Question

$\underset{0}{\overset{1}{\int }}{\left({\cos }^{-1}x\right)}^{2}dx$

Answer 1

Solution :-

$I={\int }_0^1(co{s}^{-1}x{)}^{2}dx$

Let $co{s}^{-1}x=y$

$\Rightarrow x=cosy$

$\Rightarrow dx=-sinydy$

$I={\int }_0^1-y^{2}sin\left(y\right)dy$

Now integrating by parts, assume $u=y^2$

$\Rightarrow du=2ydy$

and $dv=-sin\left(y\right)dy$

$\Rightarrow v=cos\left(y\right)$

Now $\int udv=uv=\int vdu$

$I=[y^{2}cosy{]}_0^1-2{\int }_0^{1}ycos(y)dy$

$I=cos\left(1\right)-2{\int }_0^{1}ycos\left(y\right)dy$

$=cos\left(1\right)-2\left[ysin\right(y)+cosy{]}_0^1$

$=cos\left(1\right)-2sin\left(1\right)-cos1+2.$

$=2-2sin\left(1\right)$