Question

Evaluate the following : ${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{1}{6-\cos x}\cdot dx}$

Answer 1

Let I = ${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{1}{6-\cos x}\cdot dx}$

Put ${\displaystyle \tan \left(\frac{x}{2}\right)}$ = t

∴ x = 2 tan^–1 t

∴ dx = ${\displaystyle \frac{2dt}{1+t}}$
and
cos x = ${\displaystyle \frac{1-t^2}{1+t^2}}$

When x =  ${\displaystyle \frac{\pi }{2},t=\tan \left(\frac{\pi }{2}\right)}$ = 1

When x = 0, t = tan 0 = 0

∴ I = ${\displaystyle \frac{\frac{2dt}{1+t^2}}{6-\cos \left(\frac{1-t^2}{1+t^2}\right)}}$

= ${\displaystyle {\int }_0^1\frac{2dt}{6(1+t^2)+1(1-t^2)}}$

= ${\displaystyle 2{\int }_0^1\frac{1}{t^2+7}\cdot dt}$

= ${\displaystyle 2{\left[\frac{1}{35}{\tan }^{-1} \frac{t}{5}\right]}_0^1}$

= ${\displaystyle 2[\frac{1}{35}{\tan }^{-1} \frac{1}{3}-\frac{1}{5}{\tan }^{-1}0]}$

= ${\displaystyle \frac{2}{35}{\tan }^{-1} \frac{1}{3}-\frac{7}{5}\times 0}$

= ${\displaystyle \frac{2}{\sqrt{35}}{\tan }^{-1} \sqrt{\frac{7}{5}}}$.