Specifically, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle's trigonometric ratios.
Dengan kata lain, fungsi invers trigonometri adalah fungsi invers suatu fungsi sinus, kosinus, tangen, kotangen, sekan dan kosekan, dan digunakan untuk mencari suatu sudut dari rasio trigonometri sudut yang lain.
Inverse trigonometric functions are widely used in engineering, navigation, physics, and geometry.
Fungsi invers trigonometri deretng digunakan di bidang teknik, navigasi, fisika dan geometri.
{ arctan ⁡ ( x ) + π k ∣ k ∈ Z } .
d zz arctan ⁡ ( z ) − 1 2 ln ⁡ ( 1 + z 2 ) + C ∫ arccot ⁡ ( z )
As mentioned above, if arccos ⁡ xπ (which by definition only happens when xcos ⁡ π− 1 )
Sebagai contoh, arccos ⁡ ( x )π / 2 − arcsin ⁡ ( x ) , arccsc ⁡ ( x )arcsin ⁡ ( 1 / x ) (x)=\arcsin(1/x)} , dan seterusnya.
Having considered the cases arccos ⁡ x0 and arccos ⁡ xπ , we now focus on the case where arccos ⁡ x ≠ 0 and arccos ⁡ x ≠ π , So assume this from now on.
Sebagai contoh, arccos ⁡ ( x )π / 2 − arcsin ⁡ ( x ) , arccsc ⁡ ( x )arcsin ⁡ ( 1 / x ) (x)=\arcsin(1/x)} , dan seterusnya.
The substitution arcsin ⁡ xπ 2 − arccos ⁡ x {2}}-\arccos x\;} may be used express the right hand side of the above formula in terms of arccos ⁡ x instead of arcsin ⁡ x .
Sebagai contoh, arccos ⁡ ( x )π / 2 − arcsin ⁡ ( x ) , arccsc ⁡ ( x )arcsin ⁡ ( 1 / x ) (x)=\arcsin(1/x)} , dan seterusnya.
π 2 − arcsin ⁡ ( x ) arccot ⁡ ( x )π 2 − arctan ⁡ ( x ) arccsc ⁡ ( x )
Sebagai contoh, arccos ⁡ ( x )π / 2 − arcsin ⁡ ( x ) , arccsc ⁡ ( x )arcsin ⁡ ( 1 / x ) (x)=\arcsin(1/x)} , dan seterusnya.
x\leq 1\\\arcsin &\left({\sqrt {1-x^{2}}}\right)={\frac {\pi }{2}}-\operatorname {sgn}(x)\arcsin(x)\\\arctan(x)&=\arcsin \left({\frac {x}{\sqrt {1+x^{2}}}}\right)\\\operatorname {arccot}(x)&=\arccos \left({\frac {x}{\sqrt {1+x^{2}}}}\right)\end{aligned}}}
ln ⁡ ( | x + x 2 − 1 | ) + C \int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\operatorname {sgn}(x)\ln \left(\left|x+{\sqrt {x^{2}-1}}\right|\right)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\operatorname {sgn}(x)\ln \left(\left|x+{\sqrt {x^{2}-1}}\right|\right)+C\end{aligned}}} Nilai mutlak diperlukan untuk mengimbangi nilai negatif dan positif dari fungsi arcsekan dan arckosekan.
