How do you integrate $\int \frac{1}{\sqrt{16 - x^{2}}}$ $int 1/sqrt(16-x^2)$ by trigonometric substitution?

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Answer 1 · 2016-11-21T04:56:44

$= {sin}^{- 1} (\frac{1}{4} x) + C$ $=sin^(-1)(1/4x)+C$

$\int \frac{1}{\sqrt{16 - x^{2}}} d x$ $int1/(sqrt(16-x^2))dx$

$\int \frac{1}{\sqrt{16 (1 - \frac{x^{2}}{16})}} d x$ $int1/(sqrt(16(1-x^2/16)))dx$

$\int \frac{1}{4 \sqrt{1 - {(\frac{x}{4})}^{2}}} d x$ $int1/(4sqrt(1-(x/4)^2))dx$

$\frac{1}{4} \int \frac{1}{\sqrt{1 - {(\frac{x}{4})}^{2}}} d x$ $1/4int1/(sqrt(1-(x/4)^2))dx$

Let $u = \frac{1}{4} x$ $u = 1/4x$
$\frac{d u}{d x} = \frac{1}{4}$ $(du)/(dx)=1/4$
$4 d u = d x$ $4du=dx$

$\frac{1}{4} \int \frac{1}{\sqrt{1 - u^{2}}} 4 d u$ $1/4int1/sqrt(1-u^2)4du$

$\int \frac{1}{\sqrt{1 - u^{2}}} d u$ $int1/sqrt(1-u^2)du$

$= {sin}^{- 1} u + C$ $=sin^(-1)u+C$

Substitute $u$ $u$ back in:
$= {sin}^{- 1} (\frac{1}{4} x) + C$ $=sin^(-1)(1/4x)+C$

How do you integrate ∫1√16−x2int 1/sqrt(16-x^2) by trigonometric substitution?