Koriste se supstitucija ili drugi oblici algebarskih manipulacija kako bi se dosegli integrali izlistani u tablici.
∫ arcsin x d x = x arcsin x + 1 − x 2 + C {\displaystyle \int \arcsin x\,dx=x\arcsin x+{\sqrt {1-x^{2}}}+C}
∫ arccos x d x = x arccos x − 1 − x 2 + C {\displaystyle \int \arccos x\,dx=x\arccos x-{\sqrt {1-x^{2}}}+C}
∫ arctg x d x = x arctg x − 1 2 ln | 1 + x 2 | + C {\displaystyle \int \operatorname {arctg} x\,dx=x\operatorname {arctg} x-{\frac {1}{2}}\ln |1+x^{2}|+C}
∫ arccsc x d x = x arccsc x + ln | x + x x 2 − 1 x 2 | + C {\displaystyle \int \operatorname {arccsc} x\,dx=x\operatorname {arccsc} x+\ln \left|x+x{\sqrt {{x^{2}-1} \over x^{2}}}\right|+C}
∫ arcsec x d x = x arcsec x − ln | x + x x 2 − 1 x 2 | + C {\displaystyle \int \operatorname {arcsec} x\,dx=x\operatorname {arcsec} x-\ln \left|x+x{\sqrt {{x^{2}-1} \over x^{2}}}\right|+C}
∫ arcctg x d x = x arcctg x + 1 2 ln | 1 + x 2 | + C {\displaystyle \int \operatorname {arcctg} x\,dx=x\operatorname {arcctg} x+{\frac {1}{2}}\ln |1+x^{2}|+C}