The derivatives for complex values of z are as follows:
Turunan untuk nilai kompleks z adalah sebagai berikut:
d d z arcsin ⁡ ( z )1 1 − z 2 ; z ≠ − 1 , + 1 d d z arccos ⁡ ( z )− 1 1 − z 2 ; z ≠ − 1 , + 1 d d z arctan ⁡ ( z )1 1 + z 2 ; z ≠ − i , + i d d z arccot ⁡ ( z )− 1 1 + z 2 ; z ≠ − i , + i d d z arcsec ⁡ ( z )1 z 2 1 − 1 z 2 ; z ≠ − 1 , 0 , + 1 d d z arccsc ⁡ ( z )− 1 z 2 1 − 1 z 2 ; z ≠ − 1 , 0 , + 1 {\frac {d}{dz}}\arcsin(z)&{}={\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arccos(z)&{}=-{\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arctan(z)&{}={\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arccot}(z)&{}=-{\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcsec}(z)&{}={\frac {1}{z^{2}{\sqrt {1-{\frac {1}{z^{2}}}}}}}\;;&z&{}\neq -1,0,+1\\{\frac {d}{dz}}\operatorname {arccsc}(z)&{}=-{\frac {1}{z^{2}{\sqrt {1-{\frac {1}{z^{2}}}}}}}\;;&z&{}\neq -1,0,+1\end{aligned}}} Only for real values of x: d d x arcsec ⁡ ( x )1 | x | x 2 − 1 ; | x | > 1 d d x arccsc ⁡ ( x )− 1 | x | x 2 − 1 ; | x | > 1 {\frac {d}{dx}}\operatorname {arcsec}(x)&{}={\frac {1}{|x|{\sqrt {x^{2}-1}}}}\;;&|x|>1\\{\frac {d}{dx}}\operatorname {arccsc}(x)&{}=-{\frac {1}{|x|{\sqrt {x^{2}-1}}}}\;;&|x|>1\end{aligned}}} These formulas can be derived in terms of the derivatives of trigonometric functions.
d d z arcsin ⁡ ( z )1 1 − z 2 ; z ≠ − 1 , + 1 d d z arccos ⁡ ( z )− 1 1 − z 2 ; z ≠ − 1 , + 1 d d z arctan ⁡ ( z )1 1 + z 2 ; z ≠ − i , + i d d z arccot ⁡ ( z )− 1 1 + z 2 ; z ≠ − i , + i d d z arcsec ⁡ ( z )1 z 2 1 − 1 z 2 ; z ≠ − 1 , 0 , + 1 d d z arccsc ⁡ ( z )− 1 z 2 1 − 1 z 2 ; z ≠ − 1 , 0 , + 1 {\frac {d}{dz}}\arcsin(z)&{}={\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arccos(z)&{}=-{\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arctan(z)&{}={\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arccot}(z)&{}=-{\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcsec}(z)&{}={\frac {1}{z^{2}{\sqrt {1-{\frac {1}{z^{2}}}}}}}\;;&z&{}\neq -1,0,+1\\{\frac {d}{dz}}\operatorname {arccsc}(z)&{}=-{\frac {1}{z^{2}{\sqrt {1-{\frac {1}{z^{2}}}}}}}\;;&z&{}\neq -1,0,+1\end{aligned}}} Hanya untuk nilai riil x :
z 2 n + 1 2 n + 1∑ n0 ∞ ( 2 n ) ! ( 2 n n ! ) 2 z 2 n + 1 2 n + 1 ; | z | ≤ 1 \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 \\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}{\frac {z^{2n+1}}{2n+1}}\\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n)!}{(2^{n}n!)^{2}}}{\frac {z^{2n+1}}{2n+1}}\,;\qquad |z|\leq 1\end{aligned}}} The Taylor series for arctangent is also known as the arctangent series or Gregory's series.
z 2 n + 1 2 n + 1∑ n0 ∞ ( 2 n ) ! ( 2 n n ! ) 2 z 2 n + 1 2 n + 1 ; | z | ≤ 1 \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 \\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}{\frac {z^{2n+1}}{2n+1}}\\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n)!}{(2^{n}n!)^{2}}}{\frac {z^{2n+1}}{2n+1}}\,;\qquad |z|\leq 1\end{aligned}}} arctan ⁡ ( z )
z − z 3 3 + z 5 5 − z 7 7 + ⋯∑ n0 ∞ ( − 1 ) n z 2 n + 1 2 n + 1 ; | z | ≤ 1 z ≠ i , − i }{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} Series for the other inverse trigonometric functions can be given in terms of these according to the relationships given above.
z − z 3 3 + z 5 5 − z 7 7 + ⋯∑ n0 ∞ ( − 1 ) n z 2 n + 1 2 n + 1 ; | z | ≤ 1 z ≠ i , − i }{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} Deret untuk fungsi trigonometri terbalik lainnya dapat diberikan dalam hal ini sesuai dengan hubungan yang diberikan di atas.
