反三角函数

反三角函數示意圖 在数学中，反三角函数是三角函数的反函数。 符号 sin − 1 , cos − 1 {\displaystyle \sin ^{-1},\cos ^{-1}} 等常用于 arcsin , arccos {\displaystyle \arcsin ,\arccos } 等。但是这种符号有时在 sin − 1 ⁡ x {\displaystyle \sin ^{-1}x} 和 1 sin ⁡ x {\displaystyle {\frac {1}{\sin x}}} 之间易造成混淆。 在编程中，函数 arcsin {\displaystyle \arcsin } , arccos {\displaystyle \arccos } , arctan {\displaystyle \arctan } 通常叫做 a s i n {\displaystyle \mathrm {asin} } , a c o s {\displaystyle \mathrm {acos} } , a t a n {\displaystyle \mathrm {atan} } 。很多编程语言提供两自变量atan2函数，它计算给定 y {\displaystyle y} 和 x {\displaystyle x} 的 y x {\displaystyle {\frac {y}{x}}} 的反正切，但是值域为 [ − π , π ] {\displaystyle [-\pi ,\pi ]} 。 在笛卡尔平面上 f ( x ) = arcsin ⁡ x {\displaystyle f(x)=\arcsin x} (紅)和 f ( x ) = arccos ⁡ x {\displaystyle f(x)=\arccos x} (綠)函数的常用主值的图像。 在笛卡尔平面上 f ( x ) = arctan ⁡ x {\displaystyle f(x)=\arctan x} (紅)和 f ( x ) = arccot ⁡ x {\displaystyle f(x)=\operatorname {arccot} x} (綠)函数的常用主值的图像。 在笛卡尔平面上 f ( x ) = arccsc ⁡ x {\displaystyle f(x)=\operatorname {arccsc} x} (紅)和 f ( x ) = arcsec ⁡ x {\displaystyle f(x)=\operatorname {arcsec} x} (綠)函数的常用主值的图像。 下表列出基本的反三角函数。 （注意：某些數學教科書的作者將 arcsec {\displaystyle \operatorname {arcsec} } 的值域定為 [ 0 , π 2 ) ∪ [ π , 3 π 2 ) {\displaystyle [0,{\frac {\pi }{2}})\cup [\pi ,{\frac {3\pi }{2}})} 因為當 tan {\displaystyle \tan } 的定義域落在此區間時， tan {\displaystyle \tan } 的值域 ≥ 0 {\displaystyle \geq 0} ，如果 arcsec {\displaystyle \operatorname {arcsec} } 的值域仍定為 [ 0 , π 2 ) ∪ ( π 2 , π ] {\displaystyle [0,{\frac {\pi }{2}})\cup ({\frac {\pi }{2}},\pi ]} ，將會造成 tan ⁡ ( arcsec ⁡ x ) = ± x 2 − 1 {\displaystyle \tan(\operatorname {arcsec} x)=\pm {\sqrt {x^{2}-1}}} ，如果希望 tan ⁡ ( arcsec ⁡ x ) = x 2 − 1 {\displaystyle \tan(\operatorname {arcsec} x)={\sqrt {x^{2}-1}}} ，那就必須將 arcsec {\displaystyle \operatorname {arcsec} } 的值域定為 [ 0 , π 2 ) ∪ [ π , 3 π 2 ) {\displaystyle [0,{\frac {\pi }{2}})\cup [\pi ,{\frac {3\pi }{2}})} ，基於類似的理由 arccsc {\displaystyle \operatorname {arccsc} } 的值域定為 ( − π , − π 2 ] ∪ ( 0 , π 2 ] {\displaystyle (-\pi ,-{\frac {\pi }{2}}]\cup (0,{\frac {\pi }{2}}]} ） 如果 x {\displaystyle x} 允许是复数，则 y {\displaystyle y} 的值域只适用它的实部。 余角： arccos ⁡ x = π 2 − arcsin ⁡ x {\displaystyle \arccos x={\frac {\pi }{2}}-\arcsin x} arccot ⁡ x = π 2 − arctan ⁡ x {\displaystyle \operatorname {arccot} x={\frac {\pi }{2}}-\arctan x} arccsc ⁡ x = π 2 − arcsec ⁡ x {\displaystyle \operatorname {arccsc} x={\frac {\pi }{2}}-\operatorname {arcsec} x} 负数参数： arcsin ⁡ ( − x ) = − arcsin ⁡ x {\displaystyle \arcsin(-x)=-\arcsin x\!} arccos ⁡ ( − x ) = π − arccos ⁡ x {\displaystyle \arccos(-x)=\pi -\arccos x\!