# Integrating $\int _0^1\frac{\ 1}{\sqrt{\ 1-x^2}}\sin ^{-1}\left(2x\sqrt{\ 1-x^2}\ \right)dx$

I'm sorry if this is a simple question. if anyone can correct me, I'd greatly appreciate it. So first i used substitution $$x = sin\theta$$ so we have

$$\int _0^{\frac{\pi }{2}}\frac{\ 1}{\sqrt{\ 1-\sin ^2\theta }}\sin ^{-1}\left(2\sin \theta \cos \theta \ \right)\cos \theta d\theta$$

$$\int _0^{\frac{\pi }{2}}\ \sin ^{-1}\left(\sin 2\theta \right)d\theta$$ cancelling out $$\sin^{-1}(\sin 2\theta)$$ we are only left with $$2\theta$$ so aren't we supposed to get $$\int _0^{\frac{\pi }{2}}\ 2\theta d\theta$$ which leaves us with $$2\left[\frac{\ \theta ^2}{2}\right]$$ for bound from 0 till $$\frac{\pi }{2}$$ so I get $$\frac{\pi ^2}{4}$$ but my answer is wrong according to my text book it says it should be $$\frac{\pi ^2}{8}$$ but how? where did i go wrong?

I even shaded the area we are interested in finding area of

• In $\int _0^{\frac{\pi }{2}}\ \sin ^{-1}\left(\sin 2\theta \right)d\theta$ perform the change of variable $x=2\theta$ and look at the interval of definition of $arcsin$ function.
– FDP
Oct 5 '20 at 8:32
• Note that $\arcsin(\sin x)=x$ hold true for all $0\leq x\leq \frac{\pi}{2}$ however, $0\leq 2x \leq \pi$ so the $\arcsin(\sin2x)\neq 2x$ Oct 5 '20 at 8:33
• Ohhh Noted @Naren Thank you.
– RiRi
Oct 5 '20 at 8:41

Your error is in your attempt to simplify $$\sin^{-1}\bigl(\sin(2\theta)\bigr)$$ to $$2\theta$$.

For $$0\le\theta\le{\large{\frac{\pi}{2}}}$$, we get $$\sin^{-1}\bigl(\sin(2\theta)\bigr) = \begin{cases} 2\theta&\text{if}\;0\le\theta\le\frac{\pi}{4}\\[4pt] \pi-2\theta&\text{if}\;\frac{\pi}{4}\le\theta\le\frac{\pi}{2}\\ \end{cases}$$ so you need to sum two integrals, one for each of the above cases, yielding $$\int_0^{\large{\frac{\pi}{4}}} (2\theta)\,d\theta + \int_{\large{\frac{\pi}{4}}}^{\large{\frac{\pi}{2}}} (\pi-2\theta)\,d\theta = \frac{\pi^2}{16} + \frac{\pi^2}{16} = \frac{\pi^2}{8}$$

In $$\displaystyle \int _0^{\frac{\pi }{2}}\ \sin ^{-1}\left(\sin 2\theta \right)d\theta$$ perform the change of variable $$x=2\theta$$ and look at the interval of definition of $$\displaystyle \arcsin$$ function.

If you plot the integrand $$f(x) = \frac{\sin^{-1} 2x \sqrt{1-x^2}}{\sqrt{1-x^2}}$$ on $$x \in [0,1]$$, you will discover that there is a cusp around $$x = 0.7$$. Here is a picture:

What's going on? Well, on $$[0,1]$$, the function $$\sqrt{1-x^2}$$ is smooth, so that isn't the issue. This suggests that there is something going on with the inverse sine. Specifically, how do the plots of $$g(\theta) = \sin^{-1} (\sin 2\theta), \quad h(\theta) = 2\theta$$ compare? If $$\theta = \pi/4$$, then $$\sin 2\theta = 1$$, the local maximum. So when we take the inverse sine, we don't get $$2\theta$$ back; we get an angle in $$[-\pi/4, \pi/4]$$.

So to handle this issue, we have to be a bit more careful with the substitution. On $$\theta \in [0,\pi/4]$$, which corresponds to $$x \in [0, 1/\sqrt{2}]$$ (because the substitution is $$x = \sin \theta$$, hence $$\theta = \pi/4$$ means $$x = 1/\sqrt{2}$$), the integrand does indeed simplify to $$\int_{x=0}^{1/\sqrt{2}} f(x) \, dx = \int_{\theta = 0}^{\pi/4} 2\theta \, d\theta.$$ But on $$x \in [1/\sqrt{2}, 1]$$, we have instead $$\int_{x=1/\sqrt{2}}^1 f(x) \, dx = \int_{\theta = \pi/4}^{\pi/2} 2(\pi/2- \theta) \, d\theta.$$ These two pieces are equal; their sum is $$\pi^2/8$$ as claimed.

Unfortunately the arcsin is a function and will only output values that are less than or equal to $$\frac{\pi}{2}$$. In the case of $$\sin^{-1}(\sin(2\theta))$$ for $$0 \le \theta \le \frac{\pi}{2}$$ the cancellation that you made of $$\sin^{-1}(\sin(2\theta))=2\theta$$ is invalid. As you can see on the graph: For this you must integrate from $$0 \le \theta \le \frac{\pi}{4}$$ and multiply the answer by two.

Not the most rigours explanation but hope this helps out a bit.

$$I=\int_{0}^{\pi/2} \frac{dx}{\sqrt{1-x^2}} \sin^{-1} 2x\sqrt{1-x^2}$$ Let $$x=\sin t, t=2u$$ then $$I=\frac{1}{2}\int_{0}^{\pi} \sin^{-1}\sin u du =\int_{0}^{\pi/2} \frac{u}{2}+\int_{\pi/2}^{\pi} \frac{\pi-u}{2}du =\frac{\pi^2}{8}.$$