For example, arccos ⁡ ( x )π / 2 − arcsin ⁡ ( x ) , arccsc ⁡ ( x )
Sebagai contoh, arccos ⁡ ( x )π / 2 − arcsin ⁡ ( x ) , arccsc ⁡ ( x )arcsin ⁡ ( 1 / x ) (x)=\arcsin(1/x)} , dan seterusnya.
Another series is given by: 2 ( arcsin ⁡ ( x 2 ) ) 2∑ n1 ∞ x 2 n n 2 ( 2 n n ) .
Deret lainnya diberikan oleh: 2 ( arcsin ⁡ ( x 2 ) ) 2∑ n1 ∞ x 2 n n 2 ( 2 n n ) .
{2}}\right)\right)^{2}=\sum _{n=1}^{\infty }{\frac {x^{2n}}{n^{2}{\binom {2n}{n}}}}.} Leonhard Euler found a series for the arctangent that converges more quickly than its Taylor series: arctan ⁡ ( z )
{2}}\right)\right)^{2}=\sum _{n=1}^{\infty }{\frac {x^{2n}}{n^{2}{\binom {2n}{n}}}}.} Leonhard Euler menemukan deret untuk invers tangen yang menyatu lebih cepat daripada deret Taylor: arctan ⁡ ( z )z 1 + z 2 ∑ n0 ∞ ∏ k1 n 2 k z 2 ( 2 k + 1 ) ( 1 + z 2 ) .
z 1 + z 2 ∑ n0 ∞ ∏ k1 n 2 k z 2 ( 2 k + 1 ) ( 1 + z 2 ) .
z 2 n + 1 ( 1 + z 2 ) n + 1 .
{1+z^{2}}}\sum _{n=0}^{\infty }\prod _{k=1}^{n}{\frac {2kz^{2}}{(2k+1)(1+z^{2})}}.} (The term in the sum for n0 is the empty product, so is 1.)
{1+z^{2}}}\sum _{n=0}^{\infty }\prod _{k=1}^{n}{\frac {2kz^{2}}{(2k+1)(1+z^{2})}}.} (Suku dalam jumlah untuk n0 adalah produk kosong, jadi 1.) Atau, ini dapat dinyatakan sebagai arctan ⁡ ( z )
Alternatively, this can be expressed as arctan ⁡ ( z )
Deret lain untuk fungsi arctangent diberikan oleh arctan ⁡ ( z )
^{\infty }{\frac {2^{2n}(n!)^{2}}{(2n+1)!}}{\frac {z^{2n+1}}{(1+z^{2})^{n+1}}}.} Another series for the arctangent function is given by arctan ⁡ ( z )
^{\infty }{\frac {2^{2n}(n!)^{2}}{(2n+1)!}}{\frac {z^{2n+1}}{(1+z^{2})^{n+1}}}.}
i ∑ n1 ∞ 1 2 n − 1 ( 1 ( 1 + 2 i / z ) 2 n − 1 − 1 ( 1 − 2 i / z ) 2 n − 1 ) , ^{\infty }{\frac {1}{2n-1}}\left({\frac {1}{(1+2i/z)^{2n-1}}}-{\frac {1}{(1-2i/z)^{2n-1}}}\right),} where i− 1 }} is the imaginary unit.
