 + 4x^2} + C.
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    3rd line from the bottom and the second term. What's happening there?

    I also added a second problem to the first post.

    I'll go fix my variables. I've been going a bit too long today. Thanks for your help!
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  4. #4
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    It's a standard trigonometric identity: \displaystyle \sin{2x} \equiv \frac{2\tan{x}}{1 + \tan^2{x}}
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    Man, you're fast. Any help on the second one? ( I had to stop for the night due to neck and back inflammation )
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  6. #6
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    You really should make a new thread for new questions... But anyway...

    \displaystyle \int{\frac{x^2\,dx}{(1 - x^2)^{\frac{5}{2}}}}.

    Make the substitution \displaystyle x = \sin{t} \implies dx = \cos{t}\,dt and the integral becomes

    \displaystyle \int{\frac{\sin^2{t}\cos{t}\,dt}{(1 - \sin^2{t})^{\frac{5}{2}}}}

    \displaystyle = \int{\frac{\sin^2{t}\cos{t}\,dt}{(\cos^2{t})^{\fra  c{5}{2}}}}

    \displaystyle = \int{\frac{\sin^2{t}\cos{t}\,dt}{\cos^5{t}}}

    \displaystyle = \int{\frac{\sin^2{t}\,dt}{\cos^4{t}}}

    \displaystyle = \int{\tan^2{t}\sec^2{t}\,dt}

    Now make the substitution \displaystyle u = \tan{t} \implies du = \sec^2{t}\,dt and the integral becomes

    \displaystyle \int{u^2\,du}

    \displaystyle = \frac{1}{3}u^3 + C

    \displaystyle = \frac{1}{3}\tan^3{t} + C

    \displaystyle = \frac{1}{3}\left(\pm \frac{\sin{t}}{\sqrt{1 - \sin^2{t}}}\right)^3 + C

    \displaystyle = \frac{1}{3}\left(\pm \frac{x}{\sqrt{1 - x^2}}\right)^3 + C

    \displaystyle = C \pm \frac{x^3}{3\sqrt{(1-x^2)^3}}.
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    Thanks! I'm afraid of cluttering up the forums.

    So, what's happening to the tangent between:

    \displaystyle = \frac{1}{3}\tan^3{t} + C

    and

    \displaystyle = \frac{1}{3}\left(\pm \frac{\sin{t}}{\sqrt{1 - \sin^2{t}}}\right)^3 + C ?

    What rule is being used there?

    P.S., what's the easiest way to write Latex for this forum?
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    Another standard identity:

    \displaystyle \tan{x} \equiv \frac{\sin{x}}{\cos{x}} \equiv \frac{\sin{x}}{\pm \sqrt{1 - \sin^2{x}}}...
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    Okay, I knew that one just not that I'd be applying it in that way. My instructor sent us in a bit blind here. I think it might have been unintentional.
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  10. #10
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    Quote Originally Posted by Wolvenmoon View Post
    Okay, I knew that one just not that I'd be applying it in that way. My instructor sent us in a bit blind here. I think it might have been unintentional.
    I always just use trigonometric identites to write whatever I find the integral to be in terms of whatever I had originally substituted, so that I can back substitute.

    List of trigonometric identities - Wikipedia, the free encyclopedia

    This page will be your best friend :P
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  11. #11
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    Alright, on the second problem. I'm supposed to be using a triangle to substitute back.

    So I take x=sint, this means that sint=x/1. So the side opposite on my triangle becomes X and my hypotenuse becomes 1.

    Then I come back to the original problem and I take part of the denominator. From what he told us I got the gist that we ignored the square root at times when substituting back but not at other times.

    What's the rule here? My formula sheet talks about integrands of (a^2-x^2)^(1/2) using a substitution of x=asint, one that's the root of a^2+x^2 that's x=atant, and one of the square root of x^2-a^2 and x=secT

    When do I keep the root and when do I throw it out?
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