Deriving Maclaurin series for $\frac{\arcsin x}{\sqrt{1-x^2}}$.

Solution 1:

Note that $$\int_0^1\frac{dt}{1-x^2+x^2t^2}=\frac{1}{x\sqrt{1-x^2}}\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)=\frac{\arcsin(x)}{x\sqrt{1-x^2}}$$ so we can write $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\left(\int_0^1(1-t^2)^n\,dt\right)x^{2n+1}.$$ But $$\int_0^1(1-t^2)^n\,dt=\int_0^1\sum_{k=0}^n(-1)^k\binom{n}{k}t^{2k}\,dt=\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{2k+1}=\frac{(2n)!!}{(2n+1)!!}.$$ Hence, $$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{(2n)!!}{(2n+1)!!}x^{2n+1}.$$ Also see here for proof of $\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{2k+1}=\frac{(2n)!!}{(2n+1)!!}$.

Solution 2:

Another but similar proof which does not need to use the summation formula above is this one. Start with defining

$$I(t)= \frac{1}{\sqrt{1-x^2}}\arctan{\frac{x\sin{t}}{\sqrt{1-x^2}}}$$

Then by the Fundamental theorem of calculus

$$\frac{\arcsin{x}}{\sqrt{1-x^2}}=I\left(\frac{\pi}{2}\right)-I(0)=\int_{0}^{\pi/2} \frac{\partial I}{\partial t}\mathrm{d}t=\int_{0}^{\pi/2}\frac{x\cos t}{1-x^2\cos^2 t }\mathrm{d}t$$

Ergo

$$\frac{\arcsin{x}}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}x^{2n+1}\int_0^{\pi/2}\cos^{2n+1}\! t\,\mathrm{d}t$$

Denote $J_n:=\int_0^{\pi/2}\cos^{2n+1}\! t\,\mathrm{d}t$, by per partes we have

$$J_n = \int_0^{\pi/2}\cos^{2n+1}\! t\,\mathrm{d}t = 2n\int_0^{\pi/2}\cos^{2n-1}\sin^2 t\!\,\mathrm{d}t=2n\left(J_{n-1}-J_{n}\right)$$

So $$J_n = \frac{2n}{2n+1}J_{n-1} =\frac{2n}{2n+1}\frac{2n-2}{2n-1}J_{n-2}=\dots = \frac{(2n)!!}{(2n+1)!!}J_0=\frac{(2n)!!}{(2n+1)!!}$$

since $J_0 = \int_0^{\pi/2}\cos\! t\,\mathrm{d}t =1$. Over all we get desired result

$$\frac{\arcsin{x}}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}x^{2n+1}$$

Note: Similar integral would have been also...

$$\int_{0}^{\pi/2}\frac{\mathrm{d}t}{1-x\sin t}$$