It is possible to find positive integers A,B,C,D,E such that,
$$\int_0^{\frac{2a}{a^2+1}}\arcsin \left(\frac{|1-ax|}{\sqrt{1-x^2}}\right)dx=\frac{A}{\sqrt{a^2+1}}\sin^{-1}\left(\frac{1}{a^B}\right)-C\sin^{-1}\left(\frac{1}{a^D}\right)+\frac{E\cdot a\pi}{a^2+1}$$
for all real numbers $a \geq 3$. What is the value of $ A + B + C + D + E $?
Answer: I have seen answer to this question given by math expert on another website,but I didn't understand some of its steps. If any member knows the answer to this question, may answer this question.
Hint:- Questioner has solved this question first using integration by parts, then applying the substitution $(a^2+1)*x-a= a*sin \theta$ and the applying the substitution $t= tan \frac{\theta}{2}$
I am reproducing the answer to the above question given on Internet.
First integrating by parts, then applying the substitution $(a^2+1)x-a=a \sin\theta$ and then applying the substitution $t=\tan \frac{\theta}{2}$ gives
$$F(x)=\displaystyle\int \sin^{-1}\left(\frac{1-a*x}{\sqrt{1-x^2}}\right)dx$$
$$F(x)=x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right) + \displaystyle\int\frac{x(a-x)}{(1-x^2)*\sqrt{2ax-(a^2+1)x^2}}dx$$
$$F(x)=x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right)+ \sqrt{a^2+1}\displaystyle\int\frac{a(a-x)}{(1-x^2)\sqrt{a^2-((a^2+1)x-a)^2}}dx$$
$$F(x)=x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right)+ \frac{a^2}{\sqrt{a^2+1}}\displaystyle\int\frac{(1+\sin\theta)(a^2-\sin\theta)}{(a^2-a+1-a\sin\theta)(a^2+a+1+a\sin\theta)}dx$$
$$F(x)=x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right) +\frac{a^2}{\sqrt{a^2+1}}\displaystyle\int\frac{1}{a^2}+\frac{(a^2+1)(a-1)}{2a^2(a^2-a+1-a\sin\theta)}-\frac{(a^2+1)(a+1)}{(a^2+a+1+a\sin\theta)}d\theta$$
$$F(x)=x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right) +\frac{\theta}{\sqrt{a^2+1}}+\sqrt{a^2+1}\displaystyle\int\frac{(a-1)}{(a^2-a+1)(t^2+1)-2at}-\frac{(a-1)}{(a^2+a+1)(t^2+1)+2at}dt$$
$$F(x)=x\sin^{-1}\left(\frac{1-ax}{\sqrt{1-x^2}}\right) +\frac{\theta}{\sqrt{a^2+1}} + \tan^{-1}\left(\frac{(a^2-a+1)t-a}{(a-1)\sqrt{a^2+1}}\right) -\tan^{-1}\left(\frac{(a^2+a+1)t+a}{(a+1)\sqrt{a^2+1}}\right)$$
For $0\leq x\leq\left(\frac{2a}{a^2+1}\right)$ ignoring the constant of integration, thus
$$\displaystyle\int_0^{\frac{2a}{a^2+1}}\sin^{-1}\left(\frac{|(1-ax)|}{\sqrt{1-x^2}}\right)=2*F(a^{-1})-F(0)-F\left(\frac{2a}{a^2+1}\right)$$
