Let x = 10 sin θ {\displaystyle x=10\sin \theta } so that d x = 10 cos θ d θ {\displaystyle dx=10\cos \theta d\theta } . Then we have
∫ x 2 ( 100 − x 2 ) 3 / 2 d x = ∫ 100 sin 2 θ ( 10 cos θ d θ ) ( 100 − 100 sin 2 θ ) 3 / 2 = ∫ 1000 ( sin 2 θ ) ( cos θ ) d θ ( 100 cos 2 θ ) 3 / 2 = ∫ 1000 ( sin 2 θ ) ( cos θ ) d θ 1000 cos 3 θ = ∫ sin 2 θ d θ cos 2 θ = ∫ tan 2 θ d θ = ∫ ( sec 2 θ − 1 ) d θ = tan θ − θ + C {\displaystyle {\begin{aligned}\int {\frac {x^{2}}{(100-x^{2})^{3/2}}}\,dx&=\int {\frac {100\sin ^{2}\theta (10\cos \theta d\theta )}{(100-100\sin ^{2}\theta )^{3/2}}}\\&=\int {\frac {1000(\sin ^{2}\theta )(\cos \theta )d\theta }{(100\cos ^{2}\theta )^{3/2}}}\\&=\int {\frac {1000(\sin ^{2}\theta )(\cos \theta )d\theta }{1000\cos ^{3}\theta }}\\&=\int {\frac {\sin ^{2}\theta d\theta }{\cos ^{2}\theta }}\\&=\int \tan ^{2}\theta d\theta \\&=\int (\sec ^{2}\theta -1)d\theta \\&=\tan \theta -\theta +C\end{aligned}}}
Now we isolate for the original variables. Using the diagram below (created from noting the original condition on x and by using Pythagorean theorem), we have
∫ x 2 ( 100 − x 2 ) 3 / 2 d x = tan θ − θ + C = x 100 − x 2 − arcsin ( x 10 ) + C {\displaystyle {\begin{aligned}\int {\frac {x^{2}}{(100-x^{2})^{3/2}}}\,dx&=\tan \theta -\theta +C\\&={\frac {x}{\sqrt {100-x^{2}}}}-\arcsin \left({\frac {x}{10}}\right)+C\\\end{aligned}}}
completing the proof.
so that 
. Then we have
