Answer
a) $-\frac{1}{2}\int{\frac{du}{u^{\frac{1}{2}}}}$
b) $\int{sin^{2}θcos^{2}θdθ}$
c) $-\frac{1}{2}\int{(1-u)u^{\frac{1}{2}}}du$
d) $\int{sin^{4}θ}dθ$
Work Step by Step
a)
let
$u$ = $1-x^{2}$
$du$ =$-2xdx$
$\int{\frac{x}{\sqrt {1-x^{2}}}dx}$ = $-\frac{1}{2}\int{\frac{du}{u^{\frac{1}{2}}}}$
b)
$x$ = $sinθ$
$dx$ = $cosθdθ$
$\int{x^{2}\sqrt {1-x^{2}}}dx$ = $\int{sin^{2}θ(cosθ)cosθdθ}$ = $\int{sin^{2}θcos^{2}θdθ}$
c)
$u$ = $1-x^{2}$
$du$ =$-2xdx$
$\int{x^{3}\sqrt {1-x^{2}}}dx$ = $-\frac{1}{2}\int{x^{2}\sqrt {1-x^{2}}}(-2x)dx$ = $-\frac{1}{2}\int{(1-u)u^{\frac{1}{2}}}du$
d)
$x$ = $sinθ$
$dx$ = $cosθdθ$
$\int{\frac{x^{4}}{\sqrt {1-x^{2}}}dx}$ =$\int{\frac{sin^{4}θ}{cosθ}}cosθdθ$ = $\int{sin^{4}θ}dθ$
