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Math Help - Trig Substitution (Calc. 2)

  1. #1
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    Trig Substitution (Calc. 2)

    Hey guys would love any help on this problem!

    Integral of 4x^2 / (25-x^2)^1/2

    x=5sintheta
    dx = 5costheta

    doing some plugging in and cancelling I get...

    integral (100sin^2theta)

    sin^2theta = 1-2costheta / 2

    So now I have...

    50theta - 100sintheta .... Now I'm stuck, I know I have to eventually use a triangle but I'm not sure to make the 50theta go away.

    Would appreciate any help !
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  2. #2
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    Re: Trig Substitution (Calc. 2)

    The next step is to split this one integral into two (remember the integral is a linear operator) ... see the pdf. I assume you can integrate the last line. If not, write back.

    You don't make the 50 theta go away. It's an integral, you integrate d(theta). Is that clear?
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    Last edited by zhandele; April 1st 2013 at 02:10 PM. Reason: afterthought
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    Re: Trig Substitution (Calc. 2)

    Mmmm ok so I'm kind of getting it, the dtheta is confusing me though still. I've been looking for examples similar to this to see how they solve it but can't find any.

    So would the solved integral be ---> 50theta - 100sintheta

    So I'm guessing now we need to put X's back in and set up a triangle where the hypotenuse is X, the base is (25-x^2) and the leg opposite of theta is 5?
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    Re: Trig Substitution (Calc. 2)

    \sin^{2}\theta = \frac{1}{2}(1-\cos 2\theta).
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    Re: Trig Substitution (Calc. 2)

    Right I think I used that identity already to get rid of sin^2theta but then integrating that and trying to put the X's back into theta is the part that I'm way confused on
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    Re: Trig Substitution (Calc. 2)

    No, the identity you used was incorrect.
    You're integral, (when integrated) should contain a \sin 2\theta term.
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    Re: Trig Substitution (Calc. 2)

    Ok thanks so I reworked it and got 50theta - 25sin2theta. Now to plug in the values from our triangle how would we deal with the with the equation? I can plug in the values from sin from the triangle but have no idea what to do with the 50theta
    Last edited by llama; April 1st 2013 at 03:26 PM.
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    Re: Trig Substitution (Calc. 2)

    Yes, that's correct. Now you have to go back to x as the variable.
    For the \theta, simply invert the original substitution.
    For the \sin 2\theta term you, after using another trig identity, have to deal with a \cos \theta.
    This is where your triangle comes into it. Draw a right-angled triangle with base angle \theta, fill in the lengths of the opposite side and hypotenuse, (from the original substitution) and use pythagoras to find the 'adjacent' length. You can then read off the expression for \cos \theta.
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    Re: Trig Substitution (Calc. 2)

    ok So I used sin2theta = 2sinthetacostheta, from my triangle I can just plug in the values from the triangle

    what do you mean for the theta, invert the original substitution? So we have the sin2theta taken care of I think I can just plug in the values.

    I'm guessin when you say invert the original substitution its x=5sintheta (the same thing we used on our triangle) but i'm not quite getting the inversion part.
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  10. #10
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    Re: Trig Substitution (Calc. 2)

    If you know \displaystyle x = 5\sin{(\theta)}, surely you can get this as \displaystyle \theta in terms of x...
    Last edited by Prove It; April 1st 2013 at 04:03 PM.
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    Re: Trig Substitution (Calc. 2)

    wouldit just be the sin-1(x/5) ?
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    Re: Trig Substitution (Calc. 2)

    Yes, so now can you get your final answer in terms of x?
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    Re: Trig Substitution (Calc. 2)

    Yes, thanks a lot guys. Appreciate it <3
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