} arctan ⁡ ( − x ) = − arctan ⁡ x {\displaystyle \arctan(-x)=-\arctan x\!} arccot ⁡ ( − x ) = π − arccot ⁡ x {\displaystyle \operatorname {arccot}(-x)=\pi -\operatorname {arccot} x\!} arcsec ⁡ ( − x ) = π − arcsec ⁡ x {\displaystyle \operatorname {arcsec}(-x)=\pi -\operatorname {arcsec} x\!} arccsc ⁡ ( − x ) = − arccsc ⁡ x {\displaystyle \operatorname {arccsc}(-x)=-\operatorname {arccsc} x\!} 倒数参数： arccos ⁡ 1 x = arcsec ⁡ x {\displaystyle \arccos {\frac {1}{x}}\,=\operatorname {arcsec} x} arcsin ⁡ 1 x = arccsc ⁡ x {\displaystyle \arcsin {\frac {1}{x}}\,=\operatorname {arccsc} x} arctan ⁡ 1 x = π 2 − arctan ⁡ x = arccot ⁡ x ,   {\displaystyle \arctan {\frac {1}{x}}={\frac {\pi }{2}}-\arctan x=\operatorname {arccot} x,\ }   x > 0 {\displaystyle \ x>0} arctan ⁡ 1 x = − π 2 − arctan ⁡ x = − π + arccot ⁡ x ,   {\displaystyle \arctan {\frac {1}{x}}=-{\frac {\pi }{2}}-\arctan x=-\pi +\operatorname {arccot} x,\ }   x < 0 {\displaystyle \ x<0} arccot ⁡ 1 x = π 2 − arccot ⁡ x = arctan ⁡ x ,   {\displaystyle \operatorname {arccot} {\frac {1}{x}}={\frac {\pi }{2}}-\operatorname {arccot} x=\arctan x,\ }   x > 0 {\displaystyle \ x>0} arccot ⁡ 1 x = 3 π 2 − arccot ⁡ x = π + arctan ⁡ x ,   {\displaystyle \operatorname {arccot} {\frac {1}{x}}={\frac {3\pi }{2}}-\operatorname {arccot} x=\pi +\arctan x,\ }   x < 0 {\displaystyle \ x<0} arcsec ⁡ 1 x = arccos ⁡ x {\displaystyle \operatorname {arcsec} {\frac {1}{x}}=\arccos x} arccsc ⁡ 1 x = arcsin ⁡ x {\displaystyle \operatorname {arccsc} {\frac {1}{x}}=\arcsin x} 如果有一段正弦表： arccos ⁡ x = arcsin ⁡ 1 − x 2 , {\displaystyle \arccos x=\arcsin {\sqrt {1-x^{2}}},}   0 ≤ x ≤ 1 {\displaystyle \ 0\leq x\leq 1} arctan ⁡ x = arcsin ⁡ x x 2 + 1 {\displaystyle \arctan x=\arcsin {\frac {x}{\sqrt {x^{2}+1}}}} 注意只要在使用了复数的平方根的时候，我们选择正实部的平方根(或者正虚部，如果是负实数的平方根的话)。 从半角公式 tan ⁡ θ 2 = sin ⁡ θ 1 + cos ⁡ θ {\displaystyle \tan {\frac {\theta }{2}}={\frac {\sin \theta }{1+\cos \theta }}} ，可得到： arcsin ⁡ x = 2 arctan ⁡ x 1 + 1 − x 2 {\displaystyle \arcsin x=2\arctan {\frac {x}{1+{\sqrt {1-x^{2}}}}}} arccos ⁡ x = 2 arctan ⁡ 1 − x 2 1 + x , {\displaystyle \arccos x=2\arctan {\frac {\sqrt {1-x^{2}}}{1+x}},} − 1 < x ≤ + 1 {\displaystyle -1<x\leq +1} arctan ⁡ x = 2 arctan ⁡ x 1 + 1 + x 2 {\displaystyle \arctan x=2\arctan {\frac {x}{1+{\sqrt {1+x^{2}}}}}} 通過定義可知： 每个三角函数都周期于它的参数的实部上，在每个 2 π {\displaystyle 2\pi } 区间内通过它的所有值两次。正弦和余割的周期开始于 2 π k − π 2 {\displaystyle 2\pi k-{\frac {\pi }{2}}} 结束于 2 π k + π 2 {\displaystyle 2\pi k+{\frac {\pi }{2}}} （这里的 k {\displaystyle k} 是一个整数），在 2 π k + π 2 {\displaystyle 2\pi k+{\frac {\pi }{2}}} 到 2 π k + 3 π 2 {\displaystyle 2\pi k+{\frac {3\pi }{2}}} 上倒过来。