Problem with computing $\int{\frac{dx}{2\sin^2 x+3\cos^2 x}}$

Question

How computing this integral:

$$\int{\frac{dx}{2\sin^2 x+3\cos^2 x}}$$

I tried replace $\cos^2 x$ for $1-\sin^2 x$ and solve, but no sucess.

Answer 1

Use your idea of replacing $\cos^2 x= 1 - \sin^2 x$. Then multiply by $1$ using $\sec^2 x/\sec^2 x$ to obtain $$ \int \dfrac{\sec^2 x}{3\sec^2 x - \tan^2 x} \;dx $$ Then use the $u$-sub of $u= \tan x$ and replace $\sec^2 x= \tan^2 x + 1$. This will give you $$ \int \dfrac{du}{2u^2+3} $$ This is handled the 'normal' way, multiplying by $3/3$ then rewriting the denominator as $(\sqrt{2}/3u)^2+1$. After, possibly another $u$-sub, you will find $$ \int \dfrac{1}{2 \sin^2 x + 3 \cos^2 x} \;dx= \dfrac{1}{\sqrt{6}} \arctan\left(\sqrt{\dfrac{2}{3}} \tan x \right) + C $$