Question

Solve:$\int \frac{dx}{4+9{\cos }^{2}x}$

Answer 1

$\int \frac{dx}{4+9{\cos }^{2}x}=I$

Let $\cos x=\frac{1}{u}$

$-\sin xdx=-\frac{1}{u^2}du$

$\sqrt{1-\frac{1}{u^2}}dx=\frac{1}{u^2}du$

$dx=\frac{\frac{1}{u^2}du}{\sqrt{1-\frac{1}{u^2}}}$

$I=\int \frac{\frac{1}{u^2}}{(4+\frac{9}{u^2})\sqrt{1-\frac{1}{u^2}}}$du

$I=\int \frac{u}{(4u^2+9)\sqrt{u^2-1}}$du

Let $u^2-1=t^2$

$udv=tdt$

so $I=\int \frac{1}{4t^2+13}dt$

$I=\frac{1}{2\sqrt{1}3}{\tan }^{-1}\left(\frac{2t}{\sqrt{1}3}\right)+c$

as $t=\sqrt{u^2-1}=\sqrt{{\sec }^2-1}$

$=\tan x$

$\therefore I=\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{13}}\right)+c$