Integral exercise Level 2

Exercise:

Integrate, $\int{\frac{dx}{\pi +\cos x}}$

Solution: Let $y=\tan \frac{x}{2}\Rightarrow x=2\arctan y\Rightarrow dx=\frac{2}{1+{{y}^{2}}}dy$ where $-\pi <x<\pi $

And we know that $\cos x=\frac{1-{{y}^{2}}}{1+{{y}^{2}}}$

So $\int{\frac{dx}{\pi +\cos x}=\int{\frac{2}{\pi +\frac{1-{{y}^{2}}}{1+{{y}^{2}}}}}\times \frac{dy}{1+{{y}^{2}}}}=\int{\frac{2dy}{\frac{\pi +\pi {{y}^{2}}+1-{{y}^{2}}}{1+{{y}^{2}}}}\times \frac{1}{1+{{y}^{2}}}}$

$=\int{\frac{2dy}{\pi +\pi {{y}^{2}}+1-{{y}^{2}}}}=2\int{\frac{dy}{\left( \pi -1 \right){{y}^{2}}+\left( \pi +1 \right)}}=\frac{2}{\pi -1}\int{\frac{dy}{{{y}^{2}}+\frac{\pi +1}{\pi -1}}}$

Let $y=\frac{\sqrt{\pi +1}}{\sqrt{\pi -1}}t\Leftrightarrow dy=\frac{\sqrt{\pi +1}}{\sqrt{\pi -1}}dt$

So $\int{\frac{dy}{{{y}^{2}}+\sqrt{\frac{\pi +1}{\pi -1}}}}=\int{\frac{1}{\left( \frac{\pi +1}{\pi -1} \right){{t}^{2}}+\frac{\pi +1}{\pi -1}}}\times \frac{\sqrt{\pi +1}}{\sqrt{\pi -1}}dt$

$=\frac{\sqrt{\pi +1}}{\sqrt{\pi -1}}\times \frac{{{\left( \sqrt{\pi -1} \right)}^{2}}}{{{\left( \sqrt{\pi +1} \right)}^{2}}}\int{\frac{dt}{{{t}^{2}}+1}}=\sqrt{\frac{\pi -1}{\pi +1}}\arctan t+c=\sqrt{\frac{\pi -1}{\pi +1}}\arctan \left( \frac{y\sqrt{\pi -1}}{\sqrt{\pi +1}} \right)+c$

Thus $\int{\frac{dx}{\pi +\cos x}=\frac{2}{\pi -1}\sqrt{\frac{\pi -1}{\pi +1}}\arctan \left( \frac{\sqrt{\pi -1}}{\sqrt{\pi +1}}\tan \left( \frac{x}{2} \right) \right)+c}$