Home/Calculus II/A.3. Inverse trigonometric integrals/FormulaA.3. Inverse trigonometric integralsFormulaIn A.3. Inverse trigonometric integralsFormulaCopy LaTeX∫u arctanu du=u2+12 arctanu−u2+C\int u\,\arctan u\,du = \frac{u^2 + 1}{2}\,\arctan u - \frac{u}{2} + C∫uarctanudu=2u2+1arctanu−2u+CTagsA 3 integrals trigonometric inverseDifferential calculusDerivativeDerivative formulaTableVariablesCCCConstante real de integraciónuuuVariable auxiliar de sustitucióndududuDiferencial respecto de uuuFound an error? Contact us.Related formulasRelatedA.3.2 Product with uuu∫u arcsinu du=2u2−14 arcsinu+u1−u24+C\int u\,\arcsin u\,du = \frac{2u^2 - 1}{4}\,\arcsin u + \frac{u\sqrt{1 - u^2}}{4} + C∫uarcsinudu=42u2−1arcsinu+4u1−u2+COpen formulaOpen in sectionRelatedA.3.3 General powers (n≠−1n \neq -1n=−1)∫un arcsinu du=1n+1[un+1 arcsinu−∫un+1 du1−u2]\int u^n\,\arcsin u\,du = \frac{1}{n+1}\left[u^{n+1}\,\arcsin u - \int \frac{u^{n+1}\,du}{\sqrt{1 - u^2}}\right]∫unarcsinudu=n+11[un+1arcsinu−∫1−u2un+1du]Open formulaOpen in sectionRelatedFormula∫arctanu du=u arctanu−12 ln(1+u2)+C\int \arctan u\,du = u\,\arctan u - \frac{1}{2}\,\ln\left(1 + u^2\right) + C∫arctanudu=uarctanu−21ln(1+u2)+COpen formulaOpen in sectionRelatedFormula∫un arccosu du=1n+1[un+1 arccosu+∫un+1 du1−u2]\int u^n\,\arccos u\,du = \frac{1}{n+1}\left[u^{n+1}\,\arccos u + \int \frac{u^{n+1}\,du}{\sqrt{1 - u^2}}\right]∫unarccosudu=n+11[un+1arccosu+∫1−u2un+1du]Open formulaOpen in sectionRelatedFormula∫arccosu du=u arccosu−1−u2+C\int \arccos u\,du = u\,\arccos u - \sqrt{1 - u^2} + C∫arccosudu=uarccosu−1−u2+COpen formulaOpen in sectionRelatedFormula∫un arctanu du=1n+1[un+1 arctanu−∫un+1 du1+u2]\int u^n\,\arctan u\,du = \frac{1}{n+1}\left[u^{n+1}\,\arctan u - \int \frac{u^{n+1}\,du}{1 + u^2}\right]∫unarctanudu=n+11[un+1arctanu−∫1+u2un+1du]Open formulaOpen in sectionRelatedA.3.1 Direct integrals∫arcsinu du=u arcsinu+1−u2+C\int \arcsin u\,du = u\,\arcsin u + \sqrt{1 - u^2} + C∫arcsinudu=uarcsinu+1−u2+COpen formulaOpen in section