Integrate $\int_0^1 \frac{x-\arcsin x}{x^3}$ Indefinite integral is pretty easy to solve, I did it by substitution and I'm pretty sure it can be done relatively easy via integration by parts.
The problem are boundaries.
After substitution $arcsin x=t$ we get
$$\int_0^\frac{\pi}{2} \frac{(\sin t-t)\cos t}{(\sin t)^3}dt$$
so we have
$$\int_0^\frac{\pi}{2} \frac{\cos t}{(\sin t)^2} dt+ \int_0^\frac{\pi}{2} \frac{t \cos t}{(\sin t)^3}dt$$
Now for example first integral is easy to compute and I get $\frac{-1}{\sin t}$ from 0 to $\frac{\pi}{2}$.But in 0 the value is $\infty$.
The second integral can also be solved using partial integration with $t$ and $\frac{cost}{(sint)^3}$.
The only method to bypass this that I know of is linear substitution (I'm not sure how you call it) but that doesn't give anything useful.
Any hints?
 A: We can also try to evaluate the integral simply integrating by part.
$$I=\int_0^1 \frac{x-\arcsin x}{x^3}dx=-\frac{1}{2}\int_0^1 (x-\arcsin x)d\Big(\frac{1}{x^2}\Big)$$
$$=-\frac{x-\arcsin x}{2x^2}\Big|_0^1+\frac{1}{2}\int_0^1\frac{dx}{x^2}\Big(1-\frac{1}{\sqrt{1-x^2}}\Big)$$
Making the substitution $t=x^2$ in the second integral
$$I=\frac{\pi}{4}-\frac{1}{2}+\frac{1}{4}\int_0^1\Big(t^{-\frac{1}{2}-1}-t^{-\frac{1}{2}-1}(1-t)^{\frac{1}{2}-1}\Big)dt$$
Now we can use the analytical continuation of Beta-function $\Bigl (B(\gamma,\alpha)=\int_0^1s^{\gamma-1}(1-s)^{\alpha-1}ds$, if $\gamma, \alpha >0\,\Bigr)$ for negative $\gamma\in(-1,0)$:
$$B(\gamma,\alpha)=-\frac{1}{(\exp(2\pi{i}\alpha)-1)(\exp(2\pi{i}\gamma)-1)}\oint_Ps^{\gamma-1}(1-s)^{\alpha-1}ds$$
where $P$ is Pochhammer contour in the complex plane.
It can be shown (for example, here) that for $\gamma\in(-1;0)$ and $\alpha>0$ $B(\gamma,\alpha)=\lim_{r\to0}(\int_r^1s^{\gamma-1}(1-s)^{\alpha-1}ds+\frac{r^\gamma}{\gamma})$,
so $$J(\gamma,\alpha,\alpha')=\int_0^1s^{\gamma-1}\left((1-s)^{\alpha-1}-(1-s)^{\alpha'-1}\right)ds$$$$=\lim_{r\to0}\int_r^1s^{\gamma-1}\left((1-s)^{\alpha-1}-(1-s)^{\alpha'-1}\right)ds=\lim_{r\to0}\left(B(\gamma,\alpha)-\frac{r^\gamma}{\gamma}-B(\gamma,\alpha')+\frac{r^\gamma}{\gamma}\right)$$ $$J(\gamma,\alpha,\alpha')=B(\gamma,\alpha)-B(\gamma,\alpha')\,,\,\gamma\in(-1;0)\,,\,\, \alpha,\alpha'>0 $$
It can also be proved that analytically continued Beta-function is expressed in the usual way in terms of Gamma-function: $B(\gamma,\alpha)=\frac{\Gamma(\gamma)\Gamma(\alpha)}{\Gamma(\gamma+\alpha)}$. This expression is valid for all complex $\alpha, \gamma$.
Coming back to the initial integral
$$I=\frac{\pi}{4}-\frac{1}{2}+\frac{1}{4}\Big(B\big(-\frac{1}{2};1\big)-B\big(-\frac{1}{2};-\frac{1}{2}\big)\Big)$$
$$=\frac{\pi}{4}-\frac{1}{2}+\frac{1}{4}\bigg(\frac{\Gamma\big(-\frac{1}{2}\big)\Gamma(1)}{\Gamma\big(\frac{1}{2}\big)}-\frac{\Gamma\big(-\frac{1}{2}\big)\Gamma\big(\frac{1}{2}\big)}{\Gamma (0)}\bigg)$$
Given than $\Gamma(0)=\infty$ and $\Gamma\big(-\frac{1}{2}\big)=-2\Gamma\big(\frac{1}{2}\big)$
$$I=\frac{\pi}{4}-\frac{1}{2}-\frac{1}{2}=\frac{\pi}{4}-1$$
WolframAlpha gives the same result.
A: You may not split the integral into two divergent integrals. Instead, integrate as a whole as shown below
\begin{align}\int_0^\frac{\pi}{2} \frac{(\sin t-t)\cos t}{\sin^3t}dt
&=\frac12\int_0^\frac{\pi}{2} (t-\sin t)d\left(\frac{1}{\sin^2t}\right)\\
&= \frac12\frac{t-\sin t}{\sin^2t}\bigg|_0^{\frac\pi2}-\frac12
\int_0^\frac{\pi}{2} \frac{1-\cos t}{\sin^2t} dt\\
 &= \frac12(\frac\pi2-1)-\frac12\lim_{t\to0}\frac{t-\sin t}{\sin^2t}-\frac14
\int_0^\frac{\pi}{2} \sec^2\frac t2dt\\
&= \frac12(\frac\pi2-1)-\frac12\cdot 0-
\frac12=\frac\pi4-1
\end{align}
where the lower limit term evaluates to zero per L’Hopital rule.
A: $$\begin{align*}
I &= \int_0^1 \frac{x - \arcsin(x)}{x^3} \, dx \\[1ex]
&= -\frac12\left(1 - \frac\pi2\right) + \frac12 \int_0^1 \left(1-\frac1{\sqrt{1-x^2}}\right) \, \frac{dx}{x^2} \tag{1} \\[1ex]
&= \frac\pi4-\frac12 + \frac12 \int_0^1 \frac{\sqrt{1-x^2}-1}{x^2 \sqrt{1-x^2}} \, dx \\[1ex]
&= \frac\pi4 - \frac12 + \frac12 \int_0^{\frac\pi2} (\cot(x)\csc(x) - \csc^2(x)) \, dx \tag{2} \\[1ex]
&= \boxed{\frac\pi4-1}
\end{align*}$$


*

*$(1)$ : integrate by parts

*$(2)$ : substitute $x\mapsto\sin(x)$
