Evaluate $\int_0^1 \cos^{-1} x\ dx$ by first finding $\frac{d}{dx}(x\cos^{-1} x)$ Question
Evaluate $$\int_0^1 \cos^{-1} x\ dx$$ by first finding the value of $$\frac{d}{dx}(x\cos^{-1} x).$$
My Working
As the question said to evaluate $$\frac{d}{dx}(x\cos^{-1} x),$$ I used the product rule to differentiate. We first have to let $u=x$ and $v=\cos^{-1} x$, so
\begin{align}
& \quad\frac{d}{dx}(x\cos^{-1} x)\\
&=u^\prime v+v^\prime u\\
&=1\cdot \cos^{-1}x + x\cdot\frac{-1}{\sqrt{1-x^2}}\\
&=\cos^{-1}x-\frac{x}{\sqrt{1-x^2}}
\end{align}
Unfortunately, after that, I have no idea about how to proceed, but I think that we should go somewhere from
$$\int\left[\frac{d}{dx}(x\cos^{-1}x)\right]\ dx+\int\frac{x}{\sqrt{1-x^2}}\ dx=\int \cos^{-1}x???$$
Thank you for your help!

After seeing the helpful comments, I can start using $u$-substitution for the solution. Let $u=1-x^2$, then
\begin{align}
\frac{du}{dx}&=-2x\\
du&=-2x\ dx\\
\int_0^1\frac{x}{\sqrt{1-x^2}}&=-\frac{1}{2}\int^1_0\frac{-2x}{\sqrt{1-x^2}}\ dx\\
&=-\frac{1}{2}\int^1_0 \frac{1}{\sqrt{u}}\ du\\
&=-\frac{1}{2}\int^1_0 u^{-\frac{1}{2}}\ du\\
&=-\frac{1}{2}\left[2u^\frac{1}{2}\right]^1_0\\
&=-\frac{1}{2}\left[2(1-x^2)^\frac{1}{2}\right]^1_0\\
&=-\frac{1}{2}\left(2\sqrt{1-1^2}-2\sqrt{1-0^2}\right)\\
&=-\frac{1}{2}\cdot-2\\
&=1\\
\int^1_0\left[\frac{d}{dx}(x\cos^{-1}x)\right]\ dx&=\left[x\cos^{-1}x\right]^1_0\\
&=0\\
\int^1_0\cos^{-1}x\ dx&= 0+1=1
\end{align}
