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
= -$\frac 12(\frac {\pi^2}{2}+x^2-\pi x)+C$
= -$\frac {\pi^2}{8}-\frac {x^2}{2}+\frac 12πx+C$
= $\frac {\pi x}{2}-\frac {x^2}{2} +(C-\frac {\pi^2}{8})$
= $\frac {\pi x}{2}-\frac {x^2}{2} +C_1$