7.1 Integration By Parts/27

$f'(x)=\int _{0}^{\frac {1}{2}}\cos ^{-1}(x)\cdot dx$

$\int _{0}^{\frac {1}{2}}\cos ^{-1}(x)dx$ = $x\cos ^{-1}(x){\bigg |}_{0}^{\frac {1}{2}}+\int _{0}^{\frac {1}{2}}{\frac {x}{\sqrt {1-x^{2}}}}dx$ = $({\frac {1}{2}}\cdot {\frac {\pi }{3}}){-{\frac {1}{2}}}\int _{1}^{\frac {3}{4}}{\frac {1}{\sqrt {u}}}du$ = ${\frac {\pi }{6}}-{\frac {1}{2}}\int _{1}^{\frac {3}{4}}u^{-{\frac {1}{2}}}du$

= [change the sign and flip the limits] ${\frac {\pi }{6}}+{({\frac {1}{2}}}(2u^{\frac {1}{2}})){\bigg |}_{\frac {3}{4}}^{1}$ = ${\frac {\pi }{6}}+{\frac {1}{2}}(2-{\sqrt {3}})$ = ${\frac {\pi }{6}}-{\frac {\sqrt {3}}{2}}+1$ = ${\frac {1}{6}}(\pi +6-3{\sqrt {3}})$

${u}$ = ${1-x^{2}}$

${du}$ = ${-2x}$

${-{\frac {1}{2}}du}$ = ${x}dx$

${u}$ = ${\cos ^{-1}(x)}$ , ${dv}$ = $dx$

${du}$ = ${-{\frac {1}{\sqrt {1-x^{2}}}}dx}$ , ${v}$ = ${x}$