余弦和正割的周期开始于 2 π k {\displaystyle 2\pi k} 结束于 2 π k + π {\displaystyle 2\pi k+\pi } ，在 2 π k + π {\displaystyle 2\pi k+\pi } 到 2 π k + 2 π {\displaystyle 2\pi k+2\pi } 上倒过来。正切的周期开始于 2 π k − π 2 {\displaystyle 2\pi k-{\frac {\pi }{2}}} 结束于 2 π k + π 2 {\displaystyle 2\pi k+{\frac {\pi }{2}}} ，接着(向前)在 2 π k + π 2 {\displaystyle 2\pi k+{\frac {\pi }{2}}} 到 2 π k + 3 π 2 {\displaystyle 2\pi k+{\frac {3\pi }{2}}} 上重复。余切的周期开始于 2 π k {\displaystyle 2\pi k} 结束于 2 π k + π {\displaystyle 2\pi k+\pi } ，接着（向前）在 2 π k + π {\displaystyle 2\pi k+\pi } 到 2 π k + 2 π {\displaystyle 2\pi k+2\pi } 上重复。 这个周期性反应在一般反函数上： sin ⁡ y = x   ⇔   (   y = arcsin ⁡ x + 2 k π   ∀   k ∈ Z   ∨   y = π − arcsin ⁡ x + 2 k π   ∀   k ∈ Z   ) {\displaystyle \sin y=x\ \Leftrightarrow \ (\ y=\arcsin x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=\pi -\arcsin x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )} cos ⁡ y = x   ⇔   (   y = arccos ⁡ x + 2 k π   ∀   k ∈ Z   ∨   y = 2 π − arccos ⁡ x + 2 k π   ∀   k ∈ Z   ) {\displaystyle \cos y=x\ \Leftrightarrow \ (\ y=\arccos x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=2\pi -\arccos x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )} tan ⁡ y = x   ⇔     y = arctan ⁡ x + k π   ∀   k ∈ Z {\displaystyle \tan y=x\ \Leftrightarrow \ \ y=\arctan x+k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} } cot ⁡ y = x   ⇔     y = arccot ⁡ x + k π   ∀   k ∈ Z {\displaystyle \cot y=x\ \Leftrightarrow \ \ y=\operatorname {arccot} x+k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} } sec ⁡ y = x   ⇔   (   y = arcsec ⁡ x + 2 k π   ∀   k ∈ Z   ∨   y = 2 π − arcsec ⁡ x + 2 k π   ∀   k ∈ Z   ) {\displaystyle \sec y=x\ \Leftrightarrow \ (\ y=\operatorname {arcsec} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=2\pi -\operatorname {arcsec} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )} csc ⁡ y = x   ⇔   (   y = arccsc ⁡ x + 2 k π   ∀   k ∈ Z   ∨   y = π − arccsc ⁡ x + 2 k π   ∀   k ∈ Z   ) {\displaystyle \csc y=x\ \Leftrightarrow \ (\ y=\operatorname {arccsc} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ \lor \ y=\pi -\operatorname {arccsc} x+2k\pi {\text{ }}\forall {\text{ }}k\in \mathbb {Z} \ )} 对于实数 x {\displaystyle x} 的反三角函數的导函数如下： d d x arcsin ⁡ x = 1 1 − x 2 ; | x | < 1 d d x arccos ⁡ x = − 1 1 − x 2 ; | x | < 1 d d x arctan ⁡ x = 1 1 + x 2 d d x arccot ⁡ x = − 1 1 + x 2 d d x arcsec ⁡ x = 1 | x | x 2 − 1 ; | x | > 1 d d x arccsc ⁡ x = − 1 | x | x 2 − 1 ; | x | > 1 {\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} x}}\arcsin x&{}={\frac {1}{\sqrt {1-x^{2}}}};\qquad |x|<1\\{\frac {\mathrm {d} }{\mathrm {d} x}}\arccos x&{}={\frac {-1}{\sqrt {1-x^{2}}}};\qquad |x|<1\\{\frac {\mathrm {d} }{\mathrm {d} x}}\arctan x&{}={\frac {1}{1+x^{2}}}\\{\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {arccot} x&{}={\frac {-1}{1+x^{2}}}\\{\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {arcsec} x&{}={\frac {1}{|x|\,{\sqrt {x^{2}-1}}}};\qquad |x|>1\\{\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {arccsc} x&{}={\frac {-1}{|x|\,{\sqrt {x^{2}-1}}}};\qquad |x|>1\\\end{aligned}}} 舉例說明，设 θ = arcsin ⁡ x {\displaystyle \theta =\arcsin x\!} ，得到： d arcsin ⁡ x d x = d θ d sin ⁡ θ = 1 cos ⁡ θ = 1 1 − sin 2 ⁡ θ = 1 1 − x 2 {\displaystyle {\frac {d\arcsin x}{dx}}={\frac {d\theta }{d\sin \theta }}={\frac {1}{\cos \theta }}={\frac {1}{\sqrt {1-\sin ^{2}\theta }}}={\frac {1}{\sqrt {1-x^{2}}}}} 因為要使根號內部恆為正，所以在條件加上 | x | < 1 {\displaystyle |x|<1} ，其他導數公式同理可證。 积分其导数并固定在一点上的值给出反三角函数作为定积分的表达式： arcsin ⁡ x = ∫ 0 x 1 1 − z 2 d z , | x | ≤ 1 arccos ⁡ x = ∫ x 1 1 1 − z 2 d z , | x | ≤ 1 arctan ⁡ x = ∫ 0 x 1 z 2 + 1 d z , arccot ⁡ x = ∫ x ∞ 1 z 2 + 1 d z , arcsec ⁡ x = ∫ 1 x 1 z z 2 − 1 d z , x ≥ 1 arccsc ⁡ x = ∫ x ∞ 1 z z 2 − 1 d z , x ≥ 1 {\displaystyle {\begin{aligned}\arcsin x&{}=\int _{0}^{x}{\frac {1}{\sqrt {1-z^{2}}}}\,dz,\qquad |x|\leq 1\\\arccos x&{}=\int _{x}^{1}{\frac {1}{\sqrt {1-z^{2}}}}\,dz,\qquad |x|\leq 1\\\arctan x&{}=\int _{0}^{x}{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arccot} x&{}=\int _{x}^{\infty }{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arcsec} x&{}=\int _{1}^{x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\\\operatorname {arccsc} x&{}=\int _{x}^{\infty }{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\end{aligned}}} 当 x {\displaystyle x} 等于1时，在有极限的域上的积分是瑕积分，但仍是良好定义的。 如同正弦和余弦函数，反三角函数可以使用无穷级数计算如下： arcsin ⁡ z = z + ( 1 2 ) z 3 3 + ( 1 ⋅ 3 2 ⋅ 4 ) z 5 5 + ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) z 7 7 + ⋯ = ∑ n = 0 ∞ [ ( 2 n ) ! 2 2 n ( n ! ) 2 ] z 2 n + 1 ( 2 n + 1 ) ; | z | ≤ 1 {\displaystyle {\begin{aligned}\arcsin z&{}=z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots \\&{}=\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{2n+1}}{(2n+1)}};\qquad |z|\leq 1\end{aligned}}} arccos ⁡ z = π 2 − arcsin ⁡ z = π 2 − [ z + ( 1 2 ) z 3 3 + ( 1 ⋅ 3 2 ⋅ 4 ) z 5 5 + ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) z 7 7 + ⋯ ] = π 2 − ∑ n = 0 ∞ [ ( 2 n ) ! 