i ∑ n1 ∞ 1 2 n − 1 ( 1 ( 1 + 2 i / z ) 2 n − 1 − 1 ( 1 − 2 i / z ) 2 n − 1 ) , ^{\infty }{\frac {1}{2n-1}}\left({\frac {1}{(1+2i/z)^{2n-1}}}-{\frac {1}{(1-2i/z)^{2n-1}}}\right),} dimana
z 1 + ( 1 z ) 2 3 − 1 z 2 + ( 3 z ) 2 5 − 3 z 2 + ( 5 z ) 2 7 − 5 z 2 + ( 7 z ) 2 9 − 7 z 2 + ⋱z 1 + ( 1 z ) 2 3 + ( 2 z ) 2 5 + ( 3 z ) 2 7 + ( 4 z ) 2 9 + ⋱ {1+{\cfrac {(1z)^{2}}{3-1z^{2}+{\cfrac {(3z)^{2}}{5-3z^{2}+{\cfrac {(5z)^{2}}{7-5z^{2}+{\cfrac {(7z)^{2}}{9-7z^{2}+\ddots }}}}}}}}}}={\frac {z}{1+{\cfrac {(1z)^{2}}{3+{\cfrac {(2z)^{2}}{5+{\cfrac {(3z)^{2}}{7+{\cfrac {(4z)^{2}}{9+\ddots }}}}}}}}}}} The second of these is valid in the cut complex plane.
z − z 3 3 + z 5 5 − z 7 7 + ⋯∑ n0 ∞ ( − 1 ) n z 2 n + 1 2 n + 1 ; | z | ≤ 1 z ≠ i , − i }{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} Deret untuk fungsi trigonometri terbalik lainnya dapat diberikan dalam hal ini sesuai dengan hubungan yang diberikan di atas.
The first was developed by Leonhard Euler, the second by Carl Friedrich Gauss utilizing the Gaussian hypergeometric series.
Yang pertama dikembangkan oleh Leonhard Euler; yang kedua oleh Carl Friedrich Gauss menggunakan deret hipergeometrik Gaussian.
+ ln ⁡ [ z ( 1 + z 2 − 1 z 2 ) ] + C \int \arcsin(z)\,dz&{}=z\,\arcsin(z)+{\sqrt {1-z^{2}}}+C\\\int \arccos(z)\,dz&{}=z\,\arccos(z)-{\sqrt {1-z^{2}}}+C\\\int \arctan(z)\,dz&{}=z\,\arctan(z)-{\frac {1}{2}}\ln \left(1+z^{2}\right)+C\\\int \operatorname {arccot}(z)\,dz&{}=z\,\operatorname {arccot}(z)+{\frac {1}{2}}\ln \left(1+z^{2}\right)+C\\\int \operatorname {arcsec}(z)\,dz&{}=z\,\operatorname {arcsec}(z)-\ln \left[z\left(1+{\sqrt {\frac {z^{2}-1}{z^{2}}}}\right)\right]+C\\\int \operatorname {arccsc}(z)\,dz&{}=z\,\operatorname {arccsc}(z)+\ln \left[z\left(1+{\sqrt {\frac {z^{2}-1}{z^{2}}}}\right)\right]+C\end{aligned}}}
+ ln ⁡ [ z ( 1 + z 2 − 1 z 2 ) ] + C \int \arcsin(z)\,dz&{}=z\,\arcsin(z)+{\sqrt {1-z^{2}}}+C\\\int \arccos(z)\,dz&{}=z\,\arccos(z)-{\sqrt {1-z^{2}}}+C\\\int \arctan(z)\,dz&{}=z\,\arctan(z)-{\frac {1}{2}}\ln \left(1+z^{2}\right)+C\\\int \operatorname {arccot}(z)\,dz&{}=z\,\operatorname {arccot}(z)+{\frac {1}{2}}\ln \left(1+z^{2}\right)+C\\\int \operatorname {arcsec}(z)\,dz&{}=z\,\operatorname {arcsec}(z)-\ln \left[z\left(1+{\sqrt {\frac {z^{2}-1}{z^{2}}}}\right)\right]+C\\\int \operatorname {arccsc}(z)\,dz&{}=z\,\operatorname {arccsc}(z)+\ln \left[z\left(1+{\sqrt {\frac {z^{2}-1}{z^{2}}}}\right)\right]+C\end{aligned}}} untuk riil x ≥ 1: ∫ arcsec ⁡ ( x ) d xx arcsec ⁡ ( x ) − ln ⁡ ( x + x 2 − 1 ) + C ∫ arccsc ⁡ ( x ) d xx arccsc ⁡ ( x ) + ln ⁡ ( x + x 2 − 1 ) + C \int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\end{aligned}}} untuk semua riil x tidak antara -1 dan 1: ∫ arcsec ⁡ ( x ) d xx arcsec ⁡ ( x ) − sgn ⁡ ( x )
For real x ≥ 1: ∫ arcsec ⁡ ( x ) d xx arcsec ⁡ ( x ) − ln ⁡ ( x + x 2 − 1 ) + C ∫ arccsc ⁡ ( x )
