Question

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Answer 1

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

Answer 2

Again sorry, \[\sqrt{16-x ^{2}}\]

Answer 3

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 ^_^

Answer 4

is it clearer now ice man? :)

Answer 5

Yupp, Thank you!

Answer 6

np ^_^