7.1 Integration By Parts/27

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

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

= ${\frac {\pi }{6}}-{({\frac {1}{2}}}(2u^{\frac {1}{2}})){\bigg |}_{1}^{\frac {3}{4}}$ = ${\frac {\pi }{6}}-{\frac {1}{2}}({\sqrt {3}}-1)$ = ${\frac {\pi }{6}}-{\frac {\sqrt {3}}{2}}+1$ = ${\frac {1}{6}}(\pi +6-3{\sqrt {3}})$

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

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

${-du}$ = ${sint}dt$

${u}$ = ${u}$ , ${dv}$ = $e^{u}du$

${du}$ = ${du}$ , ${v}$ = $e^{u}$