Answer: (a) \(\sin^{-1}x=x+\frac{1}{6}x^3\).
(b) \(\displaystyle \int_0^{\frac15}\frac{1}{\sqrt{1-u^2}}\,du\approx \frac{151}{750}\).
(a) Let \(f(x)=\sin^{-1}x\). Its Maclaurin series is
\(f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots\)
Now
\(f'(x)=(1-x^2)^{-1/2}\),
\(f''(x)=x(1-x^2)^{-3/2}\),
\(f'''(x)=(1-x^2)^{-3/2}+3x^2(1-x^2)^{-5/2}\).
At \(x=0\),
\(f(0)=0,\quad f'(0)=1,\quad f''(0)=0,\quad f'''(0)=1\).
So
\(\sin^{-1}x=0+1\cdot x+\frac{0}{2!}x^2+\frac{1}{3!}x^3+\cdots=x+\frac{1}{6}x^3\).
(b) Using part (a),
\(\sin^{-1}u\approx u+\frac{1}{6}u^3\).
Also, \(\dfrac{d}{du}(\sin^{-1}u)=\dfrac{1}{\sqrt{1-u^2}}\), so
\(\displaystyle \int_0^{1/5}\frac{1}{\sqrt{1-u^2}}\,du=\sin^{-1}\left(\frac15\right)\approx \frac15+\frac{1}{6}\left(\frac15\right)^3\).
Hence
\(\frac15+\frac{1}{750}=\frac{150}{750}+\frac{1}{750}=\frac{151}{750}\).
Therefore
\(\displaystyle \int_0^{\frac15}\frac{1}{\sqrt{1-u^2}}\,du\approx \frac{151}{750}\).