d xx arcsec ⁡ ( x ) − arcosh ⁡ ( | x | ) + C ∫ arccsc ⁡ ( x )
d xx arccsc ⁡ ( x )
d xx arcsin ⁡ ( x )
+ ln ⁡ ( x + x 2 − 1 ) + C \int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\end{aligned}}}
ln ⁡ ( | x + x 2 − 1 | ) + C \int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\operatorname {sgn}(x)\ln \left(\left|x+{\sqrt {x^{2}-1}}\right|\right)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\operatorname {sgn}(x)\ln \left(\left|x+{\sqrt {x^{2}-1}}\right|\right)+C\end{aligned}}} Nilai mutlak diperlukan untuk mengimbangi nilai negatif dan positif dari fungsi arcsekan dan arckosekan.
d xx arcsec ⁡ ( x ) − sgn ⁡ ( x )
d xx arcsin ⁡ ( x )
ln ⁡ | x + x 2 − 1 | + C ∫ arccsc ⁡ ( x ) d xx arccsc ⁡ ( x ) + sgn ⁡ ( x )
ln ⁡ ( | x + x 2 − 1 | ) + C ∫ arccsc ⁡ ( x ) d xx arccsc ⁡ ( x ) + sgn ⁡ ( x )
ln ⁡ | x + x 2 − 1 | + C \int \operatorname {arcsec}(x)\,dx&{}=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\end{aligned}}} The absolute value is necessary to compensate for both negative and positive values of the arcsecant and arccosecant functions.
ln ⁡ ( | x + x 2 − 1 | ) + C \int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\operatorname {sgn}(x)\ln \left(\left|x+{\sqrt {x^{2}-1}}\right|\right)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\operatorname {sgn}(x)\ln \left(\left|x+{\sqrt {x^{2}-1}}\right|\right)+C\end{aligned}}} Nilai mutlak diperlukan untuk mengimbangi nilai negatif dan positif dari fungsi arcsekan dan arckosekan.
The signum function is also necessary due to the absolute values in the derivatives of the two functions, which create two different solutions for positive and negative values of x. These can be further simplified using the logarithmic definitions of the inverse hyperbolic functions: ∫ arcsec ⁡ ( x )
Fungsi signum juga diperlukan karena nilai mutlak dalam turunan dari kedua fungsi tersebut, yang membuat dua solusi berbeda untuk nilai positif dan negatif x. Ini dapat disederhanakan lebih lanjut menggunakan definisi logaritmik dari fungsi hiperbolik invers: ∫ arcsec ⁡ ( x )
d xx arccsc ⁡ ( x ) + arcosh ⁡ ( | x | ) + C \int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\operatorname {arcosh} (|x|)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\operatorname {arcosh} (|x|)+C\\\end{aligned}}} The absolute value in the argument of the arcosh function creates a negative half of its graph, making it identical to the signum logarithmic function shown above.
d xx arccsc ⁡ ( x ) + arcosh ⁡ ( | x | ) + C \int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\operatorname {arcosh} (|x|)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\operatorname {arcosh} (|x|)+C\\\end{aligned}}} Nilai mutlak dalam argumen fungsi arcosh menciptakan setengah negatif grafiknya, membuatnya identik dengan fungsi logaritmik signum yang ditunjukkan di atas.
