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Math Help - trig substitution integral

  1. #1
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    trig substitution integral

    integral of sqrt(4-x^2)dx

    I've gathered that x = 4sin(theta) so I write it out as:

    sqrt(4 - 2sin(theta))2cos(theta)

    that's supposed to equal 4cos^2(theta) but I am getting stuck in the middle here
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  2. #2
    Member Abu-Khalil's Avatar
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    Let 4\sin z=x \Rightarrow \cos z dz = dx so \int\sqrt{4-x^2}dx=\int2\sqrt{1-\sin^2 z}\cos zdz=\int2\cos^2 zdz =\int1+\cos 2zdz=z+\frac{\sin 2z}{2}+C=\arcsin \frac{x}{4}+\frac{x}{4}\left(1-\frac{x^2}{16}\right)+C
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    Quote Originally Posted by TYTY View Post
    integral of sqrt(4-x^2)dx

    I've gathered that x = 4sin(theta) so I write it out as:

    sqrt(4 - 2sin(theta))2cos(theta)

    that's supposed to equal 4cos^2(theta) but I am getting stuck in the middle here
    Letting x = 2\sin\theta, for -\frac\pi2\leq\theta\leq\frac\pi2, we have dx=2\cos\theta\,d\theta and

    \int\sqrt{4-x^2}\,dx=\int2\cos\theta\sqrt{4-4\sin^2\theta}\,d\theta

    =4\int\cos\theta\sqrt{1-\sin^2\theta}\,d\theta

    =4\int\cos\theta\lvert\cos\theta\rvert\,d\theta

    =4\int\cos^2\theta\,d\theta\qquad\bigg(since \cos x\geq0\;\forall x\in\left[-\frac\pi2,\frac\pi2\right]\bigg)

    =4\int\frac{1+\cos2\theta}2\,d\theta

    =2\theta+\sin2\theta+C

    =2\left(\arcsin\frac x2\right)+2\sin\theta\cos\theta+C

    =2\arcsin\frac x2+\frac12x\sqrt{4-x^2}+C

    The reason we restrict \theta to the interval \left[-\frac\pi2,\frac\pi2\right] is so that there is a one-to-one correspondence between the values of x and the values of 2\sin\theta.

    Also, be sure to notice that Abu-Khalil's solution is not quite correct: if you substitute x=4\sin\theta, then the radicand becomes 4-16\sin^2\theta. However, I'm sure he only made the error because he was carrying over the faulty substitution from your original post.
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  4. #4
    Member Abu-Khalil's Avatar
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    Quote Originally Posted by Abu-Khalil View Post
    Let 4\sin z=x \Rightarrow \cos z dz = dx so \int\sqrt{4-x^2}dx=\int2\sqrt{1-\sin^2 z}\cos zdz=\int2\cos^2 zdz =\int1+\cos 2zdz=z+\frac{\sin 2z}{2}+C=\arcsin \frac{x}{4}+\frac{x}{4}\left(1-\frac{x^2}{16}\right)+C
    Oops, I forgot 4 when replacing 4\cos zdz and the square root at the end.
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