Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$?

Question

I am evaluating

$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$

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Using Weierstrass substitution:

$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv =\int \frac{1}{v^2+2}dv \quad (2) $$

And then $\:v=\sqrt{2}w$

$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)^2+2}\sqrt{2} dw$$$$= \frac{1}{2} \int \frac{1}{\sqrt{2}\left(w^2+1\right)}dw$$$$ = \frac{1}{2\sqrt{2}}\arctan \left(w\right) + C$$$$= \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C$$

Therefore,

$$\int \frac{1}{\cos 2x+3} dx = \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C $$

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That's a decent solution but I am wondering if there are any other simpler ways to solve this (besides Weierstass). Can you come up with one?

Answer 1

$$\int \frac{1}{\cos2x+3}dx=\int \frac{1}{\frac{1-\tan^2x}{1+\tan^2x}+3}dx$$ $$=\int \frac{1+\tan^2x}{2\tan^2x+4}dx$$ $$=\frac12\int \frac{\sec^2x\ dx}{\tan^2x+2}$$ $$=\frac12\int \frac{d(\tan x)}{(\tan x)^2+(\sqrt2)^2}$$ $$=\frac12\frac{1}{\sqrt2}\tan^{-1}\left(\frac{\tan x}{\sqrt2}\right)+C$$ $$=\bbox[15px,#ffd,border:1px solid green]{\frac{1}{2\sqrt2}\tan^{-1}\left(\frac{\tan x}{\sqrt2}\right)+C}$$

Answer 2

HINT:

Using Euler's formula, we have $\cos(2x)=\frac12(e^{i2x}+e^{-i2x})$.

Now make the substitution $z=e^{i2x}$ with $dx=\frac1{i2z}\,dz$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int{\dd x \over \cos\pars{2x} + 3} & = \int{\dd x \over \bracks{2\cos^{2}\pars{x} - 1} + 3} = {1 \over 2}\int{\sec^{2}\pars{x}\,\dd x \over 1 + \sec^{2}\pars{x}} \\[5mm] & = {1 \over 2}\int{\sec^{2}\pars{x}\,\dd x \over \tan^{2}\pars{x} + 2} = {1 \over 2}\,{1 \over 2}\,\root{2}\int{\bracks{\sec^{2}\pars{x}/\root{2}} \,\dd x \over \bracks{\tan\pars{x}/\root{2}}^{2} + 1} \\[5mm] & = \bbx{{\root{2} \over 4}\arctan\pars{{\root{2} \over 2}\,\tan\pars{x}} + \mbox{a constant}} \end{align}

Answer 4

Long way but doable

$$I=\int \frac{dx}{\cos (2x)+3}$$

Let $$\cos(2x)=t \implies x=\frac{1}{2} \cos ^{-1}(t)\implies dx=-\frac{1}{2 \sqrt{1-t^2}}$$ $$I=-\frac{1}{2}\int \frac{dx}{(t+3) \sqrt{1-t^2}}=-\frac{1}{2}\int \frac{\sqrt{1-t^2}}{(t+3) (1-t^2)}\,dt$$ $$\frac{1}{(t+3) (1-t^2)}=-\frac{1}{8 (t+1)}+\frac{1}{16 (t+3)}+\frac{1}{16 (t-1)}$$ and we face three integrals $$J_a=\int \frac {\sqrt{1-t^2} }{t+a}$$ $$J_a=-\sqrt{1-a^2} \log \left(\sqrt{1-a^2} \sqrt{1-t^2}+a t+1\right)+\sqrt{1-a^2} \log (a+t)+a \sin ^{-1}(t)+\sqrt{1-t^2}$$ This leads to $$I=\frac{i \left(\log (t+3)-\log \left(2 i \sqrt{2-2 t^2}+3 t+1\right)\right)}{4 \sqrt{2}}=-\frac{i}{4 \sqrt{2}}\log \left(1+i\frac{2 \sqrt{2} \sqrt{1-t^2}}{t+3}\right)$$

Answer 5

$$\cos(2x)=\cos (x+x) =\cos x \cos x-\sin x \sin x=\cos^2x-\sin^2x$$
$$I =\int \frac{1}{\cos 2x+3}dx= \int \frac{\sec^2x}{\sec^2x(\cos^2x-\sin^2x+3)}dx = \int \frac{\sec^2x}{1-\tan^2x+3\sec^2x}dx $$
Substitute $t=\tan x$ so that $dt=\sec^2x dx$
$$I=\int \frac{1}{1-t^2+3(1+t^2)}dt=\int \frac{1}{4+2t^2} dt=\frac{1} {2\sqrt 2} \tan^{-1}\frac{\sqrt 2 t} {2}+c$$ Now substitute back $t=\tan x$