Using ∫ u d vu v − ∫ v d u (i.e. integration by parts), set uarcsin ⁡ ( x )
Menggunakan ∫ u d vu v − ∫ v d u (yaitu integrasi oleh bagian), uarcsin ⁡ ( x )
d vd x d ud x 1 − x 2 vx u&=\arcsin(x)&dv&=dx\\du&={\frac {dx}{\sqrt {1-x^{2}}}}&v&=x\end{aligned}}} Then ∫ arcsin ⁡ ( x )
d vd x d ud x 1 − x 2 vx u&=\arcsin(x)&dv&=dx\\du&={\frac {dx}{\sqrt {1-x^{2}}}}&v&=x\end{aligned}}} Kemudian ∫ arcsin ⁡ ( x )
d xx arcsin ⁡ ( x ) − ∫ x 1 − x 2 d x , {\sqrt {1-x^{2}}}}\,dx,} which by the simple substitution w1 − x 2 , d w− 2 x d x ,\ dw=-2x\,dx} yields the final result: ∫ arcsin ⁡ ( x )
d xx arcsin ⁡ ( x ) − ∫ x 1 − x 2 d x , {\sqrt {1-x^{2}}}}\,dx,} yang dengan sederhana substitusi w1 − x 2 ,
π 2 − arcsin ⁡ ( z ) z ≠ − 1 , + 1 {2}}-\arcsin(z)\quad z\neq -1,+1} which has the same cut as arcsin; arccot ⁡ ( z )π 2 − arctan ⁡ ( z ) z ≠ − i ,
d zz arcsin ⁡ ( z ) + 1 − z 2 + C ∫ arccos ⁡ ( z ) d zz arccos ⁡ ( z ) − 1 − z 2 + C ∫ arctan ⁡ ( z )
i ln ⁡ ( 1 − z 2 − i z )arccsc ⁡ ( 1 z ) arccos ⁡ ( z )− i ln ⁡ ( z + i 1 − z 2 )
d zz arctan ⁡ ( z ) − 1 2 ln ⁡ ( 1 + z 2 ) + C ∫ arccot ⁡ ( z )
i ln ⁡ ( 1 − 1 z 2 − i z )arcsin ⁡ ( 1 z ) \arcsin(z)&{}=-i\ln \left({\sqrt {1-z^{2}}}+iz\right)=i\ln \left({\sqrt {1-z^{2}}}-iz\right)&{}=\operatorname {arccsc} \left({\frac {1}{z}}\right)\\[10pt]\arccos(z)&{}=-i\ln \left(z+i{\sqrt {1-z^{2}}}\right)=i\ln \left(z-i{\sqrt {1-z^{2}}}\right)={\frac {\pi }{2}}-\arcsin(z)&{}=\operatorname {arcsec} \left({\frac {1}{z}}\right)\\[10pt]\arctan(z)&{}=-{\frac {i}{2}}\ln \left({\frac {i-z}{i+z}}\right)=-{\frac {i}{2}}\ln \left({\frac {1+iz}{1-iz}}\right)&{}=\operatorname {arccot} \left({\frac {1}{z}}\right)\\[10pt]\operatorname {arccot}(z)&{}=-{\frac {i}{2}}\ln \left({\frac {z+i}{z-i}}\right)=-{\frac {i}{2}}\ln \left({\frac {iz-1}{iz+1}}\right)&{}=\arctan \left({\frac {1}{z}}\right)\\[10pt]\operatorname {arcsec}(z)&{}=-i\ln \left({\frac {1}{z}}+i{\sqrt {1-{\frac {1}{z^{2}}}}}\right)=i\ln \left({\frac {1}{z}}-i{\sqrt {1-{\frac {1}{z^{2}}}}}\right)={\frac {\pi }{2}}-\operatorname {arccsc}(z)&{}=\arccos \left({\frac {1}{z}}\right)\\[10pt]\operatorname {arccsc}(z)&{}=-i\ln \left({\sqrt {1-{\frac {1}{z^{2}}}}}+{\frac {i}{z}}\right)=i\ln \left({\sqrt {1-{\frac {1}{z^{2}}}}}-{\frac {i}{z}}\right)&{}=\arcsin \left({\frac {1}{z}}\right)\end{aligned}}} Because all of the inverse trigonometric functions output an angle of a right triangle, they can be generalized by using Euler's formula to form a right triangle in the complex plane.