2 2 n ( n ! ) 2 ] z 2 n + 1 ( 2 n + 1 ) ; | z | ≤ 1 {\displaystyle {\begin{aligned}\arccos z&{}={\frac {\pi }{2}}-\arcsin z\\&{}={\frac {\pi }{2}}-\left[z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots \right]\\&{}={\frac {\pi }{2}}-\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{2n+1}}{(2n+1)}};\qquad |z|\leq 1\end{aligned}}} arctan ⁡ z = z − z 3 3 + z 5 5 − z 7 7 + ⋯ = ∑ n = 0 ∞ ( − 1 ) n z 2 n + 1 2 n + 1 ; | z | ≤ 1 z ≠ i , − i {\displaystyle {\begin{aligned}\arctan z&{}=z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots \\&{}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}};\qquad |z|\leq 1\qquad z\neq i,-i\end{aligned}}} arccot ⁡ z = π 2 − arctan ⁡ z = π 2 − ( z − z 3 3 + z 5 5 − z 7 7 + ⋯ ) = π 2 − ∑ n = 0 ∞ ( − 1 ) n z 2 n + 1 2 n + 1 ; | z | ≤ 1 z ≠ i , − i {\displaystyle {\begin{aligned}\operatorname {arccot} z&{}={\frac {\pi }{2}}-\arctan z\\&{}={\frac {\pi }{2}}-\left(z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots \right)\\&{}={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}};\qquad |z|\leq 1\qquad z\neq i,-i\end{aligned}}} arcsec ⁡ z = arccos ⁡ ( z − 1 ) = π 2 − [ z − 1 + ( 1 2 ) z − 3 3 + ( 1 ⋅ 3 2 ⋅ 4 ) z − 5 5 + ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) z − 7 7 + ⋯ ] = π 2 − ∑ n = 0 ∞ [ ( 2 n ) ! 2 2 n ( n ! ) 2 ] z − ( 2 n + 1 ) ( 2 n + 1 ) ; | z | ≥ − 4 {\displaystyle {\begin{aligned}\operatorname {arcsec} z&{}=\arccos \left(z^{-1}\right)\\&{}={\frac {\pi }{2}}-\left[z^{-1}+\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{-7}}{7}}+\cdots \right]\\&{}={\frac {\pi }{2}}-\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{-(2n+1)}}{(2n+1)}};\qquad \left|z\right|\geq -4\end{aligned}}} arccsc ⁡ z = arcsin ⁡ ( z − 1 ) = z − 1 + ( 1 2 ) z − 3 3 + ( 1 ⋅ 3 2 ⋅ 4 ) z − 5 5 + ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) z − 7 7 + ⋯ = ∑ n = 0 ∞ [ ( 2 n ) ! 2 2 n ( n ! ) 2 ] z − ( 2 n + 1 ) 2 n + 1 ; | z | ≥ 1 {\displaystyle {\begin{aligned}\operatorname {arccsc} z&{}=\arcsin \left(z^{-1}\right)\\&{}=z^{-1}+\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{-7}}{7}}+\cdots \\&{}=\sum _{n=0}^{\infty }\left[{\frac {(2n)!}{2^{2n}(n!)^{2}}}\right]{\frac {z^{-(2n+1)}}{2n+1}};\qquad \left|z\right|\geq 1\end{aligned}}} 欧拉发现了反正切的更有效的级数： arctan ⁡ x = ∞ x 1 + x 2 ∑ n = 0 ∞ ∏ k = 1 n 2 k x 2 ( 2 k + 1 ) ( 1 + x 2 ) {\displaystyle \arctan x=\infty {x}{1+x^{2}}\sum _{n=0}^{\infty }\prod _{k=1}^{n}{\frac {2kx^{2}}{(2k+1)(1+x^{2})}}} 。 （注意对 x = 0 {\displaystyle x=0} 在和中的项是空积1。） ∫ arcsin ⁡ x d x = x arcsin ⁡ x + 1 − x 2 + C , x ≤ 1 ∫ arccos ⁡ x d x = x arccos ⁡ x − 1 − x 2 + C , x ≤ 1 ∫ arctan ⁡ x d x = x arctan ⁡ x − 1 2 ln ⁡ ( 1 + x 2 ) + C ∫ arccot ⁡ x d x = x arccot ⁡ x + 1 2 ln ⁡ ( 1 + x 2 ) + C ∫ arcsec ⁡ x d x = x arcsec ⁡ x − sgn ⁡ ( x ) ln ⁡ | x + x 2 − 1 | + C = x arcsec ⁡ x + sgn ⁡ ( x ) ln ⁡ | x − x 2 − 1 | + C ∫ arccsc ⁡ x d x = x arccsc ⁡ x + sgn ⁡ ( x ) ln ⁡ | x + x 2 − 1 | + C = x arccsc ⁡ x − sgn ⁡ ( x ) ln ⁡ | x − x 2 − 1 | + C {\displaystyle {\begin{aligned}\int \arcsin x\,dx&{}=x\,\arcsin x+{\sqrt {1-x^{2}}}+C,\qquad x\leq 1\\\int \arccos x\,dx&{}=x\,\arccos x-{\sqrt {1-x^{2}}}+C,\qquad x\leq 1\\\int \arctan x\,dx&{}=x\,\arctan x-{\frac {1}{2}}\ln \left(1+x^{2}\right)+C\\\int \operatorname {arccot} x\,dx&{}=x\,\operatorname {arccot} x+{\frac {1}{2}}\ln \left(1+x^{2}\right)+C\\\int \operatorname {arcsec} x\,dx&{}=x\,\operatorname {arcsec} x-\operatorname {sgn}(x)\ln \left|x+{\sqrt {x^{2}-1}}\right|+C=x\,\operatorname {arcsec} x+\operatorname {sgn}(x)\ln \left|x-{\sqrt {x^{2}-1}}\right|+C\\\int \operatorname {arccsc} x\,dx&{}=x\,\operatorname {arccsc} x+\operatorname {sgn}(x)\ln \left|x+{\sqrt {x^{2}-1}}\right|+C=x\,\operatorname {arccsc} x-\operatorname {sgn}(x)\ln \left|x-{\sqrt {x^{2}-1}}\right|+C\\\end{aligned}}} 使用分部积分法和上面的简单导数很容易得出它们。 使用 ∫ u d v = u v − ∫ v d u {\displaystyle \int u\,\mathrm {d} v=uv-\int v\,\mathrm {d} u} ，設 u = arcsin ⁡ x d v = d x d u = d x 1 − x 2 v = x {\displaystyle {\begin{aligned}u&{}=&\arcsin x&\quad \quad \mathrm {d} v=\mathrm {d} x\\\mathrm {d} u&{}=&{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}&\quad \quad {}v=x\end{aligned}}} 則 ∫ arcsin ⁡ ( x ) d x = x arcsin ⁡ x − ∫ x 1 − x 2 d x {\displaystyle \int \arcsin(x)\,\mathrm {d} x=x\arcsin x-\int {\frac {x}{\sqrt {1-x^{2}}}}\,\mathrm {d} x} 換元 k = 1 − x 2 . {\displaystyle k=1-x^{2}.\,} 則 d k = − 2 x d x {\displaystyle \mathrm {d} k=-2x\,\mathrm {d} x} 且 ∫ x 1 − x 2 d x = − 1 2 ∫ d k k = − k {\displaystyle \int {\frac {x}{\sqrt {1-x^{2}}}}\,\mathrm {d} x=-{\frac {1}{2}}\int {\frac {\mathrm {d} k}{\sqrt {k}}}=-{\sqrt {k}}} 換元回x得到 ∫ arcsin ⁡ ( x ) d x = x arcsin ⁡ x + 1 − x 2 + C {\displaystyle \int \arcsin(x)\,\mathrm {d} x=x\arcsin x+{\sqrt {1-x^{2}}}+C} arcsin ⁡ x + arcsin ⁡ y = arcsin ⁡ ( x 1 − y 2 + y 1 − x 2 ) , x y ≤ 0 ∨ x 2 + y 2 ≤ 1 {\displaystyle \arcsin x+\arcsin y=\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),xy\leq 0\lor x^{2}+y^{2}\leq 1} arcsin ⁡ x + arcsin ⁡ y = π − arcsin ⁡ ( x 1 − y 2 + y 1 − x 2 ) , x > 0 , y > 0 , x 2 + y 2 > 1 {\displaystyle \arcsin x+\arcsin y=\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),x>0,y>0,x^{2}+y^{2}>1} arcsin ⁡ x + arcsin ⁡ y = − π − arcsin ⁡ ( x 1 − y 2 + y 1 − x 2 ) , x < 0 , y < 0 , x 2 + y 2 > 1 {\displaystyle \arcsin x+\arcsin y=-\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),x<0,y<0,x^{2}+y^{2}>1} arcsin ⁡ x − arcsin ⁡ y = arcsin ⁡ ( x 