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Math Help - Integral sin^2(x) + 2sin(t) +1

  1. #1
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    Integral sin^2(x) + 2sin(t) +1

    How did people get the integral of 1 to be 3/2(x)?
    I'm confused.

    what is the integral of sin^2(x)?
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  2. #2
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    Do you mean: \int \left(\sin^{2} x + 2\sin x + 1 \right) dx

    First off, \int 1 \: dx = \int 1x^{0} dx = \frac{x^{1}}{1} = x + c

    As for integrating \sin^{2} x, use the double angle identity: \sin^{2} x = \frac{1 - \cos (2x)}{2}
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  3. #3
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    ok

    I got 1/2x -1/4sin(2x)

    how do I put this back in terms of sin(t) = x-1
    (the original problem is integral of (x^2)/(squareroot of (1-(x-1)^2)
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    ok

    I got 1/2x -1/4sin(2x)

    how do I put this back in terms of sin(t) = x-1
    (the original problem is integral of (x^2)/(squareroot of (1-(x-1)^2)
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    Quote Originally Posted by khuezy View Post
    I got 1/2x -1/4sin(2x)

    how do I put this back in terms of sin(t) = x-1
    (the original problem is integral of (x^2)/(squareroot of (1-(x-1)^2)
    You made the substitution x = \sin\theta + 1,\text{ where }-\frac\pi2\leq\theta\leq\frac\pi2, yes?

    So, we have:

    \theta = \arcsin(x - 1),

    \sin2\theta = 2\sin\theta\cos\theta\text{\quad\color{red}(trig identity)},

    and

    \cos\theta = \sqrt{\cos^2\theta} = \sqrt{1 - \sin^2\theta}\text{\color{red}\quad(since the cosine is positive for our values of theta)}

    \Rightarrow\cos\theta = \sqrt{1 - (x - 1)^2}

    Now just make the appropriate substitutions.
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