Question

\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]

Answer 1

\[I = \int_0^1 ( \cos^{- 1} x )^2 d x\]

\[\text{let }co s^{- 1} x = \theta\]

\[ \Rightarrow x = \cos\theta\]

\[ \Rightarrow dx = - \sin\theta d\theta\]

\[\text{when }x = 0, \theta = \frac{\pi}{2}\text{ and when }x = 1, \theta = 0\]

\[\text{Therefore, }I = \int_\frac{\pi}{2}^0 \theta^2 ( - \sin\theta) d \theta \]

\[I = - \int_\frac{\pi}{2}^0 \theta^2 (sin\theta) d \theta\]

\[I = \int_0^\frac{\pi}{2} \theta^2 (sin\theta) d \theta\]

\[I = \left[ \theta^2 ( - cos\theta) \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2\theta \int_0^\frac{\pi}{2} \sin\theta d \theta\]

\[I = \left[ \theta^2 ( - \cos\theta) \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2\theta( - \cos\theta) d \theta\]

\[= [ - \theta^2 \cos\theta ]_0^\frac{\pi}{2} + \int_0^\frac{\pi}{2} 2\theta(\cos\theta)d\theta\]

\[ = [ - \theta^2 cos\theta ]_0^\frac{\pi}{2} + 2[\theta\sin\theta - \int_0^\frac{\pi}{2} \sin\theta d\theta]\]

\[ = [ - \theta^2 cos\theta ]_0^\frac{\pi}{2} + 2[\theta sin\theta + \cos\theta ]_0^\frac{\pi}{2} \]

\[I = 2\left[\left(\frac{\pi}{2} + 0\right) - 1\right] \]

\[I = \pi - 2\]