Nested Trigonometric Integral

$3\displaystyle \int_{0}^{1} sin(2cos^{-1}(sin(cos^{-1}(x))))dx$ $=3\displaystyle \int_{0}^{1} sin(2cos^{-1}(\sqrt{1-x^{2}}))dx$ $=6\displaystyle \int_{0}^{1} sin(cos^{-1}(\sqrt{1-x^{2}}))cos(cos^{-1}(\sqrt{1-x^{2}}))dx$ $=6\displaystyle \int_{0}^{1}(\sqrt{1-1+x^{2}})\sqrt{1-x^{2}}dx$ $=6\displaystyle \int_{0}^{1}x\sqrt{1-x^{2}}dx$ Using $x^{2}=u => dx=\frac{du}{2x}$ $3\displaystyle \int_{0}^{1}\sqrt{1-u}du=(-3\cdot \frac{2}{3}(1-u)^{\frac{3}{2}}]_{0}^{1}=2$

Let $\theta=\arccos { (x) }$

$\Longrightarrow \cos { \theta } =x$

Using $\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1$ ,

$\Longrightarrow \sin ^{ 2 }{ (\cos ^{ -1 }{ (x)) } } +{ x }^{ 2 }=1$

$\Longrightarrow \sin { (\cos ^{ -1 }{ (x)) } } =\sqrt { 1-{ x }^{ 2 } }$ .

Hence the integral becomes $3\int _{ 0 }^{ 1 }{ \sin { (2\cos ^{ -1 }{ (\sqrt { 1-{ x }^{ 2 } } )) } } } dx$ .

$=6\int _{ 0 }^{ 1 }{ \sin { (\cos ^{ -1 }{ (\sqrt { 1-{ x }^{ 2 } } )) } } \cos { (\cos ^{ -1 }{ (\sqrt { 1-{ x }^{ 2 } } )) } } } dx$ (Double angle formula)

$=6\int _{ 0 }^{ 1 }{ \sin { (\cos ^{ -1 }{ (\sqrt { 1-{ x }^{ 2 } } ))\sqrt { 1-{ x }^{ 2 } } } dx } }$ .

Now let $\alpha =\cos ^{ -1 }{ (\sqrt { 1-{ x }^{ 2 } } ) }$

$\Longrightarrow \cos { (\alpha } )=\sqrt { 1-{ x }^{ 2 } }$

$\Longrightarrow \sin ^{ 2 }{ (\alpha ) } +\cos ^{ 2 }{ (\alpha ) } =1$

$\Longrightarrow \sin ^{ 2 }{ (\cos ^{ -1 }{ (\sqrt { 1-{ x }^{ 2 } } ) } ) } +1-{ x }^{ 2 }=1$

$\Longrightarrow \sin { (\cos ^{ -1 }{ (\sqrt { 1-{ x }^{ 2 } } ) } ) } =x$ .

The integral is now equal to $6\int _{ 0 }^{ 1 }{ x\sqrt { 1-{ x }^{ 2 } } dx }$ which can be easily evaluated by letting $u=1-{ x }^{ 2 }$ :

$du=-2xdx$ .

When $x=0, u=1$ and when $x=1, u=0$ . So finally:

$-3\int _{ 1 }^{ 0 }{ \sqrt { u } du }$

$=3\int _{ 0 }^{ 1 }{ { u }^{ \frac { 1 }{ 2 } }du }$

$={ \left[ 2{ u }^{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ 1 }=\boxed { 2 }$ .