Integral exercise asked in Imad zak math group by Dan Sitaru

Exercise:

Compute, $\int_{0}^{2\pi }{\frac{dx}{2+\sin x}}$

Solution: Let $y=\pi -x\Leftrightarrow dy=-dx$

$\int_{0}^{2\pi }{\frac{dx}{2+\sin x}=-\int_{\pi }^{-\pi }{\frac{dy}{2+\sin \left( \pi -y \right)}=\int_{-\pi }^{\pi }{\frac{dy}{2+\sin y}}}}$

Let $u=\tan \left( \frac{y}{2} \right)\Leftrightarrow du=\frac{1}{2}{{\sec }^{2}}\left( \frac{y}{2} \right)dy$ and $\sin y=\frac{2u}{{{u}^{2}}+1}$

But ${{\sec }^{2}}\left( \frac{y}{2} \right)=1+{{\tan }^{2}}\left( \frac{y}{2} \right)=1+{{u}^{2}}$ $\Leftrightarrow 2du=\left( 1+{{u}^{2}} \right)dy\Leftrightarrow dy=\frac{2du}{1+{{u}^{2}}}$

So $\int_{-\pi }^{\pi }{\frac{dy}{\sin y+2}}=\int_{-\infty }^{\infty }{\frac{\frac{2du}{1+{{u}^{2}}}}{2+\frac{2u}{{{u}^{2}}+1}}}=\int_{-\infty }^{\infty }{\frac{\frac{2}{{{u}^{2}}+1}}{\frac{2{{u}^{2}}+2+2u}{{{u}^{2}}+1}}du=\int_{-\infty }^{\infty }{\frac{1}{{{u}^{2}}+u+1}du}}$

But ${{u}^{2}}+u+1={{u}^{2}}+\frac{2}{2}u+\frac{1}{4}-\frac{1}{4}+1={{\left( u+\frac{1}{2} \right)}^{2}}+\frac{3}{4}$

So $\int_{-\infty }^{\infty }{\frac{du}{{{u}^{2}}+u+1}=\int_{-\infty }^{\infty }{\frac{du}{{{\left( u+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}}}$  Let $w=u+\frac{1}{2}\Leftrightarrow dw=du$

So $\int_{-\infty }^{\infty }{\frac{du}{{{\left( u+\frac{1}{2} \right)}^{2}}+\frac{3}{4}}=\int_{-\infty }^{\infty }{\frac{dw}{{{w}^{2}}+\frac{3}{4}}}}$

Let $f\left( w \right)=\frac{1}{{{w}^{2}}+\frac{3}{4}}\Leftrightarrow f\left( -w \right)=\frac{1}{{{\left( -w \right)}^{2}}+\frac{3}{4}}=\frac{1}{{{w}^{2}}+\frac{3}{4}}=f\left( w \right)$

So $\frac{1}{{{w}^{2}}+\frac{3}{4}}$ is even function and the interval $\left] -\infty ,\infty  \right[$ is symmetry

Thus $\int_{-\infty }^{\infty }{\frac{dw}{{{w}^{2}}+\frac{3}{4}}=2\int_{0}^{\infty }{\frac{dw}{{{w}^{2}}+\frac{3}{4}}}}=2\underset{\theta \to \infty }{\mathop{\lim }}\,\int_{0}^{\theta }{\frac{dw}{{{w}^{2}}+3/4}}$

But $\int_{0}^{\infty }{\frac{dw}{{{w}^{2}}+\frac{3}{4}}}=\int_{0}^{\infty }{\frac{dw}{{{w}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}}$  Let $w=\frac{\sqrt{3}}{2}z\Leftrightarrow dw=\frac{\sqrt{3}}{2}dz$

So $\int_{0}^{\infty }{\frac{dw}{{{w}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}}=\frac{\sqrt{3}}{2}\int_{0}^{\infty }{\frac{dz}{{{\left( \frac{\sqrt{3}}{2}z \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}}=\frac{\frac{\sqrt{3}}{2}}{\frac{3}{4}}\int_{0}^{\infty }{\frac{dz}{{{z}^{2}}+1}}$

$=\frac{4\sqrt{3}}{6}\underset{\theta \to \infty }{\mathop{\lim }}\,\left[ \arctan z \right]_{0}^{\theta }=\frac{2\sqrt{3}}{3}\left( \underset{\theta \to \infty }{\mathop{\lim }}\,\arctan \frac{2}{\sqrt{3}}w \right)_{0}^{\theta }=\frac{2\sqrt{3}}{3}\left[ \underset{\theta \to \infty }{\mathop{\lim }}\,\arctan \left( \frac{2}{\sqrt{3}}\theta  \right)-\arctan \left( 0 \right) \right]$

$=\frac{2\sqrt{3}}{3}\left( \frac{\pi }{2} \right)=\frac{\pi \sqrt{3}}{3}$ thus $\int_{-\infty }^{\infty }{\frac{dw}{{{w}^{2}}+3/4}=2\left( \frac{\pi \sqrt{3}}{3} \right)=\frac{2\pi \sqrt{3}}{3}}=\frac{2\pi }{\sqrt{3}}$

Therefore, $\int_{0}^{2\pi }{\frac{dx}{2+\sin x}}=\frac{2\pi }{\sqrt{3}}$          Q.E.D