f ′ ( x ) = ∫ sin − 1 ( x ) ⋅ d x {\displaystyle f'(x)=\int _{}^{}\sin ^{-1}(x)\cdot dx} ∫ sin − 1 ( x ) d x {\displaystyle \int _{}^{}\sin ^{-1}(x)dx} = x sin − 1 ( x ) − ∫ x 1 − x 2 d x {\displaystyle x\sin ^{-1}(x)-\int _{}^{}{\frac {x}{\sqrt {1-x^{2}}}}dx} = x sin − 1 ( x ) + 1 2 ∫ 1 u d u {\displaystyle x\sin ^{-1}(x)+{\frac {1}{2}}\int _{}^{}{\frac {1}{\sqrt {u}}}du} = x sin − 1 ( x ) + 1 2 ∫ u − 1 2 d u {\displaystyle x\sin ^{-1}(x)+{\frac {1}{2}}\int _{}^{}u^{-{\frac {1}{2}}}du} = x sin − 1 ( x ) + 1 2 ( 2 u 1 2 ) {\displaystyle x\sin ^{-1}(x)+{\frac {1}{2}}(2u^{\frac {1}{2}})} = x sin − 1 ( x ) + u {\displaystyle x\sin ^{-1}(x)+{\sqrt {u}}} = x sin − 1 ( x ) + 1 − x 2 + C {\displaystyle x\sin ^{-1}(x)+{\sqrt {1-x^{2}}}+C}
u {\displaystyle {u}} = 1 − x 2 {\displaystyle {1-x^{2}}}
d u {\displaystyle {du}} = − 2 x {\displaystyle {-2x}}
− 1 2 d u {\displaystyle {-{\frac {1}{2}}du}} = x d x {\displaystyle {x}dx}
u {\displaystyle {u}} = sin − 1 ( x ) {\displaystyle {\sin ^{-1}(x)}} , d v {\displaystyle {dv}} = d x {\displaystyle dx}
d u {\displaystyle {du}} = 1 1 − x 2 d x {\displaystyle {{\frac {1}{\sqrt {1-x^{2}}}}dx}} , v {\displaystyle {v}} = x {\displaystyle {x}}
