# Where have I made a mistake?

I have been trying the integral below, but cannot get the right answer and I am not sure where I have gone wrong. Let $$I=\int_0^{1/2}\arcsin(\sqrt{x})dx$$ and make the substitution $\sqrt{x}=\sin(u)$ so $dx=2\sin(u)\cos(u)du$. Now, \begin{align} I &= \int_0^{\pi/4}2u\sin(u)\cos(u)du = \left[u\sin^2(u)\right]_0^{\pi/4}-\int_0^{\pi/4}\sin^2(u)du \\ &= \frac{\pi}{8}-\frac{1}{2}\int_0^{\pi/4}(1-\cos(2u)) \, du \\ &= \frac{\pi}{8}-\left[\frac{u}{2}-\frac{\sin(2u)}{4}\right]_0^{\pi/4} = \frac{1}{4} \end{align}

• That's the same answer Wolfram Alpha gets. – saulspatz Sep 15 '18 at 22:23
• Hmmm. The answer book I checked it against said something different but my calculator also gave this answer, I presumed it was something I had typed in wrong – Henry Lee Sep 15 '18 at 22:24
• Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Sep 15 '18 at 22:25
• What does the answer book say? – Shaun Sep 15 '18 at 22:26
• Which book are you using? – Shaun Sep 15 '18 at 22:26

Let us see: $$\begin{eqnarray*}\int_{0}^{1/2}\arcsin\sqrt{x}\,dx&\stackrel{x\mapsto z^2}{=}&\int_{0}^{1/\sqrt{2}}2z\arcsin(z)\,dz\\&\stackrel{z\mapsto\sin\theta}{=}&\int_{0}^{\pi/4}2\theta\sin\theta\cos\theta\,d\theta\\&\stackrel{\theta\mapsto\varphi/2}{=}&\frac{1}{4}\int_{0}^{\pi/2}\varphi\sin\varphi\,d\varphi\\&=&\frac{1}{4}\left[\sin\varphi-\varphi\cos\varphi\right]_{0}^{\pi/2}=\frac{1}{4}.\end{eqnarray*}$$
Alternative method. Since $\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n$ we have $$\arcsin(x) = \sum_{n\geq 0}\frac{1}{(2n+1)4^n}\binom{2n}{n}x^{2n+1}$$ for any $x\in(0,1)$ and $$\int_{0}^{1/2}\arcsin\sqrt{x}\,dx = \frac{1}{\sqrt{2}}\sum_{n\geq 0}\frac{1}{(2n+1)(2n+3)8^n}\binom{2n}{n},$$ which equals $\frac{1}{3\sqrt{2}}\,\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{5}{2};\tfrac{1}{2}\right)$, is a telescopic series in disguise.
• Is This $F_1$ a geometric function? – Henry Lee Sep 15 '18 at 22:40