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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 ?

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    $\begingroup$ Have you tried the Weierstrass substitution? $\endgroup$ Feb 21, 2015 at 16:57
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    $\begingroup$ Try $u=\tan{x}$. $\endgroup$
    – JWL
    Feb 21, 2015 at 16:58
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    $\begingroup$ Here's a thought. Try multiplying the numerator and denominator by $\sec^2x$. This would give $\int{\sec^2x\,dx\over3\tan^2x+5}$. $\endgroup$ Feb 21, 2015 at 17:03
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    $\begingroup$ Rewriting the denominator to $3+2\cos^2 x$ as a first step might not hurt. Or even to $4+\cos 2x$. $\endgroup$ Feb 21, 2015 at 17:03
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    $\begingroup$ Where you had $3{\sin^2 x} + 5{\cos^2 x}$, I changed it to $3\sin^2 x + 5\cos^2 x$. Yours was coded as 3{\sin^2 x} + 5{\cos^2 x} and mine as 3\sin^2 x + 5\cos^2 x. Notice the difference in what they look like: Yours lacked proper spacing after the $3$ and the $5$. ${}\qquad{}$ $\endgroup$ Feb 21, 2015 at 17:14

5 Answers 5

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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: $$\int \frac{dx}{3\sin^2 x+5\cos^2 x} = C+\frac{1}{\sqrt{15}}\arctan\left(\sqrt{\frac{3}{5}} t\right)=C+\frac{1}{\sqrt{15}}\arctan\left(\sqrt{\frac{3}{5}} \tan x\right).$$

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$$\begin{gathered} \int {\frac{1} {{3{{\sin }^2}x + 5{{\cos }^2}x}}dx} = \int {\frac{1} {{\left( {3\frac{{{{\sin }^2}x}} {{{{\cos }^2}x}} + 5} \right){{\cos }^2}x}}dx} = \int {\frac{1} {{3{{\tan }^2}x + 5}}d\left( {\tan x} \right)} \hfill \\ = \frac{1} {3}\int {\frac{1} {{{{\tan }^2}x + \frac{5} {3}}}d\left( {\tan x} \right)} = \frac{1} {3}.\frac{1} {{\sqrt {5/3} }}\arctan \left( {\frac{{\tan x}} {{\sqrt {5/3} }}} \right) + C = \frac{1} {{\sqrt {15} }}\arctan \left( {\sqrt {\frac{3} {5}} \tan x} \right) + C \hfill \\ \end{gathered} $$

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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.

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HINT:$$\dfrac{1}{3{\sin^2 x}+5{\cos^2x}}=\dfrac{1+{\tan^2 x}}{5+3{\tan^2 x}}$$ and $$d(\tan x)=\sec^2x dx$$

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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: $$\int \frac{dx}{3\sin^2\left(x\right)+5\cos^2\left(x\right)}dx = \frac{\arctan\left(\sqrt{\frac{3}{5}}\tan\left(x\right)\right)}{\sqrt{15}} + C$$

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