Question

Find : ${\int }_0^{\frac{\pi }{2}}\frac{dx}{4+5\cos x}dx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\frac{dx}{4+5\cos x}$

Substituting $\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$, we have

$I={\int }_0^{\frac{\pi }{2}}\frac{(1+{\tan }^2\frac{x}{2})dx}{9-{\tan }^2\frac{x}{2}}$

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

$\Rightarrow 2dt=(1+{\tan }^2\frac{x}{2})dx$

$\therefore I={\int }_0^1\frac{2dt}{9-t^2}$

$\therefore I={\int }_0^1\frac{(3+t+3-t)}{3(3+t)(3-t)}$

$={\int }_0^1\frac{dt}{3(3-t)}+{\int }_0^1\frac{dt}{3(3+t)}$

$=\left[\ln \right(3-t){]}_0^1\times \frac{-1}{3}+[\ln (3+t){]}_0^1\times \frac{1}{3}$

$=(\ln 3-\ln 2+\ln 4-\ln 3)\times \frac{1}{3}$

$=\frac{1}{3}\times \ln 2$