Pregunta: 3. Evaluate the integral (a) ∫02π2−cosθ1dθ

Answer:- Let, $\mathrm{I}={\int }_0^{2\pi }\frac{1}{\sqrt{2}-\cos \theta }\mathrm{d}\theta$

$={\int }_0^{2\pi }\frac{\sqrt{2}+\cos \theta }{2-{\cos }^2\theta }\mathrm{d}\theta$

$={\int }_0^{2\pi }\frac{\sqrt{2}+\cos \theta }{1+{\sin }^2\theta }\mathrm{d}\theta$ ( as $1-{\cos }^2\theta ={\sin }^2\theta$ )