Alternatives for evaluating $ \int \frac { 1 } { 5 + 4 \cos x} \ dx $ ??

Question

$$ \int \frac { 1 } { 5 + 4 \cos x} \ dx $$

$$ \text{The solution given in the book for solving this was to use the identity.} $$

$$ \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1+ \tan^2 \frac{x}{2}} $$

I was wondering if there was any other way for solving this ?

score 0 · Answer 1 · answered Oct 02 '18 at 09:54

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You can also solve it using the substitution $x=2\arctan t$ and $\mathrm dx=\frac2{1+t^2}\,\mathrm dt$. This will transform your problem into the computation of a primitive of a rational function.

answered Oct 02 '18 at 09:54

José Carlos Santos

440,053

score 0 · Answer 2 · answered Oct 02 '18 at 09:56

$$I=\int\frac{1}{5+4\cos(x)}dx$$ now we can use Weierstrauss substitution: $$u=\tan\left(\frac{x}{2}\right)$$ so: $$dx=\frac{2}{1+u^2}du$$ and $$\cos(x)=\frac{1-u^2}{1+u^2}$$ now we can transform our integral to: $$I=\int\frac{1}{5+4\frac{1-u^2}{1+u^2}}\frac{2}{1+u^2}du=\int\frac{1}{5(1+u^2)+4(1-u^2)}du=\int\frac{1}{9+u^2}du=\frac{1}{9}\int\frac{1}{1+\left(\frac{u}{3}\right)^2}du$$ now let:$$u=3\tan(v)$$ $$du=3\sec^2(v)dv$$ so: $$I=\frac{1}{9}\int3\frac{\sec^2(v)}{1+ \tan^2(v)}dv=\frac{1}{3}\int dv=\frac{v}{3}+C=\frac{\arctan\left(\frac{u}{3}\right)}{3}+C=\frac{\arctan\left(\frac{\tan\left(\frac{x}{2}\right)}{3}\right)}{3}+C$$

Claude Leibovici · Answer 3 · 2018-10-02T10:04:33.590

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Cheating a little, $$x=2 \cot ^{-1}\left(\frac{u}{3}\right)\implies dx=-\frac{6}{u^2+9}\,du$$ $$\int \frac { dx } { 5 + 4 \cos x} =-\frac 23\int \frac {du}{1+u^2}$$

edited Oct 02 '18 at 10:04

answered Oct 02 '18 at 09:56

Claude Leibovici

289,558

is the extra $x$ in second line a mistake? – Henry Lee Oct 02 '18 at 09:57
@HenryLee. Thanks for pointing the typo ! – Claude Leibovici Oct 02 '18 at 10:05
@ClaudeLeibovici But how did you think of that Substitution ?? – Oct 02 '18 at 10:22
@JIM. I wrote it : Cheating ! – Claude Leibovici Oct 02 '18 at 10:37
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@ClaudeLeibovici How Did you cheat ?? – Oct 02 '18 at 10:48
@ClaudeLeibovici you just manipulated and used Tangent half angle substitution !! – Oct 03 '18 at 02:01
@JIM. In fact, the first idea came from $5^2-4^2=3^2$ – Claude Leibovici Oct 03 '18 at 04:43
@ClaudeLeibovici But can we manipulate tangent half angle sub. only for this problem or for other problems also ? – Oct 03 '18 at 08:51
@JIM. The tangent half-angle substitution is a very standard way for many integration problems in which only trigonometric functions are involved. The problem is that you end with "polynomial" expressions and some of them cannot be integrated using elementary functions. But, I suppose that, for the time being, it would be a very, very good way for your problems. Good luck in your studies. Cheers – Claude Leibovici Oct 03 '18 at 08:58

score 0 · Answer 4 · answered Oct 02 '18 at 10:13

$$\int\frac{1}{5+4\cos x}dx=\int\frac{1}{5+4\frac{e^{ix}+e^{-ix}}{2}} \ dx= \int\frac{2e^{ix}}{4e^{2ix}+10e^{ix}+4} \ dx $$ let $u = e^{ix}$ and $du = ie^{ix}$ $$\int\frac{-2i}{2u^2+10u+4}\ du = -i\int\frac{1}{(u+\frac{1}{2})(u+2)}du$$ and the rest is easy!

score 0 · Answer 5 · answered Oct 02 '18 at 12:47

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Let $$\displaystyle I =\int\frac{1}{5+4\cos x}dx=\int\frac{5-4\cos x}{25-16\cos^2 x}dx$$

$$ I =5\int\frac{(\tan x)'}{9+16\tan^2 x}dx-4\int\frac{(\sin x)'}{25-16\cos^2 x}dx$$

answered Oct 02 '18 at 12:47

DXT

12,047

I do not understand how did you this from the very second step? – Oct 02 '18 at 14:52

Alternatives for evaluating $ \int \frac { 1 } { 5 + 4 \cos x} \ dx $ ??

5 Answers5