Question

Evaluate ${\int }_0^{\pi /2}\frac{dx}{5+3\cos x}$

Answer 1

${\int }_0^{\frac{\pi }{2}}\frac{dx}{5+3\cos x}$

$t=tan\frac{x}{2}$

$\therefore cosx\frac{1-t^2}{1+t^2}$

$co{s}^2\frac{x}{2}=\frac{1}{(1+t^2{)}^2}$

$\therefore \frac{dt}{dx}=\frac{1}{2se{c}^2\frac{x}{2}}$

$\frac{dt}{dx}=\frac{1+t^2}{2}$

$\therefore \int \frac{1dx}{5+3\cos x}=\int \frac{2dt}{8-2t^2}$

$=\int \frac{1dt}{4-t^2}$

$\therefore {\int }_0^{\frac{\pi }{2}}\frac{1dx}{5+3cosx}=\frac{1}{4ln\left(\frac{2+t}{2-t}\right)}+c$

$={\int }_0^{\frac{\pi }{2}}\frac{1}{4ln\left(\frac{2+tan\frac{x}{2}}{2-tan\frac{x}{2}}\right)}$

$=\frac{1}{4l{n}_{\left(\frac{3}{1}\right)}}-\frac{1}{4l{n}_{\left(\frac{2}{2}\right)}}$

$=\frac{1}{|n3}$