Find the upper bound of the definite integral $\int_{0}^{b} \pi\ - \arccos (x-1) dx$ so that it equals $\frac {\pi}{2}$.

Question

Find the upper bound of the definite integral $\int_{0}^{b} \pi\ - \arccos (x-1) dx$ so that it equals $\frac {\pi}{2}$.

We have that $$\int_0^b \pi - \arccos(x-1) dx= \left[{\pi}x-\left(x-1\right) \arccos\left(x-1\right)+\sqrt{1-\left(x-1\right)^2}\right]_0^b$$ When plugging in $0$, we get $$\pi(0) - [(-1) \arccos(-1)] + \sqrt{1 - 1}=0 + \pi + 0=\pi$$ Thus, $$\int_0^b \pi - \arccos(x-1) dx={\pi}b-\left(b-1\right) \arccos\left(b-1\right)+\sqrt{1-\left(b-1\right)^2} - \pi$$ Now if $\int_{0}^{b} \pi\ - \arccos (x-1) dx$ $=$ $\frac {\pi}{2}$, then $${\pi}b-\left(b-1\right) \arccos\left(b-1\right)+\sqrt{1-\left(b-1\right)^2} - \pi = \frac{\pi}{2}$$ that is $${\pi}b-\left(b-1\right) \arccos\left(b-1\right)+\sqrt{1-\left(b-1\right)^2} = \frac{3\pi}{2}.$$ And here is where I am stuck: how do I find $b$?

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Update: So I just used the equation solver on my calculator and it gave me $b = 1.3286741629086$. However, I need an exact answer, so the answer must have involved $\pi$ or square roots somehow.

Answer 1

$$\mathcal{I}\left(\text{b}\right)=\int_0^\text{b}\left(\pi-\arccos\left(x-1\right)\right)\space\text{d}x=\pi\int_0^\text{b}1\space\text{d}\text{b}-\int_0^\text{b}\arccos\left(x-1\right)\space\text{d}x$$

Now:

1. $$\pi\int_0^\text{b}1\space\text{d}x=\pi\left[x\right]_0^\text{b}=\pi\left(\text{b}-0\right)=\text{b}\pi$$
2. Substitute $u=x-1$ and $\text{d}u=\text{d}x$ $$\int_0^\text{b}\arccos\left(x-1\right)\space\text{d}x=\int_{-1}^{\text{b}-1}\arccos\left(u\right)\space\text{d}u$$
3. Using integration by parts: $$\int_{-1}^{\text{b}-1}\arccos\left(u\right)\space\text{d}u=\left[u\arccos\left(u\right)\right]_{-1}^{\text{b}-1}+\int_{-1}^{\text{b}-1}\frac{u}{\sqrt{1-u^2}}\space\text{d}u=$$ $$\pi+\left(\text{b}-1\right)\arccos\left(\text{b}-1\right)+\int_{-1}^{\text{b}-1}\frac{u}{\sqrt{1-u^2}}\space\text{d}u$$
4. Subsitute $s=1-u^2$ and $\text{d}s=-2u\space\text{d}u$: $$\int_{-1}^{\text{b}-1}\frac{u}{\sqrt{1-u^2}}\space\text{d}u=-\frac{1}{2}\int_0^{1-\left(\text{b}-1\right)^2}\frac{1}{\sqrt{s}}\space\text{d}s=-\left[\sqrt{s}\right]_0^{1-\left(\text{b}-1\right)^2}=-\sqrt{-\text{b}\left(\text{b}-2\right)}$$

So, we get:

$$\mathcal{I}\left(\text{b}\right)=\text{b}\pi-\left(\pi+\left(\text{b}-1\right)\arccos\left(\text{b}-1\right)-\sqrt{-\text{b}\left(\text{b}-2\right)}\right)=\frac{\pi}{2}$$

Finding an exact answer is really hard, an I think that you'll not find one.