This is the integral:
$$
a=\int_0^{\pi/2} \left(1 – 0.5(\sin x)^2\right)^{-1/2}\,\mathrm dx
$$
I tried using a binomial expansion multiple times and each time I get a sum which contains a trigonometric function raised to the power of $n$ or $2n$. I never get a power series which can be readily integrated. Here is one example:
$$
a=\sqrt{2}\int_0^{\pi/2} 1+ \sum_{n=0}^\infty \frac{(-1)^n\cdot1\cdot3\cdot5\cdot7\cdots(2n-1)}{n!\cdot2^n}(\cos x)^{2n}\,\mathrm dx
$$
What is the easiest way for me to solve the problem? I am near the end of calculus $2$.
Best Answer
I do not know what you have been precisely asked but, if it is via a power series you need to expand the integrand in terms of $x$.
If this is the case, start with the usual $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}+O\left(x^{11}\right)$$ Use the binomial expansion to square it $$\sin^2(x)=x^2-\frac{x^4}{3}+\frac{2 x^6}{45}-\frac{x^8}{315}+\frac{2 x^{10}}{14175}+O\left(x^{12}\right)$$ $$1-\frac 12\sin^2(x)=1-\frac{x^2}{2}+\frac{x^4}{6}-\frac{x^6}{45}+\frac{x^8}{630}-\frac{x^{10}}{14175}+O\left(x^{12}\right)$$ $$\sqrt{1-\frac 12\sin^2(x)}=1-\frac{x^2}{4}+\frac{5 x^4}{96}+\frac{11 x^6}{5760}-\frac{11 x^8}{129024}-\frac{18121 x^{10}}{116121600}+O\left(x^{12}\right)$$ Long division now $$\frac 1 {\sqrt{1-\frac 12\sin^2(x)}}=1+\frac{x^2}{4}+\frac{x^4}{96}-\frac{71 x^6}{5760}-\frac{2591 x^8}{645120}-\frac{23759 x^{10}}{116121600}+O\left(x^{12}\right)$$ and now, integrate termwise.
Using this truncated expansion and the given bounds, you should end with something like $1.84350$ while the exact value is $1.85407$
Edit
If you want to make the series expansion keeping the trigonometric function, considering the most general case (as @JJacquelin did), consider $$\frac 1 {\sqrt{1-a\sin^2(x)}}=\frac 1 {\sqrt{1-t}}=\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n} t^n=\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}a^n sin^{2n}(x)$$ which makes $$I=\int_0^{\frac \pi 2}\frac {dx} {\sqrt{1-a\sin^2(x)}}=\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}a^n \int_0^{\frac \pi 2}sin^{2n}(x)\,dx$$ Any table of definite integrals will give $$\int_0^{\frac \pi 2}sin^{2n}(x)\,dx=\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{1}{2}\right)}{2 \Gamma (n+1)}$$ So, as a result $$I=\frac{\sqrt{\pi }} 2\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n} \frac{ \Gamma \left(n+\frac{1}{2}\right)}{ \Gamma (n+1)}a^n$$ which is convergent if $a<1$ since $$b_n=(-1)^n \binom{-\frac{1}{2}}{n} \frac{ \Gamma \left(n+\frac{1}{2}\right)}{ \Gamma (n+1)}a^n\implies \frac{b_{n+1}}{b_n}=\frac{a (2 n+1)^2}{4 (n+1)^2}=a-\frac{a}{n}+O\left(\frac{1}{n^2}\right)$$ For sure, as JJacquelin wrote $I=K(a)$.
For $a=\frac 12$, the partial sums concerge quite fast $$\left( \begin{array}{cc} p & S_p \\ 0 & 1.57080 \\ 1 & 1.76715 \\ 2 & 1.82237 \\ 3 & 1.84154 \\ 4 & 1.84888 \\ 5 & 1.85186 \\ 6 & 1.85311 \\ 7 & 1.85364 \\ 8 & 1.85388 \\ 9 & 1.85399 \\ 10 & 1.85403 \\ 11 & 1.85406 \\ 12 & 1.85407 \end{array} \right)$$