calculus - $int_0^{pi/4}!frac{mathrm dx}{2+sin x}$ , $int_0^{2pi}!frac{mathrm dx}{2+sin x}$

Please help me integrate

$$\int_0^{\pi/4}\!\frac{\mathrm dx}{2+\sin x}$$

and

$$\int_0^{2\pi}\!\frac{\mathrm dx}{2+\sin x}$$

I've tried the standard $u = \tan \frac{x}{2}$ substitution but it looks horrible.

Thanks in advance!

Answer

Let's give another try to your failed technique...

$$\displaystyle\int \frac{dx}{2+\sin x}$$

Let $u = \tan \frac{x}{2}$

$$\int \frac{du}{u^2+u+1} = \int \frac{du}{\left(u+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}$$

Let $s=u+\frac{1}{2}$

$$\begin{align}\int \frac{ds}{s^2 + \left(\frac{\sqrt{3}}{2}\right)^2} &= \frac{2}{\sqrt{3}}\arctan\left(\frac{2s}{\sqrt{3}}\right)\\ &=\frac{2}{\sqrt{3}}\arctan\left(\frac{2u+1}{\sqrt{3}}\right)\\&=\frac{2}{\sqrt{3}}\arctan\left(\frac{2\tan\frac{x}{2}+1}{\sqrt{3}}\right)\end{align}$$

Evaluating the above from $0$ to $\dfrac{\pi}{4}$ yields approximately $0.33355$ while $0$ to $2\pi$ gives $\dfrac{2\pi}{\sqrt{3}}$.