Categoria: formulae.app / Matemáticas / Cálculo Integral / Fórmulas Trigonométricas Inversas
$$\int sin^{-1}u\:du=u\:sin^{-1}u+\sqrt{1-u^2}+C$$
$$\int cos^{-1}u\:du=u\:cos^{-1}u-\sqrt{1-u^2}+C$$
$$\int tan^{-1}u\:du=u\:tan^{-1}u-\frac{1}{2}ln(1+u^2)+C$$
$$\int u\:sin^{-1}u\:du=\frac{2u^2-1}{4}sin^{-1}u+\frac{u\sqrt{1-u^2}}{4}+C$$
$$\int u\:cos^{-1}u\:du=\frac{2u^2-1}{4}cos^{-1}u-\frac{u\sqrt{1-u^2}}{4}+C$$
$$\int u\:tan^{-1}u\:du=\frac{u^2+1}{2}tan^{-1}u-\frac{u}{2}+C$$
$$\int u^n sin^{-1}u\:du=\frac{1}{n+1}[u^{n+1}sin^{-1}u-\int \frac{u^{n+1}du}{\sqrt{1-u^2}}],n \neq 1$$
$$\int u^n cos^{-1}u\:du=\frac{1}{n+1}[u^{n+1}cos^{-1}u+\int \frac{u^{n+1}du}{\sqrt{1-u^2}}],n \neq 1$$
$$\int u^n tan^{-1}u\:du=\frac{1}{n+1}[u^{n+1}tan^{-1}u-\int \frac{u^{n+1}du}{\sqrt{1+u^2}}],n \neq 1$$