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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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sorry, find the intergal of \[\sqrt{16-^{x}}\]
Again sorry, \[\sqrt{16-x ^{2}}\]
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 ^_^

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is it clearer now ice man? :)
Yupp, Thank you!
np ^_^

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