arcsin ( sin ( x ) ) = x {\displaystyle \arcsin(\sin(x))=x}
arcsin ( x ) = arctan ( x 1 − x 2 ) {\displaystyle \arcsin(x)=\arctan \left({\frac {x}{\sqrt {1-x^{2}}}}\right)}
arcsin ( x ) = 1 2 arccos ( 1 − 2 x 2 ) , 0 ≤ x ≤ 1 {\displaystyle \arcsin(x)={\dfrac {1}{2}}\arccos \left(1-2x^{2}\right){\text{,}}\ \ 0\leq x\leq 1}
d d x arcsin ( a x + b ) = a 1 − ( a x + b ) 2 {\displaystyle {\frac {d}{dx}}\arcsin(ax+b)={\frac {a}{\sqrt {1-(ax+b)^{2}}}}}
За a = 1 {\displaystyle a=1} и b = 0 {\displaystyle b=0} :
d d x arcsin ( x ) = 1 1 − x 2 {\displaystyle {\frac {d}{dx}}\arcsin(x)={\frac {1}{\sqrt {1-x^{2}}}}}
arcsin ( − 1 ) = − π 2 {\displaystyle \arcsin(-1)=-{\frac {\pi }{2}}} arcsin ( − 1 2 2 ) = − 1 4 π {\displaystyle \arcsin \left(-{\frac {1}{2}}{\sqrt {2}}\right)=-{\frac {1}{4}}\pi } arcsin ( 0 ) = 0 {\displaystyle \arcsin(0)=0\!} arcsin ( 1 2 2 ) = 1 4 π {\displaystyle \arcsin \left({\frac {1}{2}}{\sqrt {2}}\right)={\frac {1}{4}}\pi } arcsin ( 1 ) = π 2 {\displaystyle \arcsin(1)={\frac {\pi }{2}}}
∫ 0 x d t 1 − t 2 = arcsin ( x ) {\displaystyle \int _{0}^{x}{\frac {\mathrm {d} t}{\sqrt {1-t^{2}}}}=\arcsin(x)}
∫ d x m 2 − x 2 = arcsin ( x m ) + C {\displaystyle \int {\frac {\mathrm {d} x}{\sqrt {m^{2}-x^{2}}}}=\arcsin \left({\frac {x}{m}}\right)+C}
∫ arcsin ( m x ) d x = m x arcsin ( m x ) + 1 − m 2 x 2 m + C {\displaystyle \int \arcsin(mx)\mathrm {d} x={\frac {mx\arcsin(mx)+{\sqrt {1-m^{2}x^{2}}}}{m}}+C}
Нека разгледаме следния интеграл:
I = ∫ 0 x d t 1 − t 2 {\displaystyle {\mathcal {I}}=\int _{0}^{x}{\frac {\mathrm {d} t}{\sqrt {1-t^{2}}}}}
От биномната теорема получаваме:
I = ∫ 0 x ∑ k = 0 ∞ ( 2 k ) ! t 2 k d t 2 2 k ( k ! ) 2 {\displaystyle {\mathcal {I}}=\int _{0}^{x}\sum _{k=0}^{\infty }{\frac {(2k)!t^{2k}\mathrm {d} t}{2^{2k}(k!)^{2}}}}
= ∑ k = 0 ∞ ( 2 k ) ! 2 2 k ( k ! ) 2 ∫ 0 x t 2 k d t {\displaystyle \ \ =\sum _{k=0}^{\infty }{\frac {(2k)!}{2^{2k}(k!)^{2}}}\int _{0}^{x}t^{2k}\mathrm {d} t}
Освен това знаем, че:
I = ∫ 0 x d t 1 − t 2 = arcsin ( x ) {\displaystyle {\mathcal {I}}=\int _{0}^{x}{\frac {\mathrm {d} t}{\sqrt {1-t^{2}}}}=\arcsin(x)}
Следователно:
arcsin ( x ) = ∑ k = 0 ∞ ( 2 k ) ! 2 2 k ( k ! ) 2 ∫ 0 x t 2 k d t {\displaystyle \arcsin(x)=\sum _{k=0}^{\infty }{\frac {(2k)!}{2^{2k}(k!)^{2}}}\int _{0}^{x}t^{2k}\mathrm {d} t}
= ∑ k = 0 ∞ ( 2 k ) ! 2 2 k ( k ! ) 2 t 2 k + 1 2 k + 1 | 0 x {\displaystyle \qquad \quad \ \ \ \ =\sum _{k=0}^{\infty }{\frac {(2k)!}{2^{2k}(k!)^{2}}}{\frac {t^{2k+1}}{2k+1}}{\Bigg |}_{0}^{x}}
Откъдето вече лесно се вижда, че:
arcsin ( x ) = ∑ k = 0 ∞ ( 2 k ) ! x 2 k + 1 2 2 k ( k ! ) 2 ( 2 k + 1 ) , − 1 ≤ x ≤ 1 {\displaystyle {\begin{aligned}\arcsin(x)=\sum _{k=0}^{\infty }{\frac {(2k)!x^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}{\text{,}}\ -1\leq x\leq 1\end{aligned}}}
= x + 1 6 x 3 + 3 40 x 5 + 5 112 x 7 + ⋯ {\displaystyle \qquad \quad \ \ \ \ =x+{\frac {1}{6}}x^{3}+{\frac {3}{40}}x^{5}+{\frac {5}{112}}x^{7}+\cdots }