How do I calculate: $\int \frac{dx}{3\sin^2 x+5\cos^2x}?$ How can I calculate this integral ?
$$\int \frac{dx}{3\sin^2 x+5\cos^2x}=\text{?}$$
Thank you! I've tried using universal substitution but the result was too complicated to be somehow integrated. Can you please give me a useful hint ?
 A: By setting $x=\arctan t$, so that $dx=\frac{dt}{1+t^2}$, we get:
$$\int \frac{dx}{3\sin^2 x+5\cos^2 x} = \int \frac{dx}{3+2\cos^2 x} = \int \frac{dt}{(1+t^2)\left(3+\frac{2}{1+t^2}\right)}=\int\frac{dt}{5+3t^2}$$
so:
$$\int \frac{dx}{3\sin^2 x+5\cos^2 x} = C+\frac{1}{\sqrt{15}}\arctan\left(\sqrt{\frac{3}{5}} t\right)=C+\frac{1}{\sqrt{15}}\arctan\left(\sqrt{\frac{3}{5}} \tan x\right).$$
A: $$\begin{gathered}
  \int {\frac{1}
{{3{{\sin }^2}x + 5{{\cos }^2}x}}dx}  = \int {\frac{1}
{{\left( {3\frac{{{{\sin }^2}x}}
{{{{\cos }^2}x}} + 5} \right){{\cos }^2}x}}dx}  = \int {\frac{1}
{{3{{\tan }^2}x + 5}}d\left( {\tan x} \right)}  \hfill \\
   = \frac{1}
{3}\int {\frac{1}
{{{{\tan }^2}x + \frac{5}
{3}}}d\left( {\tan x} \right)}  = \frac{1}
{3}.\frac{1}
{{\sqrt {5/3} }}\arctan \left( {\frac{{\tan x}}
{{\sqrt {5/3} }}} \right) + C = \frac{1}
{{\sqrt {15} }}\arctan \left( {\sqrt {\frac{3}
{5}} \tan x} \right) + C \hfill \\ 
\end{gathered} $$
A: First, multiply top and bottom by $\sec^2x$ to get
$$\int\frac{\sec^2x\,dx}{3\tan^2x+5}$$
Now substitute $u=\tan x$ as JWL suggested.
A: HINT:$$\dfrac{1}{3{\sin^2 x}+5{\cos^2x}}=\dfrac{1+{\tan^2 x}}{5+3{\tan^2 x}}$$ and $$d(\tan x)=\sec^2x dx$$
A: At first substitute: $$\int \frac{dx}{3\sin^2\left(x\right)+5\cos^2\left(x\right)}dx = \int \frac{1}{3\sin^2\left(\arctan\left(u\right)\right) + 5\cos^2\left(\arctan\left(u\right)\right)}\frac{1}{1+u^2}du$$ where $x=\arctan\left(u\right)$ and $dx=\frac{1}{1+u^2}du$.
Then we can write this as $$\int \frac{1}{\left(u^2+1\right)\left(\frac{3u^2}{u^2+1}+\frac{5}{u^2+1}\right)}du=\int \frac{1}{3u^2+5}$$
Lets substitue again, where $u=\frac{\sqrt{5}}{\sqrt{3}}v$ and thus $du=\sqrt{\frac{5}{3}}dv$: $$\int \frac{1}{3\left(\frac{\sqrt{5}}{\sqrt{3}}v\right)^2+5}\sqrt{\frac{5}{3}}dv=\int \frac{1}{\sqrt{15}\left(v^2+1\right)}dv=\frac{1}{\sqrt{15}}\int \frac{1}{v^2+1}dv$$
As we know this relation, we can directly write $\frac{1}{\sqrt{15}}\arctan\left(v\right)$, then going the substitutions backwards: $$\frac{\arctan\left(\sqrt{\frac{3}{5}}\tan\left(x\right)\right)}{\sqrt{15}}$$
Thus: $$\int \frac{dx}{3\sin^2\left(x\right)+5\cos^2\left(x\right)}dx = \frac{\arctan\left(\sqrt{\frac{3}{5}}\tan\left(x\right)\right)}{\sqrt{15}} + C$$
