## anonymous 5 years ago s

1. anonymous

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

2. anonymous

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

3. anonymous

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

is it clearer now ice man? :)

5. anonymous

Yupp, Thank you!

6. anonymous

np ^_^