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anonymous

  • 5 years ago

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  1. anonymous
    • 5 years ago
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    sorry, find the intergal of \[\sqrt{16-^{x}}\]

  2. anonymous
    • 5 years ago
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    Again sorry, \[\sqrt{16-x ^{2}}\]

  3. anonymous
    • 5 years ago
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    Okay let's re-write it in a better form first:\[\int\limits_{}^{}\sqrt{16 - x^2}\] Let's integrate using trigonometric substitution and you'll have the following: we know that \[ x = asin \theta\] and a in this place is 4 ^_^ so: Step one : \[x = 4\sin \theta\], between \[-\pi / 2 < \theta \le \pi/2\] Step 2, substitute x in the equation: \[\sqrt{16 - x^2} = \sqrt{16 - (4\sin \theta)^2} = \sqrt{16(1-\sin^2 \theta)} \]\[= \sqrt{16\cos^2 \theta} = \left| 4 \cos \theta \right| = 4\cos \theta\] Step 3, find dx, we know that x = 4sin (theta) now so : \[dx = 4\cos \theta d \theta\] Step 4, substitute all of this in the integral and you'll get : \[\int\limits_{}^{}\sqrt{16 - x^2} dx = \int\limits_{}^{} (4\cos \theta)(4\cos \theta) d \theta\] \[= \int\limits_{}^{} 16\cos^2 \theta d \theta = 16\int\limits_{}^{}\cos^2 \theta d \theta\] \[= 16 \int\limits_{}^{} 1/2(1+\cos 2 \theta) d \theta\] \[= 16/2 \int\limits_{}^{}1 + \cos 2 \theta d \theta = 8 \int\limits_{}^{}1 + \cos 2 \theta d \theta\] now use u substitute for 2 (theta) and you'll get : \[u = 2 \theta , du = 2 d \theta\] \[= 8/2\int\limits_{}^{} 1 + \cos u du = 4\int\limits_{}^{} 1 + cosu du\] \[= 4 [ u + sinu ] + c = 4[2 \theta + \sin2\theta] + c\] now the last step is to write theta in terms of x , we already know that: \[x = 4\sin \theta\] so : \[\sin \theta = x/4\] and \[\theta = \sin^{-1} (x/4)\] so: \[\int\limits_{}^{}\sqrt{16 - x^2 }dx = 4[2\sin^{-1} (x/4) + \sin (2x/4)] + c\] \[= 8\sin^{-1} (x/4) + 4\sin(x/2) + c\] Correct me if I'm wrong ^_^

  4. anonymous
    • 5 years ago
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    is it clearer now ice man? :)

  5. anonymous
    • 5 years ago
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    Yupp, Thank you!

  6. anonymous
    • 5 years ago
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    np ^_^

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