Question

Find the value of $\int \frac{dx}{3+2{\cos }^{2}x}$

Answer 1

Let $I=\int \frac{dx}{3+2{\cos }^{2}x}$

By dividing ${\cos }^{2}x$ in numerator and denominator.

$I=\int \frac{{\sec }^{2}xdx}{3{\sec }^{2}x+2}$

$=\int \frac{{\sec }^{2}xdx}{3(1+{\tan }^{2}x)+2}=\int \frac{{\sec }^{2}xdx}{3{\tan }^{2}x+5}$

let $\tan x=t$

$\frac{d}{dx}\tan x=\frac{dt}{dx}$

${\sec }^{2}xdx=dt$

$I=\int \frac{dt}{3t^2+5}$

$=\frac{1}{3}\int \frac{dt}{t^2+\frac{5}{3}}$

$I=\frac{1}{3}\int \frac{dt}{t^2+{\left(\sqrt{\frac{5}{3}}\right)}^2}$

$I=\frac{1}{3}\times \frac{1}{\frac{\sqrt{5}}{\sqrt{3}}}{\tan }^{-1}\frac{t}{\frac{\sqrt{5}}{\sqrt{3}}}+c$

$I=\frac{1}{3}\times \frac{\sqrt{5}}{\sqrt{3}}{\tan }^{-1}\frac{\sqrt{3}t}{\sqrt{5}}+$

$=\frac{1}{\sqrt{15}}{\tan }^{-1}\frac{\sqrt{3}\tan x}{\sqrt{5}}+c$