+ ln ⁡ [ z ( 1 + z 2 − 1 z 2 ) ] + C \int \arcsin(z)\,dz&{}=z\,\arcsin(z)+{\sqrt {1-z^{2}}}+C\\\int \arccos(z)\,dz&{}=z\,\arccos(z)-{\sqrt {1-z^{2}}}+C\\\int \arctan(z)\,dz&{}=z\,\arctan(z)-{\frac {1}{2}}\ln \left(1+z^{2}\right)+C\\\int \operatorname {arccot}(z)\,dz&{}=z\,\operatorname {arccot}(z)+{\frac {1}{2}}\ln \left(1+z^{2}\right)+C\\\int \operatorname {arcsec}(z)\,dz&{}=z\,\operatorname {arcsec}(z)-\ln \left[z\left(1+{\sqrt {\frac {z^{2}-1}{z^{2}}}}\right)\right]+C\\\int \operatorname {arccsc}(z)\,dz&{}=z\,\operatorname {arccsc}(z)+\ln \left[z\left(1+{\sqrt {\frac {z^{2}-1}{z^{2}}}}\right)\right]+C\end{aligned}}} untuk riil x ≥ 1: ∫ arcsec ⁡ ( x ) d xx arcsec ⁡ ( x ) − ln ⁡ ( x + x 2 − 1 ) + C ∫ arccsc ⁡ ( x ) d xx arccsc ⁡ ( x ) + ln ⁡ ( x + x 2 − 1 ) + C \int \operatorname {arcsec}(x)\,dx&{}=x\,\operatorname {arcsec}(x)-\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\\\int \operatorname {arccsc}(x)\,dx&{}=x\,\operatorname {arccsc}(x)+\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\end{aligned}}} untuk semua riil x tidak antara -1 dan 1: ∫ arcsec ⁡ ( x ) d xx arcsec ⁡ ( x ) − sgn ⁡ ( x )
Im ⁡ ( ln ⁡ ( z + z 2 − 1 ) ) arctan ⁡ ( z ) 1 z 1 + z 2 − i ln ⁡ ( 1 + i z 1 + z 2 )− i 2 ln ⁡ ( i − z i + z )
d zz arctan ⁡ ( z ) − 1 2 ln ⁡ ( 1 + z 2 ) + C ∫ arccot ⁡ ( z )
z 1 z 2 + 1 − i ln ⁡ ( z + i z 2 + 1 )− i 2 ln ⁡ ( z + i z − i ) Im ⁡ ( ln ⁡ ( z + i ) ) arcsec ⁡ ( z ) 1 z 2 − 1 z − i ln ⁡ ( 1 + i z 2 − 1 z )
d zz arccot ⁡ ( z ) + 1 2 ln ⁡ ( 1 + z 2 ) + C ∫ arcsec ⁡ ( z ) d zz arcsec ⁡ ( z ) − ln ⁡ [ z ( 1 + z 2 − 1 z 2 ) ] + C ∫ arccsc ⁡ ( z )