Solve integral $ \int\limits_{-\pi}^\pi \sqrt{5+4\sin\theta}\,d\theta $

Question

I need to solve the next integral:

$$ \int\limits_{-\pi}^\pi \sqrt{5+4\sin\theta}~\mathrm d\theta $$

Thanks for help

Answer 1

We have,

$$\sqrt{5} \int_{-\pi}^{\pi} \sqrt{1+\frac{4}{5}\sin \theta} d\theta$$

$$=\sqrt{5} \int_{-\pi}^{\pi} \sum_{n \geq 0} {\frac{1}{2} \choose n} (\frac{4}{5})^n \sin^n x dx$$

$$=\sqrt{5} \sum_{n \geq 0} {\frac{1}{2} \choose n} (\frac{4}{5})^n F(n)$$

Where $F(n)=\int_{-\pi}^{\pi} \sin^n x dx$.

Note $F(n)=0$ for $n$ odd as $\sin^n x$ is then odd. Also note for $n$ even,

$$F(n)=2 \int_{0}^{\pi} \sin^n x dx=4 \int_{0}^{\frac{\pi}{2}} \sin^n x dx$$

Show that for $n \geq 2$,

$$\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx=\frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \sin^{n-2} dx$$

Use this to show for $n \geq 2$ even we have,

$$F(n)=\frac{1 \cdot 3 \cdots (n-1)}{2 \cdot 4 \cdots n} (2\pi)$$

That gives,

$$\int_{-\pi}^{\pi} \sqrt{5+4\sin \theta} d\theta$$

$$=2\pi \sqrt{5} (1+ \sum_{n=2}^{\infty} \frac{(-1)^n+1}{2} {\frac{1}{2} \choose n} (\frac{4}{5})^n \frac{(n-1)!!}{n!!})$$

Or equally with the definition $(-1)!!=1$ and $(0)!!=1$:

$$=2\pi \sqrt{5} \sum_{n=0}^{\infty}\frac{(-1)^n+1}{2} {\frac{1}{2} \choose n} (\frac{4}{5})^n \frac{(n-1)!!}{n!!}$$

With some terms one can get a good approximation to what Wolfram Alpha gets.