trig substitution

$\int 1 - x^{2} d x$
$x = sin t$
$d x = cos t d t$

$= \int cos^{2} t d t$

$cos 2 θ = cos^{2} θ - sin^{2} θ$
$= cos^{2} θ - (1 - cos^{2} θ)$
$= 2 cos^{2} θ - 1$

$cos^{2} θ = \frac{c o s 2 θ + 1}{2}$

$= \frac{1}{2} \int [cos 2 t + 1] d t$

$u = 2 t$
$d u = 2 d t$
$d t = \frac{d u}{2}$

$= \frac{1}{2} \int [cos u + 1] \frac{d u}{2}$
$= \frac{1}{4} \int [cos u + 1] d u$

$= \frac{1}{4} (sin u + u)$
$= \frac{1}{4} (sin 2 t + 2 t)$

$t = sin^{- 1} x$

$= \frac{1}{4} (sin (2 sin^{- 1} x) + 2 sin^{- 1} x)$

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