Question

Find the integrals of the function: \(sin^{-1} (cos x)\)

Answer 1

sin^-1 (cos x)
Let cos x = t
Then, sin x = \(\sqrt {1-t^2}\)
⇒ (-sin x)dx = dt
dx = -\(\frac {dt}{sin\  x}\)
dx = -\(\frac {dt}{\sqrt {1-t^2}}\)
∴ ∫sin^-1(cos x)dx = ∫sin^-1t\((-\frac {dt}{\sqrt {1-t^2}})\)
= - ∫\(\frac {sin^{-1}t}{\sqrt{1-t^2 }}dt\)
Let sin^-1t = u
⇒ \((\frac {1}{\sqrt {1-t^2}})\) = du
∴ ∫sin^-1(cos x)dx = ∫4du
= - \(\frac {u^2}{2}\)+ C
= -\(\frac {(sin^{-1}t)^2}{2}\) + C
= -\([\frac {(sin^{-1}(cos\ x)^2}{2}]\) + C         …....(1)
It is known that, 
sin^-1 x + cos^-1 x = \(\frac \pi2\)
∴ sin^-1(cos x)= \(\frac \pi2\) - cos^-1(cos x) =  \((\frac \pi2-x)\)
Substituting in equation (1), we obtain
∫sin^-1(cos x)dx = -\(\frac {[\frac \pi2-x]^2}{2}\) + C
= -\(\frac 12(\frac {\pi^2}{2}+x^2-\pi x)+C\)
= -\(\frac {\pi^2}{8}-\frac {x^2}{2}+\frac 12πx+C\)
= \(\frac {\pi x}{2}-\frac {x^2}{2} +(C-\frac {\pi^2}{8})\)
= \(\frac {\pi x}{2}-\frac {x^2}{2} +C_1\)