Question

$\int {\cos }^{-1}xdx$

Answer 1

$\int {\cos }^{-1}xdxPutx=\cos tHence,dx=-\sin t-\int {\cos }^{-1}\left(\cos t\right)\sin tdt-\int t\sin tdt-[t\int \sin tdt-\int \frac{dt}{dt}\int \sin tdtdt]+CwhereCisintegratingconstant.-[-t\cos t-\int -\cos tdt]+C-[-t\cos t+\sin t]+Ct\cos t-\sin t+CNowputt={\cos }^{-1}xinaboveequation{\cos }^{-1}x\cos \left({\cos }^{-1}x\right)-\sin \left({\cos }^{-1}x\right)+Cx{\cos }^{-1}x-\sqrt{1-{\cos }^2\left({\cos }^{-1}x\right)}+C\because \sin \theta =\sqrt{1-{\cos }^2\theta }x{\cos }^{-1}x-\sqrt{1-x^2}+C$