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Thread: 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 $\displaystyle 4\sin z=x \Rightarrow \cos z dz = dx$ so $\displaystyle \int\sqrt{4-x^2}dx=\int2\sqrt{1-\sin^2 z}\cos zdz=\int2\cos^2 zdz$ $\displaystyle =\int1+\cos 2zdz=z+\frac{\sin 2z}{2}+C=\arcsin \frac{x}{4}+\frac{x}{4}\left(1-\frac{x^2}{16}\right)+C$
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  3. #3
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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 $\displaystyle x = 2\sin\theta,$ for $\displaystyle -\frac\pi2\leq\theta\leq\frac\pi2,$ we have $\displaystyle dx=2\cos\theta\,d\theta$ and

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

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

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

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

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

    $\displaystyle =2\theta+\sin2\theta+C$

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

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

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

    Also, be sure to notice that Abu-Khalil's solution is not quite correct: if you substitute $\displaystyle x=4\sin\theta,$ then the radicand becomes $\displaystyle 4-16\sin^2\theta.$ 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 $\displaystyle 4\sin z=x \Rightarrow \cos z dz = dx$ so $\displaystyle \int\sqrt{4-x^2}dx=\int2\sqrt{1-\sin^2 z}\cos zdz=\int2\cos^2 zdz$ $\displaystyle =\int1+\cos 2zdz=z+\frac{\sin 2z}{2}+C=\arcsin \frac{x}{4}+\frac{x}{4}\left(1-\frac{x^2}{16}\right)+C$
    Oops, I forgot $\displaystyle 4$ when replacing $\displaystyle 4\cos zdz$ and the square root at the end.
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