By expanding the integrand in a power series, calculate
By expanding the integrand in a power series, calculate $\displaystyle\int_0^{1/5}\frac{\sin x}{\sqrt{1-x}}\,dx$ with accuracy $10^{-3}$.
Explanation of Method
The integrand is a product of the $\sin x$ series and the binomial series for $(1-x)^{-1/2}$; multiplying them out and integrating the resulting series term by term over the small interval $[0,\frac15]$ converges very quickly.
- Expand $\sin x$ and $(1-x)^{-1/2}$ separately.
- Multiply the two series to get the low-order coefficients of the product.
- Integrate term by term over $[0,0.2]$.
Step 1 · Expand both factors
$$ \sin x = x-\frac{x^3}6+\cdots, \qquad (1-x)^{-1/2} = 1+\frac x2+\frac{3x^2}8+\frac{5x^3}{16}+\cdots $$
Step 2 · Multiply the series
$$ \frac{\sin x}{\sqrt{1-x}} = x+\frac{x^2}2+\frac{5x^3}{24}+\frac{11x^4}{48}+\cdots $$
Step 3 · Integrate over $[0,0.2]$
$$ \int_0^{1/5}\frac{\sin x}{\sqrt{1-x}}\,dx = \left[\frac{x^2}2+\frac{x^3}6+\frac{5x^4}{96}+\frac{11x^5}{240}\right]_0^{0.2} \approx 0.020000+0.001333+0.000083+0.000015 \approx 0.021431. $$ The next term is well below $10^{-3}$.
Summary Table
| Property | Value |
|---|---|
| Product series | $x+x^2/2+5x^3/24+11x^4/48+\cdots$ |
| Result | $0.0214$ |
| Terms needed | $4$ |
Additional Notes
- Multiplying two power series (a Cauchy product) is unavoidable here since neither factor alone has a simple closed antiderivative; the small upper limit $0.2$ ensures the product series converges rapidly despite the extra bookkeeping.
- This is the same Cauchy-product technique used in Problem 275's $\ln^2(1-x)$, but here truncated to just a handful of terms since only $10^{-3}$ accuracy over a short interval is required.
- The binomial factor $(1-x)^{-1/2}$ is the same one used throughout the arcsine and arcsinh derivations of Problems 268–270, here playing a purely computational role rather than producing a named special function.