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Math Help - integration - urgent help please

  1. #1
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    Question integration - urgent help please

    2 times the integral from 0 to 1 of (2t-3t^2) *sqrt(1-9t^2) dt.

    Thank you very much.
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    Quote Originally Posted by kittycat View Post
    2 times the integral from 0 to 1 of (2t-3t^2) *sqrt(1-9t^2) dt.

    Thank you very much.
    \int_0^1(2t - 3t^2)\sqrt{1 - 9t^2}~dt

    First let's get that radicand simplified a bit:
    Let x = 3t \implies dx = 3dt
    \int_0^1(2t - 3t^2)\sqrt{1 - 9t^2}~dt = \int_0^3  \left [ 2 \left ( \frac{x}{3} \right ) - 3 \left ( \frac{x}{3} \right ) ^2 \right ] \sqrt{1 - x^2} ~ \frac{dx}{3}

    = \frac{1}{9} \int_0^3 (2x - x^2) \sqrt{1 - x^2} ~ dx

    = \frac{1}{9} \int_0^3 2x \sqrt{1 - x^2} ~ dx - \frac{1}{9} \int_0^3 x^2 \sqrt{1 - x^2} ~ dx

    The first integral may be done using a simple substitution. I'll let you deal with that. For the second integral:
    \int_0^3 x^2 \sqrt{1 - x^2} ~ dx

    Take a moment and look at the graph I've posted below. You can easily see that this integrand does not exist beyond x = 1. So
    \int_0^3 x^2 \sqrt{1 - x^2} ~ dx = \int_0^1 x^2 \sqrt{1 - x^2} ~ dx <-- We need to do this, else the substitution below is illegal.

    Now Let x = sin(\theta) \implies dx = cos( \theta) d \theta
    \int_0^1 x^2 \sqrt{1 - x^2} ~ dx = \int_0^{\pi/2}sin^2(\theta) \sqrt{1 - sin^2(\theta) } ~cos(\theta) ~ d \theta

    = \int_0^{\pi/2}sin^2(\theta) ~ cos^2(\theta) ~ d \theta

    I'll let you do this one also and then collect all the pieces.

    -Dan
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    Hello, kittycat!

    A variation on Dan's solution . . .


    2 \int^1_0\,(2t-3t^2)\cdot\sqrt{1-9t^2}\,dt
    Let: 3t \:= \:\sin\theta\quad\Rightarrow\quad dt \:=\:\frac{1}{3}\cos\theta\,d\theta
    . . and: . \sqrt{1-9t^2} \:=\:\cos\theta

    Substitute: . 2\int\left[2\left(\frac{1}{3}\sin\theta\right) - 3\left(\frac{1}{3}\sin\theta\right)^2\right]\cdot\cos\theta \left(\frac{1}{3}\cos\theta\,d\theta\right)

    . . = \;\;\frac{2}{9}\int(2\sin\theta - \sin^2\!\theta)\cos^2\!\theta\,d\theta \;\;=\;\;\frac{4}{9}\int\cos^2\!\theta\sin\theta\,  d\theta - \frac{2}{9}\int\sin^2\!\theta\cos^2\!\theta\,d\the  ta


    For the first integral, use: u \,=\,\cos\theta

    For the second integral, we have:
    . . \frac{1}{4}\left(4\sin^2\!\theta\cos^2\!\theta\rig  ht) \;=\;\frac{1}{4}\left(2\sin\theta\cos\theta\right)  ^2 \;=\;\frac{1}{4}\!\cdot\!\sin^2\!2\theta \;=\;\frac{1}{8}(1 - \cos4\theta)

    Can you finish it?

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    Quote Originally Posted by Soroban View Post
    Hello, kittycat!

    A variation on Dan's solution . . .


    Let: 3t \:= \:\sin\theta\quad\Rightarrow\quad dt \:=\:\frac{1}{3}\cos\theta\,d\theta
    . . and: . \sqrt{1-9t^2} \:=\:\cos\theta

    Substitute: . 2\int\left[2\left(\frac{1}{3}\sin\theta\right) - 3\left(\frac{1}{3}\sin\theta\right)^2\right]\cdot\cos\theta \left(\frac{1}{3}\cos\theta\,d\theta\right)
    The only problem I have with this method is the integration limits on this last integral. I suppose you can restrict it at this point, as I did later on in my solution?

    -Dan
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    what about for this integral,

    the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much.
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    Quote Originally Posted by kittycat View Post
    what about for this integral,

    the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much.
    Instead of using the sine function, use the tangent function. ( 1 + tan^2(x) = sec^2(x).)

    -Dan
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    Quote:
    Originally Posted by kittycat
    what about for this integral,

    the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much.


    Instead of using the sine function, use the tangent function. (.)

    -Dan

    Hi Dan,

    Could you please show me how to let the tangent function in this question? Sorry to ask you because I forgot the trig substitution. Thank you very much.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kittycat View Post
    Quote:
    Originally Posted by kittycat
    what about for this integral,

    the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much.


    Instead of using the sine function, use the tangent function. (.)

    -Dan

    Hi Dan,

    Could you please show me how to let the tangent function in this question? Sorry to ask you because I forgot the trig substitution. Thank you very much.
    The same way that Soroban did, though in this case there are no restrictions due to the limits. So
    \int t^2\sqrt{1 + 9t^2}~dt

    Let 3t = tan(\theta) \implies 3 dt = sec^2(\theta)~d \theta

    So
    \int t^2\sqrt{1 + 9t^2}~dt = \int \frac{1}{9}tan^2(\theta) \sqrt{1 + tan^2(\theta)} ~ sec^2(\theta)~d\theta

    = \frac{1}{9} \int tan^2(\theta) sec^3(\theta)~d\theta

    -Dan
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    help me again please



    Hi,

    I might be wrong in the process of calculation. But don't know why can't get the answer (given by the textbook) from your approach.

    The answer given by the textbook is -11/108 *sqrt(10) -1/36*ln(sqrt(10)-3) -4/27

    help me again please
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  10. #10
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    Quote Originally Posted by kittycat View Post


    Hi,

    I might be wrong in the process of calculation. But don't know why can't get the answer (given by the textbook) from your approach.

    The answer given by the textbook is -11/108 *sqrt(10) -1/36*ln(sqrt(10)-3) -4/27

    help me again please
    That's because your book isn't right. I get \frac{32 - 3\pi}{216}.

    -Dan
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  11. #11
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    he same way that Soroban did, though in this case there are no restrictions due to the limits. So


    Let

    So




    ========================
    Referring to : the integral from 0 to 1 of t^2 *sqrt(1+9t^2) dt.

    I got to the same answer as you posted. But how should i evaluate it if it is =>
    the integral from 0 to 1 of t^2 *sqrt(1+9t^2) dt.

    Could you please guide me to the final answer of this integration?

    Thank you very much.
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