Question #9abc1

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Question #9abc1

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Answer 1 · 2017-04-22T20:53:05

$8 arcsin (\frac{x}{4}) + 4 sin 2 arcsin (\frac{x}{4}) + C$ $8arcsin(x/4) + 4sin 2arcsin(x/4) + C$

Explanation:

$x = 4 sin t \Rightarrow d x = 4 cos t d t$ $x=4sint => dx=4costdt$

$t = arcsin (\frac{x}{4})$ $t = arcsin(x/4)$

$\sqrt{16 - x^{2}} = \sqrt{16 - {(4 sin t)}^{2}} = \sqrt{16 - 16 {sin}^{2} t} =$ $sqrt(16-x^2) = sqrt(16-(4sint)^2) = sqrt(16-16sin^2t) =$

$= \sqrt{16 (1 - {sin}^{2} t)} = \sqrt{16 {cos}^{2} t} = 4 cos t$ $= sqrt(16(1-sin^2t)) = sqrt(16cos^2t) = 4cost$

$\int \sqrt{16 - x^{2}} d x = \int 4 cos t \cdot 4 cos t d t = 16 \int {cos}^{2} t d t =$ $int sqrt(16-x^2)dx = int 4cost*4costdt = 16 int cos^2tdt =$

$16 \int \frac{1 + cos 2 t}{2} d t = 8 [t + \frac{1}{2} sin 2 t] + C =$ $16 int (1+cos2t)/2dt = 8[t+1/2sin2t]+C =$

$= 8 arcsin (\frac{x}{4}) + 4 sin 2 arcsin (\frac{x}{4}) + C$ $= 8arcsin(x/4) + 4sin 2arcsin(x/4) + C$

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Question #9abc1

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