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How do you integrate $$ \int \frac{4}{5+3\cos(2x)}\,dx $$ ? I tried with substitution method ($u = 2x$, $u = \cos(2x)$, ...) without success. Hints are accepted :)

AsukaMinato
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moonknight
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    Take a look at Weierstrass Substitution for a general way of integrating any rational expression of trigonometric functions. – Thomas Russell Jan 12 '17 at 17:11
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    Also see this http://math.stackexchange.com/questions/1740458/finding-int-fracdxab-cos-x-without-weierstrass-substitution/1741835 answer if you want to avoid the Weierstrass Substitution – user_of_math Jan 12 '17 at 17:22
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    Try to substitute $\tan(x)=t$. – szw1710 Jan 12 '17 at 17:25
  • $x=\arctan t$ is definitely the way. Since $\cos(2x)=2\cos^2(x)-1$ and $\cos^2\arctan x=\frac{1}{1+x^2}$, that substitution turns the integrand function in a simple rational function. – Jack D'Aurizio Jan 12 '17 at 17:26
  • @JackD'Aurizio That sub is effectively the so-called "Weierstrass" sub. – Mark Viola Jan 12 '17 at 17:43
  • @Dr.MV: I usually call $x=2\arctan\frac{t}{2}$ "the" Weierstrass substitution, $x=\arctan t$ is a sort of simplified version of it. – Jack D'Aurizio Jan 12 '17 at 17:45
  • @JackD'Aurizio Yes with $x/2$ and $t/2$ scaled by a factor of $2$. – Mark Viola Jan 12 '17 at 17:57
  • The best posts on this site include not only a problem but context: where did the problem arise? Why is it of interest? There is an unlimited number of integrals that could be posted; what led you to ask about this particular one? You can edit the post to include information of this kind. – Carl Mummert Jan 12 '17 at 18:20

2 Answers2

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I thought it might be instructive to present a way forward that relies on Euler's Formula along with straightforward partial fraction expansion. To that end, we proceed.

Another way forward is to use Euler's Formula to write $\cos(2x)=\frac{e^{i2x}+e^{-i2x}}{2}$. Then, we have

$$\begin{align} \int \frac{4}{5+3\cos(2x)}\,dx&=\int \frac{8e^{i2x}}{3e^{i4x}+10e^{i2x}+3}\,dx\\\\ &=\int\left(\frac{3e^{i2x}}{3e^{i2x}+1}-\frac{e^{i2x}}{e^{i2x}+3}\right)\,dx\\\\ &=\frac1{2i}\log\left(\frac{3e^{i2x}+1}{e^{i2x}+3}\right)+C\\\\ &=\frac12 \left(\arctan\left(\frac{3\sin2x)}{1+3\cos(2x)}\right)-\arctan\left(\frac{\sin(2x)}{3+\cos(2x)}\right)\right)+C\\\\ &=\frac12\arctan\left(\frac{4\sin(2x)}{3+5\cos(2x)}\right)+C \end{align}$$

Mark Viola
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Let's see a more general form $$I=\int \frac{1}{a+b\cos x}\,\mathrm dx$$ let $t=\tan\left(\dfrac{x}{2}\right)$ we get $$I=\int \frac{2}{(a+b)+t^2(a-b)}\,\mathrm dt$$ If $a^2 ＞ b^2$ , then we have $$I=\frac{2}{a-b}\int \frac{1}{\left(\sqrt{\dfrac{a+b}{a-b}}\right)^2+t^2}\,\mathrm dt$$ use the known formula $$\int \frac{1}{x^2+a^2}\,\mathrm dx=\frac{1}{a}\arctan\frac{x}{a}+C$$ we will get the answer of $I$. Hope you can take it from here.

Youngseok Choe
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Renascence_5.
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