If x=0,we have $\theta=-\frac{\pi}{2}$ and t=-1 so that,
$$F(0)=-\frac{\pi}{2\sqrt{a^2+1}}+\tan^{-1}\left(-\frac{\sqrt{a^2+1}}{(a-1)}\right)-\tan^{-1}\left(-\frac{\sqrt{a^2+1}}{(a+1)}\right)=-\frac{\pi}{2\sqrt{a^2+1}}-\tan^{-1}\left(\frac{\sqrt{a^2+1}}{a^2}\right)$$
If$x=\frac{2a}{a^2+1}$,then $\theta=\frac{\pi}{2}$ and t=1, so that,
$$F\left(\frac{2a}{a^2+1}\right)=-\frac{a\pi}{a^2+1}+ \frac{\pi}{2\sqrt{a^2+1}} + \tan^{-1}\left(\frac{a-1}{\sqrt{a^2+1}}\right)-\tan^{-1}\left(\frac{a+1}{\sqrt{a^2+1}}\right)$$
$$F\left(\frac{2a}{a^2+1}\right)=-\frac{a\pi}{a^2+1}+ \frac{\pi}{2\sqrt{a^2+1}}-\tan^{-1}\left(\frac{\sqrt{a^2+1}}{a^2}\right)$$
Finally if $x=a^{-1}$ then $\theta=\sin^{-1}(a^{-2})$ and $t=a^2-\sqrt{a^4-1}$ If $a\geq 3$,then,
$(a^4+a^2+1)\sqrt{a^2-1}-a^4\sqrt{a^2+1}\geq 0$
and we can show that
$$F(a^{-1})=\frac{1}{\sqrt{a^2+1}}\sin^{-1}(a^{-2})+\tan^{-1}\left(\frac{a\sqrt{a^4-1}-(a^3+1)}{\sqrt{a^2-1}}\right)-\tan^{-1}\left(\frac{a\sqrt{a^4-1}-(a^3-1)}{\sqrt{a^2-1}}\right)$$
$$F(a^{-1})=\frac{1}{\sqrt{a^2+1}}\sin^{-1}(a^{-2}) -\tan^{-1}\left(\frac{1}{(a^4+a^2+1)\sqrt{a^2-1}-a^4\sqrt{a^2+1}}\right)$$
$$F(a^{-1})=\frac{1}{\sqrt{a^2+1}}\sin^{-1}(a^{-2})-\frac{\pi}{2} +\tan^{-1}((a^4+a^2+1)\sqrt{a^2-1}-a^4\sqrt{a^2+1})$$
and hence,
$$\displaystyle\int_0^{\frac{2a}{a^2+1}}\sin^{-1}\left(\frac{|1-ax|}{\sqrt{1-x^2}}\right) =\frac{2}{\sqrt{a^2+1}}\sin^{-1}(a^{-2})-\pi +\frac{a\pi}{a^2+1}+2\tan^{-1}((a^4+a^2+1)\sqrt{a^2-1}-a^4\sqrt{a^2+1})+ 2\tan^{-1}\left(\frac{\sqrt{a^2+1}}{a^2}\right)$$
$$\displaystyle\int_0^{\frac{2a}{a^2+1}}\sin^{-1}\left(\frac{|1-ax|}{\sqrt{1-x^2}}\right) =\frac{2}{\sqrt{a^2+1}}\sin^{-1}(a^{-2})-\pi +\frac{a\pi}{a^2+1}+ 2\tan^{-1}(\sqrt{a^2-1})$$
$$\displaystyle\int_0^{\frac{2a}{a^2+1}}\sin^{-1}\left(\frac{|1-ax|}{\sqrt {1-x^2}}\right) =\frac{2}{\sqrt{a^2+1}}\sin^{-1}(a^{-2})-\pi +\frac{a\pi}{a^2+1}+ 2\cos^{-1}(a^{-1})$$
$$\displaystyle\int_0^{\frac{2a}{a^2+1}}\sin^{-1}\left(\frac{|1-ax|}{\sqrt{-x^2}}\right)=\frac{2}{\sqrt{a^2+1}}\sin^{-1}(a^{-2})-2\sin^{-1}(a^{-1})+\frac{a\pi}{a^2+1}$$
So that A=2, B=2, C=2, D=1, E=1 and hence A + B + C + D + E = 8
For futher information to the mathematical audience of this Mathematics Stack Exchange,
$x=\frac{a}{a^2+1}(1+\sin\theta)$
$a-x=\frac{a}{a^2+1}(a^2-\sin\theta)$
$1-x=\frac{1}{a^2+1}(a^2-a+1-a\sin\theta)$
$1+x=\frac{1}{a^2+1}(a^2+a+1+a\sin\theta)$
$\sqrt{a^2-((a^2+1)x-a)^2}=a\cos\theta$
$dx=\cos\theta d\theta$