Integrate It Please

We need the result of this other question. Note that the substitution $y = -x$ gives $\int_{-1}^1 \sin^{-1}x \cos^{-1}x \tan^{-1}x\,dx \; = \; \int_{-1}^1 \sin^{-1}y (\pi - \cos^{-1}y)\tan^{-1}y\,dy$ so that $\int_{-1}^1 \sin^{-1}x \cos^{-1}x \tan^{-1}x\,dx \; = \; \tfrac12\pi \int_{-1}^1 \sin^{-1}x \tan^{-1}x\,dx$ Manipulating the answer to the previous question, we see that the answer is $\tfrac18\pi^2\Big(-2\ln(24 - 16\sqrt{2}) - 4\sqrt{2} + \pi + 4\Big)$ making the answer $24+16 = \boxed{40}$ .

By substituting $\displaystyle x$ to $-x$ we find that

$\displaystyle \int _{ -1 }^{ 1 }{ \arcsin { x } \arccos { x } \arctan { x } } dx=\frac { \pi }{ 2 } \int _{ -1 }^{ 1 }{ \arcsin { x } \arctan { x } } dx$

Now,

$\displaystyle \int _{ -1 }^{ 1 }{ \arcsin { x } \arctan { x } } dx=\frac { { \pi }^{ 2 } }{ 4 } -\int _{ -1 }^{ 1 }{ \left( \frac { x\arctan { x } }{ \sqrt { 1-{ x }^{ 2 } } } +\frac { x\arcsin { x } }{ 1+{ x }^{ 2 } } \right) dx }$

$\displaystyle \int _{ -1 }^{ 1 }{ \left( \frac { x\arctan { x } }{ \sqrt { 1-{ x }^{ 2 } } } +\frac { x\arcsin { x } }{ 1+{ x }^{ 2 } } \right) dx } =\frac { \pi \ln { 2 } }{ 2 } -\int _{ -1 }^{ 1 }{ \left( \frac { \ln { 1+{ x }^{ 2 } } }{ 2\sqrt { 1-{ x }^{ 2 } } } -\frac { \sqrt { 1-{ x }^{ 2 } } }{ 1+{ x }^{ 2 } } \right) dx }$

Now comes the real beauty :D. Let

$\displaystyle J(a)=\int _{ -1 }^{ 1 }{ \left( \frac { \ln { 1+{ { a }^{ 2 }x }^{ 2 } } }{ \sqrt { 1-{ x }^{ 2 } } } \right) dx }$

Then

$\displaystyle J'(a)=\int _{ -1 }^{ 1 }{ \left( \frac { 2a{ x }^{ 2 } }{ (1+{ a }^{ 2 }{ x }^{ 2 })\sqrt { 1-{ x }^{ 2 } } } \right) dx } =\frac { 2a }{ 1+{ a }^{ 2 } } \left( \int _{ -1 }^{ 1 }{ \left( -\frac { \sqrt { 1-{ x }^{ 2 } } }{ (1+{ a }^{ 2 }{ x }^{ 2 }) } +\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) dx } \right) =\frac { 2a\pi }{ 1+{ a }^{ 2 } } -\frac { 2a }{ 1+{ a }^{ 2 } } \int _{ -1 }^{ 1 }{ \left( \frac { \sqrt { 1-{ x }^{ 2 } } }{ (1+{ a }^{ 2 }{ x }^{ 2 }) } \right) dx }$

Now let

$\displaystyle K(a)=\int _{ -1 }^{ 1 }{ \left( \frac { \sqrt { 1-{ x }^{ 2 } } }{ (1+{ a }^{ 2 }{ x }^{ 2 }) } \right) dx }$

$\displaystyle K(a)=\int _{ -1 }^{ 1 }{ \left( \frac { \sqrt { 1-{ x }^{ 2 } } }{ (1+{ a }^{ 2 }{ x }^{ 2 }) } \right) dx } =\frac { 1 }{ a } \int _{ -1 }^{ 1 }{ \left( \frac { x\arctan { (ax) } }{ \sqrt { 1-{ x }^{ 2 } } } \right) dx }$

$\displaystyle (aK)'(a)=\int _{ -1 }^{ 1 }{ \left( \frac { { x }^{ 2 } }{ (1+{ a }^{ 2 }{ x }^{ 2 })\sqrt { 1-{ x }^{ 2 } } } \right) dx } =\frac { J' }{ 2a }$

So, we have

$\displaystyle (aK)'=\frac { J' }{ 2a }$ and

$\displaystyle J'=\frac { 2a\pi }{ 1+{ a }^{ 2 } } -\frac { 2aK }{ 1+{ a }^{ 2 } }$

So,

$\displaystyle K+aK'=\frac { \pi }{ 1+{ a }^{ 2 } } -\frac { K }{ 1+{ a }^{ 2 } }$

$\displaystyle K'+K(\frac { 2 }{ a } -\frac { a }{ 1+{ a }^{ 2 } } )=\frac { \pi }{ a(1+{ a }^{ 2 }) }$

So,

$\displaystyle K=\frac { \pi \left( \sqrt { 1+{ a }^{ 2 } } +c \right) }{ { a }^{ 2 } }$

And we want

$\displaystyle K(0)=\int _{ -1 }^{ 1 }{ \sqrt { 1-{ x }^{ 2 } } dx } =\frac { \pi }{ 2 } =\frac { \pi \left( \sqrt { 1+{ a }^{ 2 } } +c \right) }{ { a }^{ 2 } }$

So, c=-1

Now,

$\displaystyle \frac { J' }{ 2 } =\pi \left( \frac { 1 }{ a } +\frac { a }{ \sqrt { 1+{ a }^{ 2 } } } -\frac { \sqrt { 1+{ a }^{ 2 } } }{ { a } } \right)$

The hardest integral here should(?) be the last term and so, I will give a clue of how I have done it.

Hint:Let $a=\tan{\theta}$

$\displaystyle \frac { J }{ 2 } =\pi \left( \ln { (1+\sqrt { 1+{ a }^{ 2 } } ) } +c \right)$

And we want $J(0)=0$

So, $\displaystyle J(a)=2\pi \left( \ln { (1+\sqrt { 1+{ a }^{ 2 } } ) } -\ln { 2 } \right)$

Combining all these together and recall that the integrals we want is when a=1, we get

$\displaystyle \int _{ -1 }^{ 1 }{ \arcsin { x } \arccos { x } \arctan { x } } dx=\dfrac{\pi^{2}}{8} \left( -2\ln(24-16\sqrt{2})-4\sqrt{2}+\pi+4 \right)$