1 − y 2 − y 1 − x 2 ) , x y ≥ 0 ∨ x 2 + y 2 ≤ 1 {\displaystyle \arcsin x-\arcsin y=\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right),xy\geq 0\lor x^{2}+y^{2}\leq 1} arcsin ⁡ x − arcsin ⁡ y = π − arcsin ⁡ ( x 1 − y 2 − y 1 − x 2 ) , x > 0 , y < 0 , x 2 + y 2 > 1 {\displaystyle \arcsin x-\arcsin y=\pi -\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right),x>0,y<0,x^{2}+y^{2}>1} arcsin ⁡ x − arcsin ⁡ y = − π − arcsin ⁡ ( x 1 − y 2 + y 1 − x 2 ) , x < 0 , y > 0 , x 2 + y 2 > 1 {\displaystyle \arcsin x-\arcsin y=-\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),x<0,y>0,x^{2}+y^{2}>1} arccos ⁡ x + arccos ⁡ y = arccos ⁡ ( x y − 1 − x 2 ⋅ 1 − y 2 ) , x + y ≥ 0 {\displaystyle \arccos x+\arccos y=\arccos \left(xy-{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),x+y\geq 0} arccos ⁡ x + arccos ⁡ y = 2 π − arccos ⁡ ( x y − 1 − x 2 ⋅ 1 − y 2 ) , x + y < 0 {\displaystyle \arccos x+\arccos y=2\pi -\arccos \left(xy-{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),x+y<0} arccos ⁡ x − arccos ⁡ y = − arccos ⁡ ( x y + 1 − x 2 ⋅ 1 − y 2 ) , x ≥ y {\displaystyle \arccos x-\arccos y=-\arccos \left(xy+{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),x\geq y} arccos ⁡ x − arccos ⁡ y = arccos ⁡ ( x y + 1 − x 2 ⋅ 1 − y 2 ) , x < y {\displaystyle \arccos x-\arccos y=\arccos \left(xy+{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),x<y} arctan x + arctan y = arctan x + y 1 − x y , x y < 1 {\displaystyle \arctan \,x+\arctan \,y=\arctan \,{\frac {x+y}{1-xy}},xy<1} arctan x + arctan y = π + arctan x + y 1 − x y , x > 0 , x y > 1 {\displaystyle \arctan \,x+\arctan \,y=\pi +\arctan \,{\frac {x+y}{1-xy}},x>0,xy>1} arctan x + arctan y = − π + arctan x + y 1 − x y , x < 0 , x y > 1 {\displaystyle \arctan \,x+\arctan \,y=-\pi +\arctan \,{\frac {x+y}{1-xy}},x<0,xy>1} arctan ⁡ x − arctan ⁡ y = arctan ⁡ x − y 1 + x y , x y > − 1 {\displaystyle \arctan x-\arctan y=\arctan {\frac {x-y}{1+xy}},xy>-1} arctan ⁡ x − arctan ⁡ y = π + arctan ⁡ x − y 1 + x y , x > 0 , x y < − 1 {\displaystyle \arctan x-\arctan y=\pi +\arctan {\frac {x-y}{1+xy}},x>0,xy<-1} arctan ⁡ x − arctan ⁡ y = − π + arctan ⁡ x − y 1 + x y , x < 0 , x y < − 1 {\displaystyle \arctan x-\arctan y=-\pi +\arctan {\frac {x-y}{1+xy}},x<0,xy<-1} arccot ⁡ x + arccot ⁡ y = arccot ⁡ x y − 1 x + y , x > − y {\displaystyle \operatorname {arccot} x+\operatorname {arccot} y=\operatorname {arccot} {\frac {xy-1}{x+y}},x>-y} arccot ⁡ x + arccot ⁡ y = arccot ⁡ x y − 1 x + y + π , x < − y {\displaystyle \operatorname {arccot} x+\operatorname {arccot} y=\operatorname {arccot} {\frac {xy-1}{x+y}}+\pi ,x<-y} arcsin ⁡ x + arccos ⁡ x = π 2 , | x | ≤ 1 {\displaystyle \arcsin x+\arccos x={\frac {\pi }{2}},|x|\leq 1} arctan ⁡ x + arccot ⁡ x = π 2 {\displaystyle \arctan x+\operatorname {arccot} x={\frac {\pi }{2}}} 设 θ = arccos ⁡ x {\displaystyle \theta =\arccos x} ，得到： d arccos ⁡ x d x = d θ d cos ⁡ θ = − 1 sin ⁡ θ = 1 1 − cos 2 ⁡ θ = − 1 1 − x 2 {\displaystyle {\frac {d\arccos x}{dx}}={\frac {d\theta }{d\cos \theta }}={\frac {-1}{\sin \theta }}={\frac {1}{\sqrt {1-\cos ^{2}\theta }}}={\frac {-1}{\sqrt {1-x^{2}}}}} 因為要使根號內部恆為正，所以在條件加上 | x | < 1 {\displaystyle |x|<1} 。 设 θ = arctan ⁡ x {\displaystyle \theta =\arctan x} ，得到： d arctan ⁡ x d x = d θ d tan ⁡ θ = 1 sec 2 ⁡ θ = 1 1 + tan 2 ⁡ θ = 1 1 + x 2 {\displaystyle {\frac {d\arctan x}{dx}}={\frac {d\theta }{d\tan \theta }}={\frac {1}{\sec ^{2}\theta }}={\frac {1}{1+\tan ^{2}\theta }}={\frac {1}{1+x^{2}}}} 设 θ = arccot ⁡ x {\displaystyle \theta =\operatorname {arccot} x} ，得到： d arccot ⁡ x d x = d θ d cot ⁡ θ = − 1 csc 2 ⁡ θ = 1 1 + cot 2 ⁡ θ = − 1 1 + x 2 {\displaystyle {\frac {d\operatorname {arccot} x}{dx}}={\frac {d\theta }{d\cot \theta }}={\frac {-1}{\csc ^{2}\theta }}={\frac {1}{1+\cot ^{2}\theta }}={\frac {-1}{1+x^{2}}}} 设 θ = arcsec ⁡ x {\displaystyle \theta =\operatorname {arcsec} x} ，得到： d arcsec ⁡ x d x = d θ d sec ⁡ θ = 1 sec ⁡ θ tan ⁡ θ = 1 | x | x 2 − 1 {\displaystyle {\frac {d\operatorname {arcsec} x}{dx}}={\frac {d\theta }{d\sec \theta }}={\frac {1}{\sec \theta \tan \theta }}={\frac {1}{\left|x\right|{\sqrt {x^{2}-1}}}}} 因為要使根號內部恆為正，所以在條件加上 | x | > 1 {\displaystyle |x|>1} ，比較容易被忽略是 sec ⁡ θ {\displaystyle \sec \theta } 產生的絕對值 sec − 1 ⁡ θ {\displaystyle \sec ^{-1}\theta } 的定義域是 0 ≤ θ ≤ π , θ ≠ π 2 {\displaystyle 0\leq \theta \leq \pi ,\theta \neq {\frac {\pi }{2}}} ，所以 tan ⁡ θ = ± x 2 − 1 {\displaystyle \tan \theta =\pm {\sqrt {x^{2}-1}}} ，所以 x {\displaystyle x} 要加绝对值。 设 θ = arccsc ⁡ x {\displaystyle \theta =\operatorname {arccsc} x} ，得到： d arccsc ⁡ x d x = d θ d csc ⁡ θ = − 1 csc ⁡ θ cot ⁡ θ = − 1 | x | x 2 − 1 {\displaystyle {\frac {d\operatorname {arccsc} x}{dx}}={\frac {d\theta }{d\csc \theta }}={\frac {-1}{\csc \theta \cot \theta }}={\frac {-1}{\left|x\right|{\sqrt {x^{2}-1}}}}} 因為要使根號內部恆為正，所以在條件加上 | x | > 1 {\displaystyle |x|>1} ，比較容易被忽略是 csc ⁡ θ {\displaystyle \csc \theta } 產生的絕對值 csc − 1 ⁡ θ {\displaystyle \csc ^{-1}\theta } 的定義域是 − π 2 ≤ θ ≤ π 2 , θ ≠ 0 {\displaystyle -{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}},\theta \neq 0} 。 数学主题 正切半角公式 平方根 埃里克·韦斯坦因. Inverse Trigonometric Functions. MathWorld.  http://mathworld.wolfram.com/InverseTangent.html（页面存档备份，存于互联网档案馆）